Geometry, Algebra,
and Trigonometry
by Vector Methods
Arthur H. Copeland, sr..
University of Michigan
ALLENDOERFER UNDERGRADUATE SERIES

GEOMETRY,
ALGEBRA,
and
TRIGONOMETRY
by VECTOR
METHODS
Arthur H. Copeland, Sr.
UNIVERSITY OF MICHIGAN
THE MACMILLAN COMPANY NEW YORK
A Division of The Crowell-Collier Publishing Company

© The Macmillan Company 1962
All rights reserved. No part of this book may be
reproduced in any form without permission in writing
from the publisher, except by a reviewer who wishes
to quote brief passages in connection with a review
written for inclusion in a magazine or newspaper.
First Printing
Library of Congress catalog card number: 62-8147
The Macmillan Company, New York
Brett-Macmillan Ltd., Gait, Ontario
Printed in the United States of America

PREFACE
The radically different elementary texts appearing recently mark a
reaction against the failure of earlier books to take advantage of the modern
developments in mathematics. In many of the new texts the novelty is in the
subject matter, whereas in the present one modern methods are used to
develop conventional subject matter. A cursory reading of this text may give
the impression that the employment of these modern techniques presupposes
a high degree of sophistication on the part of the student. This is not the case.
The reader is carefully prepared for the modern techniques and they should
cause no more difficulty than the older methods. Often a modern method has
a much broader range of applicability than the older one, although the two
methods can be applied with comparable difficulty to the development of a
given subject. In such cases modern methods are used in this text. For
example, vector algebra is used to simplify the development of analytic geom-

PREFACE
etry and college algebra and to unify the two subjects. The time saved by the
simplification is sufficient for the development of the vector algebra.
Moreover, vectors are sufficiently similar to numbers so that by studying vectors
a student can review his ordinary algebra without being subjected to a course
in pure drill.
The consensus among mathematicians is that much of the material
usually contained in a course in trigonometry should be deleted. The portion of
trigonometry remaining after such deletion is developed in this text. Vector
methods simplify the development. Further simplification is achieved by
using three- instead of five-place tables, thus allowing students to concentrate
on understanding rather than on arithmetic. The sections on trigonometry
can be used as references for students who have previously studied the
subject and (with the exception of Section 56) can be omitted without loss of
continuity. This text is designed for students who have had at least plane
euclidean geometry and a year and a half of high school algebra.
In high school geometry a student is introduced to the axiomatic method
but is often not presented with another example of the method unless he
continues his mathematical training into graduate school. The majority of
students are thus given a somewhat distorted picture of mathematics. Moreover,
in the absence of axioms, arguments are presented without explicit statement
of the assumptions. Although there is much to be said in favor of an intuitive
presentation in intermediate courses, the usual compromise between intuition
and rigor is puzzling to students, since no criterion is furnished forjudging
the legitimacy of reasoning. In the first eight sections of the present text an
intuitive background for a set of postulates for vector algebra is developed.
The postulates are presented in Section 9, and proofs are based on the
postulates in the remaining sections. This furnishes a relatively firm basis
for analytic geometry and college algebra and provides a fresh example of
axiomatic method.
It has often been observed that many students do not know how to study
mathematics. A device for training students how to study is included in the
present text, namely, a set of questions at the end of each section. Best results
will, in the opinion of the author, be achieved if the teacher insists that the
students answer all the questions, preferably before attempting to solve the
problems. For example, if a student is unable to solve a certain problem, the
teacher might ask him to pick out those questions which seem likely to have
some bearing on the solution and then to answer those questions accurately.

CONTENTS
1. VECTORS 1
1. Velocities and Vectors 1
2. Addition of Vectors 5
3. The Zero Vector and the Negative of a Vector 9
4. Multiplication of a Vector by a Number; Absolute Value 11
5. Components of Vectors 15
6. Further Properties of Components 19
7. Scalar Products 21
8. Vectors in Three-Dimensional Space 25
9. The Postulates 29

CONTENTS
2. PLANE ANALYTIC GEOMETRY 34
10. Circles and Lines 34
11. Applications of Parametric Equations 39
12. The Distance from a Point to a Line 42
13. Equations of Lines 47
14. Second-Order Determinants 50
15. Simultaneous Linear Equations 54
16. Intersection of Lines 59
17. Linear Dependence 62
18. Geometrical Properties of Linear Equations 65
19. The Product of Two Determinants 69
20. Loci 73
21. The Parabola 77
22. The Ellipse 79
23. The Hyperbola 82
3. SOLID ANALYTIC GEOMETRY 87
24. Spheres, Lines, and Planes 87
25. Third-Order Determinants 91
26. Further Properties of Determinants 94
27. Three Equations in Three Unknowns 99
28. The Product of Two Determinants 104
29. Distances from Points to Lines and Points to Planes 108
30. Vector Products 113
31. Properties of Vector Products 117
32. Linear Equations 122
33. The Intersection of Two Planes 126
34. The Intersection of Three Planes 130
35. Further Properties of Intersecting Planes 133
36. Loci in Three-Dimensional Space 136
37. Drawing Three-Dimensional Figures 137
38. Ruled Surfaces 142
39. Models of the Postulates 147
40. Determinants of Higher Order 152
41. Four Equations in Four Unknowns 158
42. Geometrical Applications of Determinants 163

CONTENTS
IX
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
4. TRANSFORMATIONS OF COORDINATES
Translations
Quadratic Equations in Two Variables
Applications of Translations
Rotations
Further Properties of Rotations
Removing the Cross-Product Term
Properties of Matrices
Recapitulation
Quadratic Equations in Three Dimensions
Three-by-Three Matrices
Cubics and Higher-Degree Equations
The Intersection of a Plane with a Cone
168
168
173
177
180
182
185
189
193
199
204
207
211
5. TRIGONOMETRY 214
55. Cosines and Sines
56. Rotation through an Angle
57. Reduction Formulas
58. Angles
59. The Laws of Sines and Cosines and the Angle Sum
60. Triangle Solution
61. Radian Measure
62. Graphs and Interpolation
63. Constructing a Slide Rule
64. The Slide Rule and Logarithms
65. Exponents
66. Triangle Solution by Logarithms
67. Identities
68. Inverse Functions
Theorem
214
218
223
225
229
232
235
238
242
244
248
254
256
261
6. ALTERNATIVE COORDINATE SYSTEMS 265
69. Polar Coordinates
70. Cylindrical Coordinates and Spherical Coordinates
265
272

CONTENTS
7. COMPLEX NUMBERS 277
71. Algebraic Operations with Complex Numbers 277
72. Multiplication and Rotation 280
73. De Moivre's Theorem 283
APPENDIX
Table 1 293
Table 2 294
INDEX
295

VECTORS
1
1. VELOCITIES AND VECTORS
Vectors furnish a means of studying geometry, algebra, and trigonometry
and the relations among these subjects. They introduce a sufficient degree
of simplification into the study to make it worth while devoting a chapter
to the task of acquainting the reader with properties of vectors. We begin
by studying velocities, since this is a way of showing how vectors are used.
An object moving in a straight line with constant speed traces a line
segment AB during an interval of 1 sec. The magnitude (i.e., length) of the
segment measured in feet is the speed of the object measured in feet per
second. The line segment itself indicates the direction of the motion. If the
object moves from A to B, then the motion is said to have the sense from A
toward В and an arrow is placed at the terminal end В of the segment (see

2
VECTORS
(Ch. 1
Fig. 1.1) to indicate this sense. A motion from В toward A is said to have
the opposite sense. A segment to which a direction and sense have been
assigned is called a vector. The segment AB with sense from A to В is
denoted AB. The French use the expression "sens unique" to indicate a
street with a unique sense, i.e., a one-way street. An ordinary street has
a direction such as north and south,
but a one-way street in which traffic
is permitted to move only to the north
has also a sense, namely, to the north.
Suppose that the moving object is
one of the airplanes in a squadron of
Figure 1.1 planes flying in formation and that
a second plane moves from С to D
during the same interval of 1 sec. The second plane then has the velocity
vector CD (see Fig. 1.1), and if the planes are flying in proper formation,
the two velocity vectors have the same magnitude, direction (i.e., they are
parallel), and sense. Two vectors that are so related are said to be equivalent,
and we express the equivalence by writing
AB = CD.
Customarily an equation is used in mathematics only when the right- and
left-hand members are symbols for the same mathematical entity. However,
to follow this convention in the case of vectors would necessitate defining
them in an abstract manner, which would make the concept difficult to
grasp.1 We shall therefore violate this convention and shall even call
1 For the sake of readers who may be curious, we shall indicate how this is done. Note
that there is one drawback in picturing a velocity vector as a segment with an arrow
attached, namely, the segment has a specific position in space, whereas velocity should be
independent of position. Thus two planes in a squadron flying in formation are regarded
as having a common velocity vector even though those planes trace different segments
during an interval of 1 sec. Neither segment is identified with the vector but rather each
segment is regarded as representing the vector. Any equivalent segment (i.e., any segment
with the same magnitude, direction, and sense) is also a representative. The vector is
interpreted as that which has all these equivalent representatives, and it is characterized
by the magnitude, direction, and sense common to its representatives. The vector has no
specific position, whereas each of its representatives does have a specific position. A
velocity vector when thus interpreted characterizes a velocity without specifying a position.
Next consider the notational problem connected with the equation AB = CD. We
interpret AB зал symbol for the vector of which the segment A B is a representative rather than
as a symbol for the segment itself; similarly for CD. Then A B and CD are two different
symbols for the same vector, and А В = CD is an equation in the usual mathematical sense.

Sec. 1]
VELOCITIES AND VECTORS
3
AB = CD an equation but we shall interpret the equation as indication
that the right- and left-hand members are equivalent but not necessarily the
same segments.
We shall often use a single letter printed in boldface type to denote a
vector. A letter written in italics will denote a number. Thus ν denotes
a vector, whereas υ denotes a number. Boldface-type distinction is not
adapted to handwritten work, but to avoid confusion it is advisable to adopt
some notation to distinguish between vectors and numbers. It is suggested
that the reader place an arrow over a letter (for example v) when that letter
represents a vector. If ν is the vector AB, then the point A is called the origin
and В is the tip or terminal point of v. The sense of a vector is from its origin
toward its tip. The magnitude of the vector AB is denoted by \AB\ and the
magnitude of ν is denoted by | ν |. Observe that a magnitude is always a non-
negative number and that | v| denotes a number and not a vector. A vector
whose magnitude is 1 is called a unit vector. Thus u is a unit vector if | u | = 1.
When we are concerned only with vectors in a plane, the directions and
senses of these vectors can be specified with reference to the unit vector u
whose direction is horizontal and whose sense is to the right. Thus, if ν is any
vector having the same origin as u, then u, ν form the sides of an angle.
Let A be the measure of this angle in degrees, where a positive number A
indicates a positive or counterclockwise rotation from u to ν through an
angle of A". Then the number A determines the direction and sense of v.
If the magnitude |v| of ν is r, then the numbers A, r determine the
direction, sense, and magnitude of v, and we call ν the vector [A°, r]. For
a = (90°,3)
Ы
90°
?. 70°
(^
e = (135°,2.5)
135°
b = (270°,2)
Figure 1.2
example, u is the vector [0°, 1]. Figure 1.2 shows the vectors a = [90°, 3],
b = [270°, 2], с = [135°, 2.5], whose magnitudes are respectively |a| = 3,
|b| = 2, |с| = 2.5.

4
VECTORS
(Ch. 1
Vectors furnish a convenient means of locating points. Thus, if we
choose an origin at a point 0, and χ is a vector with origin at 0, then the
tip of χ is the point X such that OX = x. This method of locating points
enables us to use vectors to obtain geometrical results. If χ is in the plane,
say, of a sheet of paper, then it is determined by a pair of numbers A, r, and
hence the tip X is determined by A, r.
Questions
It is advisable to answer these questions before attempting to solve the problems.
Try to answer them without consulting the text, but if there are some you cannot
answer, look them up and then phrase the answer in your own words. You can review
this section by rereading the questions and re-examining the problems. As to the
problems, if there are any you cannot solve, try to discover which questions might be
pertinent to the solution and then read the corresponding portions of the text to see
whether you have missed any point in your answers. These suggestions apply also to
the question in the subsequent sections.
1. How do the magnitude, direction and sense of a vector characterize the motion
of an object? 2. Describe the difference between direction and sense. 3. What do the
symbols |Λ2?|, |ν| mean? 4. What is a unit vector? 5. Describe the meaning of
the symbol [A°, r]. 6. How is a point located by means of a vector?
Problems
A protractor and centimeter ruler are needed for the solutions of the problems of
this text. A 30°-60°-90° triangle will be useful, and a compass will be needed in
Section 55. Choose 1 cm as a unit in all drawings unless otherwise instructed.
1. Choose a convenient origin 0 and draw the vectors u = [0°, 1], ν = [135°, 4].
2. A man walking along a straight road with constant speed has the velocity vector
ν = [135°, 4], where |v| is his speed in feet per second. Let [0°, 1] indicate
the direction and sense to the east and [90°, 1] the direction and sense to the
north. Describe the direction and sense of the man and the direction of the road
in terms of points of the compass. Let О be the position of the man at a certain
time, A his position 1 sec later, and В his position 2 sec later. Locate the points
А, В in your figure of Problem 1.
3. Draw the vectors a = [30°, 3], b = [75°, 4], с = [210°, 3], d = [255°, 4] and
label their tips respectively A, B, C, D. Draw the vectors AB, CD, ВС, AD.
4. How are the magnitudes, directions and senses of the vectors AB and CD of
Problem 3 related? Are these vectors equivalent? How are the magnitudes,

Sec. 2]
ADDITION OF VECTORS
5
directions and senses of the vectors ВС and AD related? Are these vectors
equivalent?
5. Let a, b be defined as in Problem 3. Find j a j, | b |.
6. Let A, B, C, D be defined as in Problem 3. Estimate \AB\, | BC\ by
measurement.
2. ADDITION OF VECTORS
Sections 1 to 8 are intended to acquaint the reader with the properties of
vectors, to make these properties seem reasonable, and to give some
indication of the ways in which vectors are used. There is no attempt to give
formal proofs in these sections. In Section 9 the properties of vectors are
collected together and form a set of postulates. The function of a set of
postulates is to give a concise description of the particular mathematical
system which an author is discussing. In Section 9 and subsequent sections a
number of proofs are given. As in euclidean geometry a proof consists in a
sequence of steps, each of which is justified by a postulate (axiom), definition,
or previously proved theorem. There are no such proofs in Sections 1 to 8,
since these sections precede the postulates. However, these sections make use
of some of the results of euclidean geometry in order to make the postulates
seem plausible.
Vectors are added and multiplied in a manner closely analogous to
addition and multiplication of numbers, and this fact makes it easy to
perform various computations with vectors. Such computations are used to
obtain results in both geometry and algebra. Vector addition is also used in
combining velocities. Thus consider a boat sailing in a straight line with
constant speed in a river. To simplify the picture, imagine an idealized situation
in which the velocity vector of each particle
of water is equivalent to that of every other
particle. Let 0 denote the position of the boat
at a given time. One second later a particle
of water located initially at 0 will have drifted
to a point A, and the boat will have moved to
a point B. Thus the velocity vector of the
particle of water (i.e., of the river) is u = OA,
and the velocity vector of the boat is w = OB
(see Fig. 2.1). As the boat moves, it produces a
wake which drifts with the river while it is
Figure 2.1

6
VECTORS
(Ch. 1
being formed. The segment from A to В is the portion of this wake formed
during the interval of 1 sec, and the vector ν = AB gives the direction and
sense in which the boat is heading. The magnitude |v| of this vector is the
speed which the boat would have if it were sailing in still water with the
same amount of power as it is using in the river. We call ν the velocity
vector of the boat relative to the river. The velocity vector w of the boat is the
resultant of the velocity u of the river and the velocity ν of the boat relative
to the river. We call w the sum of u and ν and write w = u + v.
In general (see PI and PI 8 of Section 9) the sum u + ν of the vectors u, ν
is the вес tor determined by the following construction: Let u be the vector OA. Construct
the vector А В which is equivalent to ν and has its origin at the tip A of the vector u.
The sum u + ν is the vector OB whose origin 0 coincides with the origin of u and
whose tip В coincides with the tip of the vector А В equivalent to v. The construction is
described by the equation
where
OB = 0A + AB,
u = OA, ν = AB, u + ν = OB.
Figure 2.2 shows two vectors u, v; Fig. 2.3 shows the construction of the
sum u + v; Fig. 2.4 shows the construction of ν + u; and Fig. 2.5 shows the
tv
Figure 2.2
Figure 2.3
Figure 2.4
two constructions combined in a single figure. It follows from well-known
properties of the parallelogram that (see P2 of Section 9)
u + ν = ν + u (commutative law).
Let us see why the equation (see P3 of Section 9)
(u + v) + w = u + (v + w) (associative law)

Sec. 2]
ADDITION OF VECTORS
7
is also valid. By the above construction, u + ν = OB, where u = OA and
where AB is equivalent to ν and has its origin at the tip of u (see Fig. 2.6).
Similarly, (u + v) + w = ОС where ВС is equivalent to w and has its
Figure 2.5
origin at the tip of u + v. Since ВС is equivalent to w and has its origin at the
tip of AB, where AB is equivalent to v, the sum ν + w is equivalent to AC.
Finally, since AC is equivalent to v +w and has its origin at the tip of u,
we have u + (v + w) = ОС = (u + ν) + w. These constructions are
described by the equations
{OA + AB) + ВС = OB + ВС = ОС = OA + AC = OA + (AB + ВС)
where
u = OA, ν = AB, w = ВС.
Note that the associative law holds for any three vectors u, v, w even when
the points О, А, В, С cannot all be contained in a plane.
A force is represented by a vector. The magnitude of the vector is the
magnitude of the force measured, say, in pounds. The direction and sense
of the vector are the direction and sense in which the force tends to pull
the object. If two forces act at a point and are represented by vectors u, v,
then the effect of these forces is the same as that of a single force represented
by the vector u + v. This result has been checked experimentally. The
motion of a rotating rigid body is also represented by a vector called an
angular velocity vector. The magnitude of this vector is the speed of the rotation

8
VECTORS
(Ch. 1
measured, say, in revolutions per minute or degrees per second, etc. The
direction of the vector is that of the axis of revolution. The sense is described
as follows: Point the thumb of your right hand in the direction and sense of
the vector and let your fingers curl. Then your fingers indicate the sense of
the rotation.
Questions
The reader is requested to follow the instructions given with the questions of
Section 1.
1. Describe how a sum of the form w = u + ν is related to the motion of a boat
in a river and how each of the vectors u, v, w is related to the motion. 2. Make a
drawing illustrating the equation OB = OA + AB. 3. Make a drawing illustrating
the construction of u + v, a second drawing illustrating the construction of ν + u,
and a third drawing combining the two constructions in a single figure. 4. State the
commutative law for vector sums and explain why it holds. 5. State the associative
law and explain why it holds. Draw figures illustrating your explanation.
Problems
1. Let a = [0°, 3] be the velocity vector of a river and b = [120°, 5] be the
velocity vector of a boat relative to the river, where speeds are measured in
knots. Construct the resultant velocity vector of the boat and estimate the speed
by measurement.
2. Let a = [30°, 3], b = [210°, 5] be interpreted respectively as in Problem 1.
Construct the resultant vector and find the exact speed.
3. Let a = [30°, 3], b = [210°, 3] be interpreted as in Problem 1. Find the
resultant speed.
4. Let a = [30°, 3], b = [90°, 5]. In separate figures construct a + b and b + a.
Make a third drawing, combining the two constructions.
5. Let a = [30°, 3], b = [90°, 5], с = [240°, 4]. In a single figure show the
construction of (a + b) + с and a + (b + c).
6. Let a = [30°, 3]. Construct a, a + a and a + a + a.
7. In Problem 3 of Section 1, one of the vectors in the figure is equal to the sum
a + AB. State which. There is a vector χ in the figure such that с + χ = d.
State which vector is equal to x.
8. Let a = [0°, 3] be the velocity vector of a river, b be the unknown velocity of
the boat relative to the river, and с = [120°, 5] be the resultant velocity.
Construct the vector b.

Sec. 3] ZERO VECTOR, NEGATIVE OF A VECTOR 9
3. THE ZERO VECTOR AND THE NEGATIVE OF A VECTOR
An object that is not moving is said to have zero speed. Since the object
occupies the same position 0 at the beginning and end of an interval of
1 sec, the velocity vector is the segment from 0 to 0, i.e., the single point 0.
This vector is called the zero vector and is denoted by 0. The magnitude 101 of
the zero vector is zero, and this is the speed of the object. If a boat is sailing in
still water, then the velocity vector of the water is 0 and the velocity vector ν
of the boat is the same as the velocity of the boat relative to the water. Even
in this case the velocity of the boat is the velocity of the water plus the velocity
of the boat relative to the water, i.e. (see P4 of Section 9)
ν = 0 + ν for every vector v.
This can also be seen as follows: If the boat sails from О to В during an
interval of 1 sec, then a particle of water located initially at О remains at 0,
and the portion of the wake formed during the interval is the segment from
О to B. Thus ν = OB is both the velocity of the boat and the velocity of the
boat relative to the water, and
OB = 00 + OB
where
OB = v,00 = 0.
Consider again a boat sailing in a river and let u, ν denote respectively
the velocity of the river and the velocity of the boat relative to the river.
If ν has the same direction and magnitude as u but the opposite sense, the
boat will stand still, and its velocity vector will be 0. In this case ν is called
the negative of u and is denoted by — u. Thus the velocity of the boat will be
0 when the velocity of the boat relative to the river is the negative of the
velocity of the river. This is expressed by the equation (see P5 of Section 9)
0 = u+ (-u).
Another way of looking at this is the following: In an interval of 1 sec a
particle of water located initially at О will move to a point A. Even though
the boat is standing still, it will produce a wake, and the portion of this wake
formed during the interval is the segment from A to 0. Thus the velocity of
the boat is 0 = 00, the velocity of the river is u = OA, the velocity of
the boat relative to the river is — u = АО, and
00 = OA + АО.

10
VECTORS
(Ch. 1
As in algebra, we use abbreviations such as
u + ( —u) = u — u, b + ( — a) = b — a.
We can probably guess that if a, b are any two vectors, then the solution
of the equation
x + a = b
for χ is b — a. To show that this guess is correct, we substitute b — a =
b + (— a) for χ in the left-hand member and obtain
χ + a = (b + (-a)) + a = Ь + ((-a) + a)
= b + (a + (-a)) = b + 0 = 0 + b = b.
Hence χ = b — a is a solution. If a, b have a common origin 0, then they
have the form a = О A, b = OB. Hence
b - a = b + (-a) = (-a) + Ь = АО + OB = AB;
i.e., b — a is the vector from the tip of a to the tip of b.
If the vectors fi, f2, · · · fn represent forces acting at a point, then these
forces are equivalent to a single force represented by the sum
fl + f2 + · · · fn-
The forces are said to be in equilibrium if and only if this sum is the vector 0.
For example, iff and —f represent forces acting at a point, then f + ( —f) = 0,
and hence these two forces are in equilibrium and are said to be equal
and opposite.
Questions
1. What is meant by the zero vector 0 and how is it related to a motion of an
object? 2. How is the equation ν = 0 + ν related to a motion of a boat in a lake?
3. What is meant by the negative — ν of a vector v, and how is the equation ν + (—ν)
= 0 related to a motion of a boat in a river? 4. Show that a + x = bifx = b — a.
5. Make a drawing illustrating how to construct χ such that a + χ = b when a, b
are given. 6. When several forces act at a point, how are they combined into a single
force? 7. When are these forces said to be in equilibrium?
Problems
1. Find -a where a = [30°, 5].
2. Let a = [0°, 3], b = [120°, 3], с = [240°, 3]. Construct a + b + с and show
that this sum is 0.

Sec. 4] MULTIPLICATION OF A VECTOR 11
3. Given χ + a = b. Add —a to both sides of this equation and show that
χ = b — a.
4. Given a = [330°, 4], b = [30°, 4], construct χ such that a + χ = b.
5. Let a = [0°, 4], b = [60°, 3] be forces acting at a point. By construction find a
force с which when added to a, b will produce equilibrium.
6. In constructing the sum a + b + с of Problem 2, you form an equilateral
triangle. Let О be the origin of a and F be the foot of the perpendicular dropped
from the tip of b to the line of a. Draw the figure and find the exact value of
\OF\.
4. MULTIPLICATION OF A VECTOR BY A NUMBER;
ABSOLUTE VALUE
Let ν be the velocity vector of an object moving in a straight line with
constant speed and let О be the origin of v. One second after the object is at
0, it will be at the tip of v; 2 sec after it is at 0, it will be at the tip of
ν + v; and 3 sec after, it will be at the tip of ν + ν + v. We call ν + ν
and ν + ν + ν respectively 2v and 3v. In general we define the product
fv = vt of a vector ν by a number / in such a manner that the object is at
the tip of fv at time /, where the origin of fv is О and where t = 0 is the
time at which the object is at this origin. Thus, if X is the position of the
object at time t, then X is the tip of fv, and hence
OX = fv.
The points X corresponding to various values of t all lie on the line of the
vector v. When t is positive, the vector fv has the same direction and sense
as ν and / times the magnitude. Since the object is at О when / = 0, we have
00 = Ov = 0.
The time — / (where / is positive) is / seconds before the object reaches the
point 0. The two points corresponding to — / and / are on opposite sides of
О and equidistant from it. Thus
(-/)v- -(fv).
We have now interpreted the product fv for any vector ν and any real
number /. Clearly this product has the properties (see P6, P7, P8, P9 of
Section 9)
(s + /)v = sv + fv, s(tv) = (st)v, tv = ν
when / = 1.

12
VECTORS
(Ch. 1
We wish to show that it also has the property (see P10 of Section 9)
/(u + v) = /u + fv.
If u = OA, ν = AB, then u + ν = OB; and if /u = ОС, tv = CD, then
tu + tv = OD. It is easily seen that the triangles OAB and OCD are similar
and that OD = /02?; i.e., that tu + tv =
t(u + v). Figure 4.1 depicts this result
for the case in which / is positive.
If X is the tip of the vector xu, where
u is the unit vector [0°, 1] considered in
Section 1 and 0 is the origin of u, then
OX = xu and χ is called the coordinate of
the point X on the line of u. Each real
' С number χ determines a point having χ as
its coordinate, and each point of the line
Figure 4.1 of the vector u determines its coordinate
x. Thus the real numbers can be pictured
as points of this line. The distance from X to the origin is the magnitude of
OX. When χ is positive, the magnitude of OX = xu is χ times that of u and
hence equal to x. In this case X lies | 0X\ = χ units to the right of 0. When
χ is negative, X lies | 0X\ units to the left of 0-and χ = — | 0X\. When x
is zero, X coincides with 0, and the distance | 0X\ is zero. The distance
from X to 0 is called the absolute value of the coordinate χ of X and is denoted by
|x|. Thus |x| = \0X\ and
χ = | χ | if χ is positive,
χ = — \x\ or |χ| = — χ if д; is negative,
101 = 0 and | — x\ = |x|.
If x, у are respectively the coordinates of Χ, Υ on the line of u and if Ζ is
the tip of the vector
OX + OY = xu + yu = (* + y)u,
then χ + у is the coordinate of Z. Thus the addition of numbers can be
pictured as vector addition in which the vectors are restricted to lie within

Sec. 4] MULTIPLICATION OF A VECTOR 13
the line of the vector u. For example, we can picture the equation 2 + (— 5)
= — 3 by constructing the sum 2u + ( —5)u and noting that this vector has
its tip three units to the left of 0. Coordinates can be assigned to the points of
any line by choosing one of the points as an origin and a unit vector to
determine the sense.
The concept of absolute value can be used to describe the magnitude of
xv. If χ is positive, then the magnitude of xv is χ times that of ν and hence
|x| times that of v. If χ is negative, then the magnitude of xv = — |x|v is
the same as that of | x\ v, and this magnitude is in turn | x\ times that of v.
If χ = 0, then the magnitude of xv is zero, and this magnitude is 0 times that
of ν or |x| times that of v. In any case
\xv\ = |x|·Iv|.
If a, b are two nonzero vectors which do not necessarily have a common
origin, then these vectors have the same direction (i.e., they are parallel) if
and only if there is a number χ such that
b = хя.
They have the same sense when χ is positive and opposite senses when χ is
negative. To find x, note that since
|b| = |xa| = |χ|·|a|,
we must have
_|b|
I a. |
that if a, b have the same sense, then
_|b|
I a. |
and that if they have opposite senses, then
|b|
I a. |
EXAMPLE 4.1
If a, b are respectively the vectors [90°, 3], [90°, 5], then they have the
same direction and same sense and |a| = 3, |b| = 5. Hence b = хя, where
χ = \x\ = |b|/|a| = %;i.e.,
b = $a and §b = f($a) = (f-|)a = a.

14
VECTORS
(Ch. 1
EXAMPLE 4.2
If a, b are respectively [90°, 3], [270°, 5], then they have the same
direction but have opposite senses. Hence b = xa, where χ = — | χ \ =
-|b|/|a| = -^;i.e.,
Ь = -fa, а = -fb.
It is often desirable to find a unit vector u having the same direction and
sense as a given vector v. If ν is a nonzero vector, then u will have the same
direction and sense as ν if there is a positive number / such that u = vt.
Moreover, u will be a unit vector if
1 = | u j = | v/1 = | ν | · | /1.
If / is the positive number satisfying this condition, then
1
| ν |
Thus
ιι \
u = ν/ = ν I 1
\ Ivl /
is the desired vector. We shall use the abbreviation
—(π)-π
\ V / V
Questions
1. What does sv = vs mean (a) when s = 2, (b) when s is any positive number,
(c) when s = 0, (d) when s = — 1, (e) when s is any negative number? 2. Describe
how the real numbers are made to correspond to the points on the line of the vector
u = [0°, 1]. In this correspondence what meaning is assigned to |x|? 3. State the
values of | — 71, 171, |0|.4. Explain the equation | xa | = | χ | · | a |. 5. How do you
find s such that a = bs when a, b are nonzero vectors with the same direction and
sense? How do you find s when a, b have opposite senses? 6. How do you find a unit
vector with the same direction and sense as a when a is a given nonzero vector?
Problems
1. Let ui = [0°, 1], u2 = [90°, 1]. Draw the vectors 2ub 3u2, — Зщ, —4u2.
Construct the vectors 2ui + Зиг, — Зщ + Зиг, 3ui — 4иг, — 3ui — 4иг.
2- Let ui = [210°, 1], u2 = [330°, 1], u3 = [90°, 1]. Draw the vectors 2щ, 5u2) 4u3.
Construct the sums 2ui + 5иг, 2щ + 4из, 5иг + 4из, (2ui + 5иг) + 4ua,

Sec. 5]
COMPONENTS OF VECTORS
15
(2щ + 4из) + 5u2 in a single figure, using parallelograms for the constructions
in each case. Your figure should look like a box drawn in perspective.
3. Construct the vectors ui — 2u2 + Зиз, — 2ui + Зиг — из, where щ, иг, из
are defined as in Problem 2.
4. Let a = [0°, 5], b = [60°, 4] have common origin at 0; let B' be the foot of
the perpendicular dropped from the tip of b to the line of a and let и = a/1 a |.
Draw a, b and locate B'. Find the exact value of у such that OB' = yu.
5. Proceed as in Problem 4, using a = [0°, 5], b = [120°, 4].
6. Let a, b, у be defined as in Problem 4; let A' be the foot of the perpendicular
dropped from the tip of a to the line of b and let ν = b/1 b |. Find χ such that
О A' = xv and show that y\ a | = χ | b |.
7. Let ν = [30°, 1] have origin 0. Locate the points A, B whose coordinates are
respectively 3 and —5 with respect to the origin О on the line of the unit vector v.
Construct the sum OA + OB. Find |CM|, \OB\, |3|, I -5|.
5. COMPONENTS OF VECTORS
Components furnish a means of specifying vectors both in the plane and
in three-dimensional space. If a = О A, b = OB are nonzero vectors with
common origin 0, then the component of b in the direction a is the coordinate
of the projection B' of the tip В of b on the line of a where this coordinate
is referred to the origin О and is positive or negative according as B' and
the tip of a are on the same or opposite sides of 0. See Figs. 5.1, 5.2.
The coordinate of a point B' on the line of a is the number у such that

16
VECTORS
(Ch. 1
OB' = yu, where u = a/|a| is the unit vector having the same direction
and sense as a. Since B' is the projection of В on the line of a, it must be
that point of the line for which B'B is perpendicular to a (abbreviated
B'B _L a), where
B'B = OB - OB' = Ь - yu.
Hence the component of b in the direction a (i.e., the coordinate of B')
is that number у such that
b — y\i _L a,
where u = a/1 a |. This component can be estimated by measurement after
constructing the vectors a, b and locating B'.
Let us see how the component of b in the direction a is related to the
component of a in the direction b; i.e., to the number χ such that
a — xv _L b,
where ν = b/1 b |. Note that χ is the coordinate of a point A' on the line of b
and hence OA' = xv and that A' is that point of this line for which
(see Figs. 5.1, 5.2)
A'A = OA - OA' = a - xv,
where
a — xv _L b.
Since the triangles OB'В and OA'A have a common angle at 0 and each
contains a right angle, they are similar, and hence
\0B'\ \0A'\
\0B\ \0A\
or
\0B'\ -\OA\ = \OA'\-\OB\,
where \OB'\ = \yu\ = \y\, |CM'| = |x|, |CM| = |a|, |ОЯ| = |Ь|
Thus
\y\-|a| = |*|-|b|
and moreover
y\a\ = x\ b|,

Sec. 5]
COMPONENTS OF VECTORS
17
since χ and у always have the same sign. That is, χ and у are both positive or
both negative according as the angle between a and b is acute as in Fig. 5.1
or obtuse as in Fig. 5.2, and they are both zero when this angle is a right
angle. Thus the component of b in the direction a multiplied by the
magnitude of a is equal to the component of a in the direction b multiplied by
the magnitude of b.
The product y\a.\ has the following physical interpretation: If an
object moves from a point 0 to a point A, then the vector a = OA is
called the displacement. If the object is acted upon by a force b throughout
this motion and у is the component of the force in the direction of the
displacement, then the product у | а | of this component times the magnitude of
the displacement is called the work done on the object by the force. The concept
of work is important in physics, but our interest in the product ji|a|
(as we shall see) is its usefulness in geometry.
We shall show how to specify a vector in a plane by means of
components. If ub U2 are respectively the mutually perpendicular unit vectors
[0°, 1], [90°, 1] with common origin 0, and χ = OX is any vector in
the plane of Ui, U2 with origin 0, then the components of χ in the
directions ub иг are respectively the numbers X\, X2 such that χ — *iUi _L
Ui, χ — X2U2 J. иг- The parallelogram used in constructing the sum
XjUj
Uj
Ο ύ, x, и," "'
Figure 5.3
*iUi + X2U2 is the rectangle shown in Fig. 5.3, and it is easily seen from
this figure that
X = XlUi + *2U2.
We call χ the vector [xit X2]. The components x\, хг determine χ and
hence its tip X. This tip is called the point (χι, Χ2) and xu *2 are called its

18
VECTORS
(Ch. 1
coordinates. Thus a pair of numbers in square
brackets represents a vector and a pair in
parentheses represents a point. Figure 5.4
shows the vectors [3, — 2], [—4, 3] and their
tips (3, — 2), ( — 4,3). A pair of mutually
perpendicular unit vectors such as ub иг
is called an orthonormal system, and the
vectors forming the pair are called respectively
the *i-axis and лгг-axis. The device of locat-
Figure 5.4 ing points by means of their coordinates
furnishes a method of studying plane
geometry by means of algebra and of picturing algebraic relations geometrically.
This is the method of plane analytic geometry.
Questions
1. Define the component of a vector b in the direction of a vector a. 2- Show that
this component is the number у such that b — yu _L a where u = a/1 a |. 3. Show
that if χ is the component of a in the direction of b, then у | a | = χ | b |. Note that the
angle between a and b may be an acute, obtuse, or a right angle. Consider all three
cases. 4. What is meant by a displacement vector? 5. What is meant by the work done
by a force? 6. What is an orthonormal system? 7. How are the vectors ui, иг chosen to
form such a system? 8. Show by a drawing that if x\, хг are the components of a vector
χ in the directions ui, иг, where χ is in the plane of ub u2, then χ = *iUi + *2u2.
9. How are the coordinates of a point defined? 10. Explain the notations [*i, *2]
and (*i, *2).
Problems
1. Let a = [30°, 5], b = [75°, 2]. Find exact values of the component у of b in
the direction a and the component χ of a in the direction b. Check that
y\a| = χ|b|. Draw figure.
2- Proceed as in Problem 1 for the vectors a = [30°, 5], b = [345°, 2].
3. Find the components of the vectors a = [45°, 3], b = [135°, 4], с = [225°, 1],
d = [315°, 2] in the directions ub u2, where щ = [0°, 1], u2 = [90°, 1]. Draw
figure.
4. Find the components of ui, u2, — ui, — u2 in the directions ui, u2.
5. Plot the points (1, 3), (2, -2), (4, -3), (-3, -2).
6. A force [90°, 5] acts on an object during a displacement [30°, 2]. Find the
work done.
7. Draw the vectors a = [2, 2], b = [ — 3, 3]. Find the component of b in the
direction of a.
(-4,3)
......
О
1
1

Sec.6] FURTHER PROPERTIES OF COMPONENTS 19
8. What point is the projection of the tip of a on the line of a? Show that the
component of a in the direction of a is |a|.
6. FURTHER PROPERTIES OF COMPONENTS
We shall study components of the sum of two vectors; to do so, we
make use of a property of perpendicular vectors. Let u, v, w be three
vectors with common origin 0 such that w _L u and w _L v. If u, ν have
different directions, they determine a plane, and w must be perpendicular
to this plane. Hence w must be perpendicular to any vector in this plane
and in particular w _L u + v, since this sum lies in the plane. This result
is depicted in Fig. 6.1. To visualize
the figure, imagine that the plane of w
u, ν is horizontal and that w is
vertical. If u, ν have the same direction,
they are contained in a line, and w
is perpendicular to this line. Again
w _L u + v, since this sum is
contained in the line. The condition that
u, v, w have a common origin was
introduced so as to simplify the
reasoning. This condition is unnecessary, Figure 6.1
since if a vector χ is perpendicular to
a vector y, then χ is perpendicular to any vector equivalent to y. Thus in any
case w _L u and w _|_ ν imply w _|_ u + v.
If a, b, с are nonzero vectors (Fig. 6.1), then the components of b, с in
the direction a are respectively the numbers y, ζ such that
b — yu _L a and с — zu _L a,
where u = a/|a|. Hence a must be perpendicular to the sum (b — yu) +
(c — zu), and this sum can be rearranged as follows:
Ъ— yu-\-c — zu = b-\-c — yu — zu
= (b + c) - (y + z)u.
That is, the sum у + ζ of the components of b, с is the number such that
(Ы- c) - (y + z)u _L a,
where u = a/1 a |, and this number is the component of b + с in the di-

20
VECTORS
ICh. 1
rection a. Therefore the component of the sum of two vectors is the sum of
their components.
Let a, b be nonzero vectors, у be the component of b in the direction
of a, and / be a nonzero real number. Clearly b — yu _L a implies
/(b - yu) _L a or (/b) - (ty)u _L a;
i.e., ty is the number such that
(Л) - (ty)u _L a,
where u = a/1 a |, and this number is the component of /b in the direction
of a. Therefore the component of /b in the direction a is / times the
component of b in the direction of a.
If two forces b, с act on an object during a displacement a, then the
work done by b is у\л\ and that done by с is z|a|, where y, ζ are
respectively the components of b, с in the direction of the displacement. If b, с
act at a single point of the object, they are equivalent to a single force
b + с The work done by b + с is
(y + z)|a| = y\a\ + z|a|,
since у + ζ is the component of b + с in the direction of the displacement.
Hence the work done by b + с is that done by b plus that done by с The
component of the force /b in the direction of the displacement is ty, and
the work done by this force is (/y)|a| = t(jr|a|). That is, the work done by
/b is / times that done by b.
Questions
1. Explain why u _|_ w and ν J_ w imply u + ν _L w. Consider both the case in
which u, ν have different directions and the case in which they have the same
direction. Is a common origin necessary? 2· Show that the component of b + с in
the direction of a is the sum of the components of b and с in that direction. 3. Show
that the component of ib is t times the component of b. 4. Show that the work done by
the force b + с is the work done by b plus the work done by с 5. Show that the
work done by ib is t times the work done by b.
Problems
1. Let a = [αϊ, аг], b = [b\, Ьг\. Recall that αϊ, аг are the components of a in the
directions ui, U2, respectively, and that the component of a sum is the sum of
the components. Show that a + b = [αϊ + b\, аг + Ьг].
2. Show that ia = [ta\, /аг] if a = [αϊ, аг].

Sec. 7]
SCALAR PRODUCTS
21
3. Draw the vectors a = [3, 4], b = [5, 0], с = [0, 2]. Find the components of a in
the directions b, с
4. Let a, b, с be defined as in Problem 3. Compute |a| by the pythagorean
theorem. Also find |b|, |c|.
5. Recall the relation between the component of a in the direction b and the
component of b in the direction a. Find the components of b and с in the
direction a where a, b, с are defined as in Problem 3.
6. Let a, b, с be defined as in Problem 3 and let d = [5, 2] = b + с Find the
component of d in the direction a.
7. Let a = [αϊ, 0], b = [Ь\, Ьг\. Show that the component of b in the direction a is
b\ when a\ is positive and —b\ when a\ is negative. Draw figures.
7. SCALAR PRODUCTS
We have been concerned with a product у | a | of the component у of a
vector b in the direction a times the magnitude of a. The number у\я.\
depends on the vectors a, b; it is called the scalar product or dot product of a and
b and is denoted by (see Pll of Section 9)
a-b = у\л\.
The reason for introducing this concept and calling it a product is that it has
properties which are analogous to the product of two numbers, and this fact
makes for ease in computation. Note that a*b = y\a.\ is the work done by
the force b during the displacement a. We wish to show that (see PI 2 of
Section 9)
a-b = b-a (commutative law).
This law follows from the fact that b · a = χ | b |, where χ is the component of
a in the direction b, and from the relation у | a | = χ | b | between the
components x, y. We show next that (see PI 3 of Section 9)
a-(b + c) = a-b + a-c (distributive law).
To show this, let ζ be the component of с in the direction a. Then у + ζ is the
component of b + с in the direction a, and hence
a-(b + c) = (ji + z)|a| =>|a| +z|a| = a-b + a-c,
since у\л\ = a-b and z|a| = a·c. We have also noted that (see Section 6)
the component of /b is ty, and hence (see PI 4 of Section 9)
a-(/b) = ty\a\ = /(a-b).

22 VECTORS (Ch. 1
In defining projections, we assumed that the vectors were nonzero. To
define the scalar product when one of the vectors is zero, we assume that the
equation a-(/b) = /(a-b) holds when / = 0; i.e., that
a-(Ob) = a-0 = 0(a-b) = 0.
We also assume that the equation a-b = b-a holds even for zero vectors.
Thus
a-0 = 0 = 0a = 00.
The above properties of scalar products are easy to remember because they
are analogous to properties of numerical products.
Let us see how scalar products are used. If a, b are nonzero vectors, then
a _L b if and only if the component у of b in the direction a is zero and,
hence, if and only if a · b = j> | a | =0. For the sake of uniformity we assume
that this result holds even for zero vectors. Then a _|_ 0 for every vector a,
since a-0 = 0. We can now state that for any two vectors a, b (zero or
otherwise) a _|_ b if and only if a-b = 0. Next let us compute a*a. The
component of a in the direction of a is | a |, since the projection of the tip of a
on the line of a is the tip of a, and this point has the coordinate |a|. As a
check on this result note that у = | a | is such that
|a|a
a — | a | u = a = a — a = 0 and 0 _L a.
| a. |
Hence
a-a = >|a| = |a|2.
This furnishes a means of computing | a |, namely,
| a. | = V a-a.
Note that the scalar product of two vectors is not a vector. The word
"scalar" is synonymous with "number" and a scalar product is a number.
We shall always use the dot to indicate a scalar product. A product of a
scalar (number) times a vector such as sa is never written with a dot, whereas a
product of two scalars is written sometimes with a dot and sometimes without.
Thus, in a product such as |^| · |a|, the dot furnishes a convenient way of
separating the two vertical bars, whereas in the product y\a\ no such
separation is needed.
We shall develop three more properties of scalar products and then show

Sec. 7]
SCALAR PRODUCTS
23
how to compute these products. We have
(a + b)-c = c-(a + b) = c-a + c-b
= а· с + b-c,
and this is another distributive law. Next
(*а)-Ь = Ь-(я) = ж(Ь-а) = ж(а-Ь).
Thus there is no ambiquity in writing
дг(а-Ь) = дга-Ь.
Finally
(дга)-(^Ь) = ха-(уЪ) = xy(a-b).
Since ui, иг is an orthonormal system, | ui | =1 and hence ui'Ui =
l2 = 1. Similarly, 112-112 = 1. Since ui _L U2, it follows that ιΐι·ιΐ2 = 0.
We thus have the multiplication table
Ui'Ui = 1 = U2*U2, Ui'U2 = 0 = U2'Ui.
If χ = *iiii + *2ii2 then
ui-x = ur(*iiii) + иг(дг2и2) = ^iui-ui + дггигиг
= *i + 0 = *i,
and similarly
иг'Х = *ιΐΐ2·ιΐι + дггиг'иг = дсг-
Thus the components of χ in the directions of ui, 112 are respectively
ui-x, u2-x. If у = >iui + y2u2, then
x-y = (дг1и1 + *2ii2)-y = дс^гу + дг2и2-у
= *1>1 + Х2У2,
since игу = >i, u2*y = уг- In particular
2 1 2 I |2
X'X = X\ + *2 = |x| .
If ДГ1, *2 are positive, then they are the lengths of the sides of a right triangle
and |x| is the length of the hypotenuse. See Fig. 5.3. Thus the equation
*!2 + *22 = |x|2 states the Pythagorean theorem. Note that we also have a
means of computing | χ |, namely,

24
VECTORS
ICh. 1
EXAMPLE 7.1
Let χ = 3ui + 4u2, у = — 4ui + 3u2- Compute |z|, |y|, and show
that χ _L y. We have
xx = 32 + 42 = 25, |x| = Vx7^ = 5,
УУ= (-4)2 + 32 = 25, |y| =5,
x.y = 3·(-4) + 4-3 = -12+ 12 = 0,
and hence χ _L y.
Questions
1. What is meant by the scalar product a· b? 2. Show that a· b = b· a. 3. Show that
a*(b + c) = a-b + a· c. 4. Show that a-(ib) = i(a*b). 5. How are the equations of
Questions 2 and 4 used to define a· 0, 0· 0? 6. Show that a _L b if and only if a· b = 0.
Consider zero vectors. 7. Show that a·a = ja|2 and |a| = л/а-а. 8. Show that
(a + b)-c = a-c + b-c. 9. Show that (дга)-Ь = дг(а-Ь) and (хя)-(уЬ) = jry(a-b).
10. Write down the multiplication table for the vectors щ, иг and explain. 11. Let
χ = *iui + *2U2, у = )Ίΐΐι + угиг. Show that urx = x\, иг·χ = дгг, χ·γ = х\у\ +
хгуг. 12. Show that | ж |2 = x\2 + дгг2 and show that this equation is a statement of
the pythagorean theorem. 13. How is |x| obtained when χ = дпщ + *2ii2?
Problems
1. Compute vpvi, V2-V2, \\·\ϊ, where
Vl = fill + |u2, V2 = — $Ul + fu2.
Show that vi, V2 is an orthonormal system.
2. Compute νι·νι, V2*V2, vi-V2, where
vi = viui + ^U2' V2= viui+ viU2'
Show that vi, V2 is an orthonormal system.
3. Show that vi, V2 is an orthonormal system if
1 , л/з -У/Ъ , 1
Vl=-UiH ^-"2, V2 = U1 + -U2·

Sec.8] VECTORS IN THREE-DIMENSIONAL SPACE 25
4. Show that the component of b in the direction a is the number у such that
a-(b — yu) = 0 where u = a/|a|. Apply a distributive law to this product
and solve fory. Show that a-u = |a| and hence that у = a-b/|a|.
5. Show that it follows from the definition of scalar product that the component
of b in the direction a is a·b/1a|.
6. Let a = 3ui + 4иг, b = ui — 112. Compute a-b and |a|. Find the
component of b in the direction a.
7. Let a = aiUi + агиг, а' = — 02111 + αιΐΐ2· Compute a-a', |a|, |a'|. Show
that a _L a' and |a| = |a'|.
8. Let b = A1U1 + A2112 and let a, a' be defined as in Problem 7. Compute a-b
and show that if a-b = 0 and a\ ^ 0, then
к Ьг ι \ к *2
b\ = — ( — аг), Ьг = — βι,
"ι αϊ
and hence b = (bt/ai)*'.
Show that if a-b = 0 and аг у* 0, then
к ~bl к ~bl 1 \
*2 = a I, *1 = (~"2)
ач ач
and hence b = ( —oi/a2)a'.
9. Compute | a |, where a is defined as in Problem 8, and show that | a | = 0
only when a\ = 0 and аг = О.
10. Let a, a', b be defined as in Problem 8. Show that if a _L b and | a | ^ 0,
then either
Ьг —b\
b = — a' or b = a'.
a\ аг
8. VECTORS IN THREE-DIMENSIONAL SPACE
Let ui, u2, 113 be mutually perpendicular unit vectors. Such a system of
vectors is again called an orthonormal system. To visualize these vectors, face a
corner of a room. Let the intersection of the floor with the wall to your left
contain ui, the intersection of the floor with the wall to your right contain
112, and the intersection of the two walls contain 113. Let these vectors have a
common origin О at the corner of the room, and let their senses be such that
they are all visible from a point inside the room.
Figure 8.1 shows the construction of a sum of the form
X = *llli + *2U2 + ДС3ИЗ

26
VECTORS
(Ch. 1
X3U3
Figure 8.1
in which *i, *2, *з are all positive. First the sum д^щ + *2U2 is constructed on
the floor of the room. The parallelogram used in the construction is a
rectangle drawn on the floor. This rectangle is the base of a rectangular
parallelepiped, one of whose vertical edges is the vector ДГ3113. The opposite
vertical edge is the vector that is equivalent to *зиз and has its origin at
the tip of *!Ui + *2ii2· The vector from О to the tip of this equivalent
vector is the desired sum χ = (*iui + ДГ2112) + ДГ3113. Let us show that
*i> *2, '3 are the components of χ in the directions щ, иг, 113. Note that the
plane of the upper face of the rectangular parallelepiped is perpendicular to
113. Hence the line drawn in this plane from the tip of χ to the tip of ДГ3113 is
perpendicular to 113. That is, the tip of ДГ3113 is the projection of the tip of χ on
the line of 113, and since дгз is the coordinate of this projection, it follows that
*з is the component of χ in the direction 113. Similarly, the line from the tip of χ
to the tip of ДГ2112 is perpendicular to 112, the tip of *2U2 is the projection of the
tip of χ on the line of 112, and the coordinate *2 of this projection is the
component of χ in the direction иг- Finally, χι is the component of χ in the
direction щ.
Figure 8.1 is called an isometric projection. Let us see how to draw such a
figure. An isometric projection is much simpler to draw than a perspective
and is a sufficiently close approximation to a perspective to be a reasonably

Sec.8] VECTORS IN THREE-DIMENSIONAL SPACE 27
convincing picture of a three-dimensional configuration viewed from an
appropriate position. An appropriate position is a point equidistant from the
floor and the two walls, and from this point the vectors ui, 112, 113 appear to
make equal angles with one another. The isometric projection of 113 is the
vertical vector [90°, 1], that of ui is [210°, 1], and that of u2 is [330°, 1] so
that between any two vectors in the drawing there is an angle of 120°. If we
make a drawing showing the construction of a sum of the form *iui +
*2"2 + *3i3, in which щ, иг, 113 are actually the vectors [210°, 1], [330°, 1],
[90°, 1], respectively, then this drawing is the isometric projection showing
the construction of the sum x-juj + *2U2 + ДГ3113 in which щ, иг, 113 are
interpreted as the vectors forming the
orthonormal system described at the
beginning of this section (see Problems 2 and
3 of Section 4). The isometric projection 3u3
of the point ta, x2, *3) is located at the tip
of the isometric projection of the vector
*iui + *2ii2 + ДГ3113. The isometric
projection of a line through two points is the
line through the isometric projections
of the points. A 30°-60°-90° triangle Figure 8.2
is useful in drawing isometric projections,
since a line parallel to ui, 112, or 113 is drawn by placing the triangle so that
one edge coincides with an edge of the paper. Figure 8.2 shows the isometric
projections of the vectors ui — 2иг + З113 and — 2ui + З112 — 113.
A vector is determined by its components. If χ has components x\, x2, *з,
it is called the vector [дгь x2, X3]. If χ has origin at О and tip at X, then X is
called the point ta, x2, *з), and x\, x2, *з are called the coordinates of this point.
The vectors ui, 112, 113 are called respectively the *i, x2, *3-axes.
Since I ui I = 1, we have ui*ui = l2 = 1; similarly for 112, 113. Since
ui -L «г, we have щ-иг = 0; similarly for up из, иг· из. Hence we have the
multiplication table
Ui'Ui = U2'U2 = U3*U3 = 1
Ul'U2 = U2'Ui = U2'U3 = U3 · U2 = U3 · Ul = Ul · U3 = 0.
If x = д^щ + *2U2 + *з"з> then
U!-X = Ill-tall!) + Ul'tau2) + Ui-(*iU3)
= *lUl'Ul + ДГ2и1-и2 + *3Ul'U3 = *Ь
and similarly иг-х = дсг, из*х = дез- Thus x has the components u^x, иг'Х,

28
VECTORS
ICh. 1
u3-x. If у = yiui + y2u2 + Узт, then
χ.γ = (*lUl + *2u2 + *з"з)'У = *iury + *2И2'У + *зиз'У
= Х\У\ + *2>2 + ХЗУЗ-
In particular
х-х= Xl2 + x22 + x32 =\x\2 and |х| = VXl2 + X22 + X32.
If дгь *2, дгз are positive, they are the lengths of three adjoining edges of a
rectangular parallelepiped and |x| is the length of a diagonal. The equation
*!2 + *22 + ДГ32 = |x|2 states the three-dimensional Pythagorean theorem.
EXAMPLE 8.1
Let χ = m + 2u2 + 2из, у = 2щ + иг — 2u2. Compute |z|, |y|,
and show that χ J. y. We have
xx = l2 + 22 + 22 = 9, |x| = л/х^х = 3,
yy = 22+l2+(-2)2 = 9, |y|=3,
x-y = 1-2 + 2-1 + 2·(-2) = 0
and hence χ J. y.
Questions
1. Describe the location of the vectors щ, иг, из when the origin is taken on the
floor in the corner of a room. 2- Describe in detail how a sum of the form χ = (*iui+
*2иг) + *зиз is constructed. You may refer to Fig. 8.1. 3. Explain why the line from
the tip of χ to the tip of дгзи3 is perpendicular to U3. 4. Explain why дг3 is the
component of χ in the direction u3. Explain why *2, *i are respectively the components of
χ in the directions иг, Щ. 6. Explain how to draw the isometric projection of a vector
*iui + дггиг + дгзиз. 7. Explain the notations [*i, *2, *з], (*i, дг2, дг3). 8. Write down
the multiplication table for the vectors ui, иг, из and explain. 9. Let χ = Xiu\ +
*2U2 + дсзЧз, у = .yiui + угиг + у3щ. Compute ui-x, иг·», и3·*, χ·γ and explain.
10. Show how to find | χ |.
Problems
1. Compute vrvi, V2*V2, уз-уз, vi'Vj, V2'V3, vj'Vi where
Vl = 5U1 + f U2 + f U3
V2 = §Ul + 5U2 - f U3,
Vj = — §Ui + |u2 - 5U3.
Show that vi, V2, V3 is an orthonormal system.

Sec. 9]
THE POSTULATES
29
2. Show that vi, V2, V3 is an orthonormal system if
V3 = v1Ul + 7IU2-^U3'
3. Draw the isometric projections of the vectors 2щ + 5иг + 4из, 5ui — 2иг +
4u3, — 2ui + 4u2 — 5u3.
4. Plot the isometric projections of the points A = (1, 2, 3), В = (3, — 1, —2).
Draw the isometric projection of AB.
5. Let χ = [xh x2, x3], γ = [y\, уг, уг\- Show that χ + у = [x\ + y\, хг + уг, хз +
уз], tx = [txi, tx2, tx3], —χ = [ — χι, —Χ2, —хз], Υ — » = lyi — χι, Уг — хг, уз —
хз\. See Problems l and 2 of Section 6.
6. Show that if χ = [xh x2, x3], γ = [yi, уг, уз], then
ly - ж| = V(yi - Xly + (y2 - Xiy + (уз - хз)2.
7. Find \AB\ where A, B are defined as in Problem 4.
8. Let a = [1, 2, 2], b = [3, —1, 1]. Compute a-b and |a|. Find the component
of b in the direction a.
9. THE POSTULATES
Postulates 1 to 14 inclusive are assumed to hold for any three vectors
a, b, с and any two real numbers x, y.
PI. a + b is a vector.
P2. a + b = b + a.
P3. a + (b + c) = (a + b) + с
P4. There exists a vector 0 such that a + 0 = a.
P5. There exists a vector —a such that a + ( —a) =0.
Postulates 1, 3, 4, 5 state that vectors form what is called a group with
respect to the operation +. Postulate 2 states that the group is commutative (or
abelian). The real numbers also form an abelian group with respect to the
operation + and so do the integers.
P6. дга is a vector.
P7. хл = a, if χ = 1.
P8. x(ya) = (xy)a.
P9. (* + у)л = дга + ул.
ΡΙΟ. х(я + Ь) = хл + хЪ.

30
VECTORS
(Ch. 1
A mathematical system that satisfies PI to P10 inclusive is called a vector
space. The following postulates define the scalar product:
PH. a-b is a number determined uniquely by я, b.
P12. a-b = b-a.
P13. a-(b + c) = a-b + a-c.
P14. я-(уЪ) = j-(a-b).
P15. If я И 0, then a-a > 0.
The following postulates determine the relations between points and
vectors:
Ρ16. For every two points А, В there is a unique vector с such that с = AB.
Ρ17. For every point A and every vector с there is a unique point В such that с = AB.
P18. AB + ВС = AC for every three points А, В, С
Ρ19. There exists at least one point.
The scalar product can be used to define the following:
D9.1. The magnitude of a vector a is the number |a| = \/a-a.
D9.2. The distance between two points А, В is the magnitude \AB\ of the vector AB.
D9.3. л Lbif and only if a-b = 0.
From now on we shall be concerned with proofs of theorems, and the
steps in the proofs are required to be justified by the postulates, definitions,
or previously proved theorems. There is a reason for this fussiness. Unless we
agree on which statements can be used in proofs, we are permitting the use of
any statement that seems plausible, and occasionally we encounter a
statement that at first glance seems plausible but which turns out to be inconsistent
with the mathematical system. However, we shall allow certain relaxations
of the requirement in order to save time and avoid boredom. Namely, there
are algebraic manipulations that require an unreasonable amount of time to
justify but which probably will almost always be performed correctly without
the justifications. We shall consider an example of such manipulations at the
end of this section.
Note that P4 and P5 are concerned with a vector denoted by 0 but that
these postulates do no explicitly state that there is only one such vector.
However, the uniqueness is established in Problem 1 by showing that if
a + χ = a and a + у = a for every a, then χ = у; i.e., there is only one
vector satisfying the conditions of P4. This vector is denoted by 0. Similarly
the vector —a is unique (see Problem 4). Note that the results of problems
are regarded as theorems.

Sec. 9]
THE POSTULATES
31
It follows from PI 2 and PI 4 that (*a)-b = *(a-b) and (хя)-(уЪ) =
дгу(а-Ь). (See Section 7.) Let us see how the latter result can be used to show
that | xa | = | χ | · | a |. We have
|*a|= V(*a)-(*a) = V*2a-a = V7V^ = V7|a|,
and it remains to show that V*2 = \x\, where V*2 is the symbol for the
non-negative number whose square is x2. If χ is non-negative, then |*| =
χ and |*|2 = x2; and if χ is negative, then |*| = — χ is positive and hence
non-negative, and again |*|2 = ( —x)2 = x2. Therefore VV = |*| and
| xa. | = | χ | · | a |.
We shall use vectors in studying both plane and solid analytic geometry.
The above postulates do not specify whether the vectors all lie in a plane or a
three-dimensional space or in a space of some other number of dimensions.
We have noted that a plane (i.e., two-dimensional space) contains an ortho-
normal system consisting of two vectors ui, иг (i.e., | ui | =1 = |иг| and
ui _L 112) with the property that every vector χ in the plane can be expressed
in the form χ = дгцц + *2ΐΐ2· Note that xu х2 are uniquely determined by
ui, иг, x, since x-ui = дгь χ·ιΐ2 = *2 (see Problem 16). At a later point we
shall prove that the condition that every χ can be so expressed is sufficient to
ensure that the space cannot contain an orthonormal system vi, V2, V3 of
three vectors, and hence it cannot contain one of more than three vectors.
A three-dimensional space contains an orthonormal system of three vectors
«i> u2, из with the property that every χ in the space can be expressed in
the form χ = дгцц + *2U2 + ДГ3113. We shall prove that when every χ can
be so expressed, the space cannot contain an orthonormal system of more
than three vectors. A line (i.e., a one-dimensional space) contains a unit
vector u (i.e., an orthonormal system of one vector) with the property that
every χ can be expressed in the form χ = *u. When every χ can be so
expressed, the space cannot contain an orthonormal system of more than
one vector. These considerations suggest the following definitions:
D9.4. An η-dimensional vector space is one which contains an orthonormal system of η
vectors ui, U2, · · · un with the property that every vector χ in the space can be expressed
in the form χ = *iiii + *2ii2 + · · ·*ηιιη.
We think of a space as being made up of points. It follows from P19 that
there exists at least one point 0; from PI 7, that to every vector χ there
corresponds a unique point X such that χ = OX. These points X constitute
what is called a set. A set of points is merely a collection of points. We have:
D9.5. An η-dimensional space is the set of points associated with an n-dimensional
vector space.

32
VECTORS
(Ch. 1
The reader will soon learn to perform algebraic manipulations with
vectors as easily as he would if the symbols stood for numbers instead
of vectors. When such manipulations are correct, they can always be justified
by the postulates. Such justifications are arduous but otherwise not difficult,
and after this section they will be omitted. However, it is instructive to see
some examples in which the justifications are given.
EXAMPLE 9.1
Show that у — χ = (>! — χ^ιΐι + (y2 — x2)u2 if x = *iui + x2u2, γ =
>iui + >2"2· It follows from the result of Problem 5 that у — χ = у +
(— 1 )x. The remainder of the proof consists in the following ten steps,
and in Problem 15 you will be asked to specify the authority for each of
these steps.
(1) (-l)x = (-lX'iU! + *2u2) = (-1)(*1И1) + (-l)(*2u2)
(2) = (-*l)ui + (-*2)U2,
and hence
(3) у + (-l)x = (у!Щ + y2u2) + ((-*i)ui + (-«2)112)
(4) = ((У1Щ + y2u2) + (-*i)ui) + (-*2)u2
(5) = (У1Щ + (y2U2 + (-*l)u!)) + (-*2)u2
(6) = (У1Щ + ((-*ι)ΐΐ! +>2U2)) + (-*2)u2
(7) = ((>lUl + (-*ΐ)ΐΐ!) + y2U2) + (-X2)u2
(8) = (У1Щ + (-*ι)ΐΐ!) + (y2U2 + (~X2)u2)
(9) = 0-1 + (-*1))U! + 0-2 + ("*2))U2
(10) = (>i - *i)ui + (y2 - x2)u2.
Questions
1. Define the magnitude of a vector and the distance between two points. 2. State
the conditions under which two vectors are perpendicular. 3. Define an n-dimensional
vector space. 4. Discuss the correspondence between points and vectors. 5. Define an
η-dimensional space.

Sec. 9]
THE POSTULATES
33
Problems
Prove the following. Postulates to be used in the proofs are given in parentheses in
some cases.
1. If a + χ = a and a + у = a for every a, then χ = у. (Р2.)
2. If χ + χ = x, then ж = 0. (P5, P3, P4.)
3. If χ = 0a, then χ + χ = χ and hence χ = 0. (P9.)
4. If a + χ = 0 and a + у = 0, then χ = χ + (a + (-a)) = y. (PI, P3,
P4, P5.)
5. a + (-l)a = 0 and hence (-l)a = -a. (P7, P9.)
6. If χ = A A, then χ = 0. (PI 8.)
7. If с = ЛЯ, then -с = ΒΑ.
8. If a·a = 0, then a = 0. (PI 5.)
9. (»)·(») = *Va).
10. If хл = 0, then χ = 0 or a = 0.
11. a + a = 2a.
12. If a = —a, then 2a = 0 and hence a = 0.
13. If AB = 0, then A = B.
14. If AB = BA, then A = B.
15. For each of the ten steps of Example 9.1, state which postulate justifies the step.
16. Prove that if щ, иг is an orthonormal system and χ = *iUi + *2ii2, у =
yiUi + угмг, then ui-ui = 1 = 112*112, χ·Μ\ = χ\, χ·να = хг and χ·γ =
х\У\ + хгуг-

PLANE
ANALYTIC
GEOMETRY
10. CIRCLES AND LINES
In plane euclidean geometry we study the constructions which can be
performed with ruler and compass. A line can be drawn with a ruler and a
circle can be drawn with a compass. In this section we study circles and lines
by the methods of analytic geometry.
Plane geometry is the study of the points of a plane, i.e., the study of the
points associated with a two-dimensional vector space. We select a point О
of the plane and call it the origin. If A" is any other point of the plane, then OX
is a vector by P16. The vector space contains an orthonormal system ui, 112
with the property that every vector OX can be expressed in the form OX =
*!Ui + *2ii2. Recall that (дгь *2) denotes the point X and that *b *2 are its

Sec. 10]
CIRCLES AND LINES
35
coordinates. The coordinates are uniquely determined by X, since ui · OX =
χι, u2OX = *2. Conversely an ordered pair Xl, *2 determines a unique point
X such that OX = ^ui + *2u2.
The distance between two points X = (*ι, *2), Υ = (y\, y2) of the plane is
|ΛΎ|, by D9.2. Since OY = OX + XY, by P18, it follows that
AT = OY - OX = (yi - Х1)щ + (y2 - *2)u2,
and hence
\XY\ = VAT·AT = V(yi - Xl)2 + (y2 - X2)2.
This is called the distance formula. We have the theorem:
T10.1. The distance between the points (xly *2), {y\,y2) is
V(^ - x{f + (y2 - x2)2.
EXAMPLE 10.1
The distance between (1, —2) and ( — 2, 2) is
V(-2 - l)2 + (2 + 2)2 = V32 + 42 = 5.
EXAMPLE 10.2
The distance between (1, —2) and (дгь *2) is
V(^ - i)2 + (X2 + if.
A circle is made up of all points in the plane at a given distance (called
the radius) from a given point (called the center). That is, the circle is the set of
such points. Consider a circle with radius 5 and center at (1, —2), and let
(*i> '2) be a point of this circle. The equation
V(Xl - l)2 + (X2 + 2)2 = 5
is equivalent to the sentence, "the distance between (1, —2) and (дгь *2) is
5" and hence is equivalent to the sentence, "(*i, *2) lies on the circle with
radius 5 and center at (1, —2)." This equation is called an equation of the
circle. The circle is the set of all points (*b *2) such that the equation is
satisfied when the values of xu дг2 are substituted in it. Any equivalent
equation is also an equation of the circle. Thus, squaring both sides of the

36 PLANE ANALYTIC GEOMETRY (Ch. 2
above equation, we obtain the equivalent equation
(*i " I)2 + (*2 + 2)2 = 25.
Other equations of the circle are
*!2 - 2*! + 1 + *22 + 4X2 + 4 = 25
and
*i2 + *22 - 2*, + 4*2 - 20 = 0.
The reader can readily check that all four equations of the circle will be
satisfied if we substitute χι = — 2, x2 = 2. Hence ( — 2, 2) is a point of
the circle.
A line is defined to be a one-dimensional space (see Section 9). If an
object moves along a line with constant speed, it will have a velocity vector v.
If the object is at a point A = (ab a2) at time / = 0, then at any other time /
it will be at the point X = (xu x2) such that AX = vt. This suggests that the
set of points X for which AX is of the form vt is a one-dimensional space and
the set of vectors χ of the form vt is a one-dimensional vector space. To prove
that this suggestion is correct, it is necessary to check that these vectors
and these points satisfy all the postulates. We prefer to postpone this checking
until a later point (Section 39) in order to progress more rapidly into analytic
geometry. At present we merely note that when / = 1/|»|, we obtain the
unit vector u = v/ = v/1 ν |, that ν = u | ν |, and that for arbitrary / we have
AX = vt = u|v|/ = их
where χ = | ν | / is the coordinate of X relative to the origin A and the ortho-
normal system u. The reader is requested to accept tentatively without proof
that the set of points X for which AX has the form vt is a one-dimensional
space. This space is called the line through A with direction v. If ν has origin at A,
then the space is also called the line containing the vector ν with origin A.
The scalar / is not given a direct geometrical interpretation, and such a
scalar is sometimes called a parameter. We have noted, however, that the
parameter / may have a physical interpretation, namely, that of time. We can
obtain a relation between / and the coordinates of A, X, and components of
ν = [»i, d2] as follows:
OX = OA + AX = OA + vt = aiui + a2u2 + (»iui + v2u2)t
= (θΐ + M)ul + (θ2 + V2t)u2
or
[*i> '2] = [αϊ + vit, a2 + v2t].

Sec. 101
CIRCLES AND LINES
37
Since the components of a vector are unique, we have
*1 = al + »l/, *2 = 02 + V2t.
These two scalar equations are called parametric equations of the line. By
substituting any value of the parameter into the parametric equations, we obtain
the coordinates дгь x2 of a point X of the line. In particular when / = 0, we
obtain the point A = (a\, a2).
EXAMPLE 10.3
Plot various points of the line containing the vector [3, 1] with origin
( — 1, 0). The parametric equations are
*i 1+3/, *2 = /.
Substituting the values —1,0, 1, 2 for t in these equations, we obtain
respectively the points ( — 4, —1), ( — 1, 0), (2, 1), (5, 2). These points are plotted
in Fig. 10.1.
(-4,-1)
Figure 10.1
Let A = (au a2), В = (bu b2) be any two points in the plane. The line
containing the vector A B with origin A is called the line through the points
A, B. Since ν = А В = OB — О A, we have
OX = OA + (OB - OA)t = OA{\ - t) + OBt
[*i, *г] = k(l - 0 + b^, a2{\ - t) + bat],
and hence the parametric equations have the form
*, = a,(l - 0 + bit, x2 = a2{\ - t) + b2t.
When / = 0, we obtain the point A = (ab a2), and when / = 1, we obtain
Β = (6ι, b2).

38 PLANE ANALYTIC GEOMETRY (Ch. 2
EXAMPLE 10.4
Plot various points of the line through (3, 2), (7, 0). The parametric
equations are
*! = 3(1 - /) + 7/, *2 = 2(1 - /).
The points corresponding to the values —1,0, H> 1, 2 of the parameter are
respectively (— 1, 4), (3, 2), (5, 1), (7, 0), (11, —2). These points are shown
in Fig. 10.2.
(-1,4)
Figure 10.2
Questions
1. Derive the distance formula. 2. Describe a circle as a set of points. 3. What is
meant by an equation of a circle? 4. What is a line? 5. What is the line through a
point A with direction v? 6. Derive parametric equations of this line. 7. Derive
parametric equations of a line through two points A, B.
Problems
1. Find the distance between the points (3, —1), ( — 1, 7).
2. Find an equation of the circle with center ( — 1, 2) and radius 3.
3. Show that *i2 + *22 — dx\ + 8*2 = 0 is an equation of the circle with center
at (3, —4) and radius 5. Show that the origin (0, 0) lies on this circle.
4. Find parametric equations of the line containing the vector [3, —1] with origin
( — 2, 2). Plot the points corresponding to the values —1, 0, 1, 2 of the
parameter and draw the line.
5. Find parametric equations of the line through (1, 2), (5, —2). Plot the points
corresponding to the values 0, И, 1 of the parameter t and draw the line.
6. Find the distance between the points for which t = 0 and t = ^i in the line of
Problem 5. Also find the distance between the points for which t = \i and

Sec. 11] PARAMETRIC EQUATIONS 39
/ = 1. Show that the two distances are equal. Note that the point for which
t = И is the midpoint of the segment joining the other two points.
7. Find an equation of the circle, one of whose diameters is the segment joining
the point for which t = 0 to the point for which t = 1 of the line of Problem 5.
11. APPLICATIONS OF PARAMETRIC EQUATIONS
Consider the problem of locating a point X so that it divides a segment
from A to В into two segments whose lengths are in a prescribed ratio. The
two segments determine the two vectors
AX = ABt, XB = AB - AX = AB - ABt = AB{\ - t).
An interior point of the segment from A to В is defined to be one for which the
corresponding / is between zero and one, and both / and 1 — / are positive.
The lengths of the two segments (i.e., the vectors) are
\AX\ = \AB\-\t\ = \AB\t, \XB\ = \AB\-\\ - t\ = \AB\{\ - t),
and the ratio of the lengths is
\AX\ \AB\t t
\XB\
\AB\{\ - t) (1 ')
To find the / which will produce a prescribed ratio m/n, we set t/(\ — t) =
m/n and solve. Thus
/
1 -
m
m
t η
mt + nt = (m + n)t = m,
m + η m η
-, 1 - / = =
/ =
m + η m + η m + η τη + η
Substituting these values of/, (1 — /) in the parametric equations, we obtain
the coordinates
η m na\ + mb\ na2 + mb2
Xl=ai—— +*!— x2 =
m + η m + η m + η τη + η
These equations state that (дгь *2) divides the segment from (αϊ, α2) to (bu b2)
in the ratio m/n.

40 PLANE ANALYTIC GEOMETRY (Ch. 2
EXAMPLE 11.1
Consider the triangle whose vertices А, В, С are respectively the points
(3, 2), (7, 0), (6, 4). The midpoint of the segment from A to В divides this
segment in the ratio 1/1. Thus we set m = η = 1. The coordinates of the
midpoint are
ai+ b1 3 + 7 2 + 0
*i = = = 5, *2 = = 1.
1 + 1 2 2
A segment joining a vertex to the midpoint of the opposite side is called a
median. The medians meet in a point that divides each median in the ratio
2/1. To find the point (yi,y2), which divides the segment from (6, 4) (i.e.,
the point C) to the midpoint (5, 1) of А В in the ratio 2/1, we set m = 2, η = 1
and obtain
1-6 + 2-5 16 1-4 + 2-1
«! = = 1 y2 = = 2.
2+1 3 2+1
We shall show how to find parametric equations of the line through a
given point A perpendicular to a given line. If the given line has direction
ν = [»i, v2] where ν И 0, and if v* = [ — v2, »i], then v* _L ν and |v*| = |v|,
since
vv* = (viUi + v2u2)-( — »2«i + »1«г) = —"iv2 + v2Vi = 0,
v*.v* = (-„^ + 02 = 0f + tf = VV.
Hence the desired line is the one through A with direction v*.
EXAMPLE 11.2
Find parametric equations of the line through (2, —3) perpendicular
to the line through (4, 1) with direction [3, —2]. The desired line contains
the vector [2, 3] with origin (2, —3). Its parametric equations are
*, = 2 + 2/, x2 = - 3 + It.
We shall show next that there is only one direction perpendicular to a
given direction ν where ν И 0; i.e., if h _L v, then there is a scalar s such
that h = jv*, i.e., h, v* have the same direction. Since h _L v, we have
h-v = hiVi + h2v2 = 0

Sec. 11] PARAMETRIC EQUATIONS 41
where »i, v2 are not both zero, since ν И 0. If»i И 0, we can solve the above
equation for hx and obtain
-h2o2 /h2\
hi = = -»21 — I = -v2s
Vi \Vi/
h2
where s = h2/v\. Hence
h = Лцц + ^2U2 = (— V2s)\li + (»ii)u2 = (—»2Ui + ViU2)s = V*S.
Similarly, if v2 И 0, we can solve for h2 and obtain
fe =
1 = * (v)= "lJ
v2 \ v2
hi = -v2s = -v2[ I
h = v*i
where s = —hi/v2. In either case h has the same direction as v*. We have
now proved the following theorem.
Tll.l. In the plane of щ, и2 there exists one and only one direction perpendicular to
a given direction. If ν = [v\, v2\ is a nonzero vector, then v* = [ — v2, V\] _L v,
Iv I = Iv* I > and for any other vector h perpendicular to v, there exists a scalar s such
that h = v*s.
Questions
1. Given that X is such that AX = ABt, where t and 1 — t are both positive.
Show that \AX\/\XB\ = t/{\ - t). 2- Solve i/(l - i) = m/n for t. Also find 1 - t.
3. Find formulas for the coordinates of the point X which divides the segment from A
to В in the ratio m/n. 4. Let ν = [»i, »j]. Find a vector v* such that v* J_ ν and
| v* | = | ν |. 5. Show that if ν ^ 0 and h is any vector perpendicular to v, then
there is a scalar s such that h = v*s. 6. Show how to find parametric equations of
the line through a given point perpendicular to a given line.
Problems
1. Let А, В, С be respectively (2, —1), (4, 3), (3, 4); let D be the midpoint of the
segment from A to B; and let X be the point which divides the segment from
С to D in the ratio 2/1. Find D and X.

42 PLANE ANALYTIC GEOMETRY (Ch. 2
2. Compute \CX\, \XD\ and show that \CX\/\XD\ = 2/1, where C, D, X are
defined as in Problem 1.
3. Let А, В, С be respectively (αϊ, аг), (Αι, Ьг), {с\, сг); let D, Ε, Fbe respectively
the midpoints of the segments from В to C, from С to A, and from Л to B; and
let Ζ divide the segment from A to D in the ratio 2/1. Find the coordinates of
D, E, F in terms of the coordinates of А, В, С Show that X is the point
(i
+ *i + ci аг + Ьг + c2
)
Show that X divides the segment from В to Ε in the ratio 2/1 and also divides
the segment from С to F in this ratio. Note that you have proved that each
median passes through X.
4. Find parametric equations of the line through (7, 2) perpendicular to the line
through (15, 8), with direction (3, 4).
5. Show that the line through (1, —2), (4, 2) is perpendicular to the line through
(1,-2), (-3,1).
12. THE DISTANCE FROM A POINT TO A LINE
Consider a point X and a line through a point С with direction a. If Υ
is the point of the line such that the distance | YX \ from Υ to X is a minimum,
then this minimum distance is defined
to be the distance from X to the line.
Since Υ lies on the line, there exists a
scalar / such that CY = a/ and hence
(see Fig. 12.1)
YX = CX - CY = CX - a/.
Let YX = h, CX = b. The above
vector equation then becomes
h = Ь - a/.
Figure 12.1
It is to be suspected that | YX\ = |h|
is a minimum when / is chosen so that h _L a; i.e., h-a = 0. We shall show
later that this is the case. We thus set
h-a = (b - a/)-a = a-b - (a-a)/ = 0

Sec. 12] THE DISTANCE FROM A POINT TO A LINE 43
and solve for /. We obtain
_ ab
a-a
h = Ь - a( —)
\a-a/
We can now compute |h| = Vh-h.
EXAMPLE 12.1
Find the distance from the point (4, 5) to the line through (1,2) with
direction [2, 3]. In Problem 1 you will be asked to supply the details of this
computation. We have
a = 2ui + Зиг, b = 3ui + Зиг,
ab - (a-a)/ = 15 - 13/ = 0, / = f£,
h = b - a/ = (Д)щ - (А) "г,
and hence the desired distance is
VhH = ^
Let us show that | h| is a minimum when h-a = 0; i.e., if У is any other
point of the line and h' = Y'X, then | h'| is greater than | h|. This is
equivalent to showing that h'*h' is greater than h*h. Since У lies on the line, there
exists /' such that СУ = a/', and hence
h' = b - a/' = b - a/ + a(/ - /')
= h + a(/ - /')·
We now have
h'-h' = (h + a(/ - /'))-(h + a(/ - /'))
= hh + 2h-a(/ - /') + a-a(/ - /')2
= h-h + a-a(/ - /')2,
since h · a = 0. Thus h' · h' is equal to h · h plus a non-negative number, and
therefore h'-h' ^ h-h. Moreover h'-h' > h-h unless /' = / and h' = h.
If the point X lies on the line, then there exists a number /' such that
CX = b = a/'. Hence
h' = b - a/' = 0, |h'|= 0.

44 PLANE ANALYTIC GEOMETRY (Ch. 2
We must then have | h | = 0, since | h' | cannot be less than the minimum
| h |. Conversely, if | h | = 0, then
h = Ь - a/ = 0, Ь = CX = at
and hence X lies on the line.
We can obtain a simpler formula for the distance from a point to a line as
follows: Theorem 11.1 implies that if a, a* are respectively the vectors
\au аг], [—02, βι], then a* _L a and |a*| = |a| and that if also h _L a, then
there exists a number s such that h = sa*. In order to compute h*h, we
substitute Ъ — at for just one of the factors h and later substitute sa* for the
other. Thus (since h-a = 0)
|h|2 = hh = h-(b - at) = hb - (ha)/
= hb = sa*-h,
|h| = |ia*| = |j|#|a*| = |j|'|a|,
|h| _ ±|h|
yhere
a a
Hence
±a*-b|h|
|h|2 = a*-bs —
We cancel | h | from the extreme members of these equalities and obtain
±а*-Ь ±а*-СЛ-
|h| =
where the sign is chosen so that the right member is non-negative. Thus we
have the theorem:
T12.1. The distance from a point X to the line through a point С with direction a is
±a*-CX
where a = [au a2], a* = [ — a2, aj.

Sec. 12] THE DISTANCE FROM A POINT TO A LINE 45
EXAMPLE 12.2
In Example 12.1 we have a = [2, 3], b = CX = [3, 3]. Hence a* =
[ — 3,2], | a | = \/Ϊ3> and
|h| =
±a*· b ± (- 3«, + 2u2) · (3u, + 3u2)
| a. |
±(-3)
лЛз
We settle the ambiguous sign ± as minus, since | h | cannot be negative.
That is, |h| = + Ъ/л/\Ъ.
We can use T12.2 to obtain the area of the parallelogram two of whose
sides are formed by the vectors a, b (see Fig. 12.2). A base of this
parallelogram is formed by a and the
corresponding altitude by h. We assume
the formula for area, namely: The
area is the length of the base times
the length of the altitude and hence
equal to
|a|(±a*-b)
h =
= ±a*-b.
Figure 12.2
We also assume that the triangle
two of whose sides are formed by a, b has half the area of the parallelogram.
Then the area of the triangle is
±£a*-b.
EXAMPLE 12.3
Find the area of the triangle whose vertices А, В, С are respectively
(1, 3), (3, -2), (2, 4). Let AB = a, AC = b. Then a - [2, -5), b =
[1,1], a* = [5, 2], and
±£a*-b = ±^(5 + 2) = ±1
Hence the area is %.
Questions
1. Let X be any point and let Y, Y' be points of the line through С with direction
a. Let
CX = b, YX = h, Y'X = h'.

46 PLANE ANALYTIC GEOMETRY (Ch. 2
Show that there exist scalars t, t' such that
h = b - ai, h' = b - ai' = h + a(i - i').
Show that h-a = 0 if t = a-b/a-a. Show that for this value of t we have
h'-h' = h-h + aa(i - i')2 > h-h
unless t' = t and Y' = Y. Show that the distance \XY\ is a minimum when t =
a-b/a-a. Show that X lies on the line if and only if |h| =0 when t = a-b/a-a.
2. Let a = [αϊ, аг], а* = [ — α2, αϊ]. Show that there exists s such that
1*1
where h is defined as in Question 1. Show that
I4..„..b-±!£U1
and hence that
, , ±a*-b
3. Find expressions for the area of the parallelogram two of whose sides are formed by
the vectors a, b, and the area of the triangle two of whose sides are formed by a, b.
Problems
1. Supply the details for the computations in Example 12.1.
2. Let C, X be respectively the points (1, 2), (3, 6), and let a be the vector [6, 2].
Let h = b — ai, where b = CX. Find t such that h-a = 0 and find the
corresponding h and |h|. Check by the formula |h| = ±a*-b/|a|.
3. Let C, a be defined as in Problem 2, but let X be a point of the line through С
with direction a. Show that
л*-СХ = -2χι + 6*2 - 10 = 0.
4. Find the component of b in the direction a where a, b are defined as in
Problem 2.
5. Find the area of the triangle two of whose sides are the vectors a, b of Problem 1.
Draw this triangle and its altitude formed by the vector h. Measure this altitude
and check with the answer to Problem 2.
6. Let η = [щ, иг], а = [щ, —гц]. Show that the distance from X = (*i, дгг) tc
the line through С with direction a is
n-CT _ n-(OX - ОС) _ Bi*i + πΐχι - η ОС
|n| _± Η ± VV + B1*

Sec. 13]
EQUATIONS OF LINES
47
7. Show that if a = [01,02], a* = [—02,01], and a-b = 0, then a*-b =
± I a I · I b I. Hint: There exists s such that b = j-a*.
13. EQUATIONS OF LINES
We shall find an equation that is satisfied by the coordinates of a point
X = (*i, x2) if and only if X is at zero distance from a given line and hence
lies on the line. By T12.1, the distance from X to the line through С with
direction a is |h| = ± a*-CX/\a\. If a = [аь a2], then a* = [ — a2,a{[ is
perpendicular to a and is called a normal vector to the line. It is sometimes
convenient to specify a line in terms of a point С and a given normal vector
η = [nun2]. Then [nun2] = а* = [-a2,a{\, a2 = -nu ax = n2, a = [n2,
— n{[ and |n| = |a*| = |a|. The distance from X to the line is
±л*-СХ ±n-CX
IЫ ^ i—i r—
|a| I n|
where
n-CX = n-(OX - ОС) = n-OX - n-OC
= ηλχι + n2x2 — n-OC,
|n| = V„i2 +
Let —n-OC = n3. Then the distance from X to the line is
*22·
, n-CX riixi + n2x2 + n3
Ihl = ± = ±
ι». V*i2 + „22
We have proved that Λ7 lies on the line if and only if this distance is zero, and
this is the case if and only if the numerator of the right-hand member is zero;
i.e., if and only if
"1*1 + "2*2 + Я3 = 0·
This latter equation is called an equation of the line.
We shall show that every equation of the form
cixi + c2x2 + c3 = 0
in which Ci and c2 are not both zero, represents a line; i.e., it is an equation
of a line. Such an equation is said to be linear in xu x2. We should expect the
line to have the normal vector η = [c\, c2] and hence the direction [c2, — cj.

48 PLANE ANALYTIC GEOMETRY (Ch. 2
It is thus only necessary to locate a point С such that the line through С with
direction [c2, —c\] is represented by the above equation. The coordinates of С
must satisfy the equation. There are many points (дгь x2) whose coordinates
satisfy the equation, and we attempt to find that one for which x2 = 0.
This is possible when c\ И 0, for in this case we can solve the resulting
equation Ci*i + 0 + сз = 0 for χλ and obtain xx — — c^/ci. It is easy to check
that the coordinates of the point C= ( — сз/ti, 0) satisfy the equation.
Moreover
-n-OC = -(ciU! + c2u2)· ( Ju! = — c,( J = c3.
When ci = 0, we must have c2 И 0, and in this case we can set χλ = 0 and
solve for x2. We obtain x2 = —сз/с2. We choose С to be the point (0, —сз/с2).
Its coordinates satisfy the equation; and moreover
—* f~cA
— n-OC= — (ciUi + c2u2)· ( lu2 = c3.
In either case the distance from a point X to the line through С with normal
η = [сь с2] is
П-СХ CiXi + C2X2 + C3
±_Й~ = ± VCl2 + c22
and the equation of this line is
CiXi + C2X2 + C3 = 0.
Thus:
T13.1. Every linear equation
c\x\ + c2x2 + c3 = 0
represents a line. If (x\, x2) is a point not necessarily on the line, then the distance
from (*!, x2) to the line is
^1*1 + C2X2 + C3
± VcJT^2
EXAMPLE 13.1
Find the distance from (4, 5) to the line — Ъх\ + 2x2 —1=0. The
distance from an arbitrary point (χχ, χ2) to the line is
— 3*1 + 2x2 — 1 — 3*1 + 2д:2 — 1
V9 + 4 ~ лДз

Sec. 13]
EQUATIONS OF LINES
49
and in particular the distance from (4, 5) to the line is
+ (-3-4 + 2-5 - 1) ^ +3
л/Тз л/Тз
The reader can easily check that the line contains the point (1, 2) and has
the direction [2, 3]; i.e., it is the line of Example 12.1.
EXAMPLE 13.2
Find the distance from (5, 3) to the line x\ — 2 = 0, where x\ — 2 = 0
is the linear equation in which ολ = 1, c-ι = 0, c3 = —2. The distance from
(x\,x-i) to the line is ±(*i — 2)/\/\ + 0 = ±(*i — 2), and the distance
from (5, 3) to the line is +(5 — 2) = +3. The equation χλ — 2 = 0
requires χι to be 2 but imposes no restriction on x2. Any point (xif X2) for
which x\ = 2 will satisfy the equation. For example, (2, 0) and (2, 7) are
points of the line. The normal is [1,0] = U! and the line has the direction
[0, 1] = иг. The line is parallel to the axis U2, and every point of the line is
two units to the right of this axis.
Questions
1. Show that the distance from X = (*i, дгг) to the line through С with normal
η = [m, m\ is
П-СХ _ niXl + П2Х2 + Я8
where m = — n· ОС. Show that X lies on the line if and only if n\x\ + пусг + пг = 0.
2. Let a = fo, — ci] where c% *■* 0. Show that the distance from {χ\,χ·ϊ) to the line
through (0, — сг/сг) with direction a is ±{c\x\ + сусг + caj/x^ci1 + C22.
Problems
1. Find the distance from (3, 6) to the line — 2x\ + 6*2 — 10 = 0 and check
your answer with that of Problem 2 of Section 12.
2. Find the distance from (2, 3) to the line x\ — 5 = 0 and the distance from
(2, —5) to this line. Draw figure.
3. Find the distance from (2, 3) to the line хг + 5 = 0 and the distance from
(5, 3) to this line. Draw figure.
4. Show that the point (J^, 0) is equidistant from the lines 5*i + 5*2 — 2 = 0,
7*i + *2 - 3 = 0.

50 PLANE ANALYTIC GEOMETRY (Ch. 2
5. Find two points А, В of the line —2jci + 6*2 — 10 = 0. Let AB = [01,02],
η = [ — 02, αϊ], Χ = (3, 6). Find η. Check that n-/LY/|n| agrees with your
answer to Problem 1.
14. SECOND-ORDER DETERMINANTS
We have seen that the area of a parallelogram two of whose sides are the
vectors a, b is ±a*-b, where a = [αλ, a2], a* = [ — a2, aj, b = [b\, b2]
(see Section 12). The product a**b is called a determinant and is denoted by
|a, b| = a*b.
The vertical bars do not indicate absolute value signs. In fact we have seen
that a*-b can be negative. In Section 15 we shall show that determinants
can be used to solve simultaneous linear equations, but in order to show this,
we need to develop some of the properties of determinants. We shall prove
the following properties:
P14.1. Iti!, tfel = 1.
P14.2. |a, b + c| = |a, b| + |a,c|.
P14.3. |a,jyb| =>|a,b|.
P14.4. |b,a| = -|a,b|.
To prove P14.1, we let a = Ul = [1, 0], b = u2 = [0, 1]. Then a* =
[0, 1] = u2, and hence |ui,u2| = u2-u2 = 1. The following is a proof of
P14.2:
I a, b + c| = a*-(b + c) = a*-b + a*-c = |a, b + |a, c|.
The proof of PI4.3 is as follows:
|а(>Ь| =а*.(>Ь)=>(а*-Ь)=>|а,Ь|.
To prove P14.4, we note that b* = [ — b2, bi] and hence
|b, a I = b*-a = (-ft2U! + fr1u2),(eiu1 + a2u2) = -b2ax + M2,
I a, b| = a*-b = ( — a2U! + aiU2)-(&iUi + *2u2) = — αφλ + αφ2.
Therefore | b, a = — |a, b|. Properties 14.1 to 14.4 are more convenient to
use than the definition | a, b | = a * · b, and we shall show that they define
the determinant. We begin by showing that they imply the additional
properties:

Sec. 14] SECOND-ORDER DETERMINANTS 51
P14.5. |a + b,c| = |a,c| + |b,c|.
P14.6. |xa, b| = χ|а, Ь|.
P14.7. |a, a| = 0.
P14.8. |a,0| = |0,a| = 0.
It follows from P14.4 and P14.2 that
|a + b,c| = -|c,a + b| = -|c,a| - |c,b| = |a,c| + |b,c|
and this establishes P14.5. It follows from P14.4 and P14.3 that
|*a, b| = — | b, xa| = -*|b, a| = *|a, b|
and this proves P14.6. By P14.4 we have
|a.a| = -|a,»l
and hence a, a =0. This is P14.7. By P14.3 and P14.4 we have
|a,0| = |а,0Ь| =0|а,Ь| =0,
|0,b| = -|Ь,0| =0
and this is P14.8.
We now show that the above properties determine the value of the
determinant; i.e., they define the determinant. By P14.5 and P14.6 we have
|a, b = !a,u, + a2u2, b| = |a,u,,b| + |a2u2, b|
= a,|u,, b| + a2|u2, b|
where the determinants | Ui, b |, | u2, b | are called the signed minors of β,, α2,
respectively. These signed minors are computed as follows. By P14.2, P14.3,
PI4.7, and PI4.1 we have
|u,, Ь| = |ub ft,U, + ft2U2| = |u,,ft,U,| + |u,,ft2U2|
= ft, |u,, U,| + ft2|u,,U2| = 0 + ft2,
and by P14.2, P14.3, P14.7, P14.1, and P14.4,
| u2, b | = ft, I u2, u, I + ft21 u2, u21 = ft, I u2, u, I +0
= -ft,|u,,u2| = -ft,.
Hence
| a, b | = a, | u,, Ь | + a2 | u2, b | = a,ft2 - a2ft,.
We shall describe a method of remembering this formula in terms of an
alternative notation for a determinant. This notation displays the compo-

52 PLANE ANALYTIC GEOMETRY (Ch. 2
nents of the vectors. It is the following:
la, Ы =
a2 b2
The elements aif b2 form what is called the. principal diagonal, and a2, bi form
the secondary diagonal. The value of the determinant is the product of the
elements of the principal diagonal minus the product of the elements of
the secondary diagonal.
EXAMPLE 14.1
Find the area of the triangle whose vertices А, В, С are respectively
(-3, 0), (1, 4), (5, 2). Let a = AB, b = AC. Then a = [4, 4], Ь = [8, 2]
and
4 1
4 :
±*a*-b = ±4|a,b|
±$(8 - 32) = ±12.
Hence the area is 12.
We have noted that |ui, b|, |u2, b| are called signed minors of αλ, α2,
respectively. Similarly a, Ui |, а, иг| are called signed minors of bif b2.
The signed minors of aly a2, bly b2 are equal to b2, — Ь1у —a2, au respectively
(see Problem 3), and the elements b2, b\, a2, a\ are called minors of a\, a2,
b\, b2, respectively. The minor of a given element is obtained as follows:
Locate the row and column in which the given element is found and remove
from the determinant all the elements in that row and column. The element
that remains is the minor of the given element. A signed minor differs at
most in sign from the corresponding minor. The elements βι, α2 belong to the
first column, and the expansion
I a, b I = a, I ub ЬI + a21 u2, Ь | = αφ2 - αφλ
is called the expansion in minors of the first column. In Problem 3 you will be
asked to show that
I a, b | = fti I a, U! I + b2 \ a, u21
I a, U! I
-02, I a, u2|
Ol
where a2 is the minor and | a, Ui | the signed minor of bu and αλ is the minor and
I a, u21 the signed minor of b2. Your expansion is called the expansion in minors
of the second column.
If in the determinant | a, b | we interchange the rows with the columns,

Sec. 14] SECOND-ORDER DETERMINANTS 53
we obtain a new determinant called the transpose of | a, b | and denoted by
|a, b|'. Thus
a, ao
|a,b|' =
bl b2
The value of the transpose is the product of the elements of the principal
diagonal minus the product of the elements of the secondary diagonal. Hence
| a, b|' = axb2 - ha2 = | a, b|.
T14.1. A second-order determinant is equal to its transpose.
We have defined second-order determinants in terms of two vectors in
two-dimensional space. Third-order determinants are defined in terms of
three vectors in three-dimensional space.
Questions
1. Let a = [αϊ, aj], b = [b\, Ьг\, a* = [ — аг, αϊ]. Show that the determinant
ι li * v. "ι bi
|a,b|=a*-b =
02 02
has the properties
|u,,u2| = 1, |a,b + c| = |a,b| + |a,c|,
|a,,b| =j|a,b|, |b, a| = - |a, b|.
2. Show that these properties imply the additional properties
|a + b,c| = |a,c| + |b,c|, |ш,Ь| -дс|а,Ь|,
|a,a|=0, |a,0| = |0,b| = 0.
3. Show that the above properties imply that
| a, b| =ai|ui, b| +a2|u2, b|
where
|m, b| = b2, | u2, b | = — bi
and hence that
| a, b| = αι^2 — аг^ь
4. Write down the minors and the corresponding signed minors of the elements
ai, "2, bi, Ьг. 5. Show that
αϊ b\
ai bi
=
αϊ аг
b\ Ьг

54 PLANE ANALYTIC GEOMETRY (Ch. 2
Problems
1. Compute |3ui — 5иг, ui + 4иг|.
2. Let a, b be any two vectors and x, у any two scalars. Expand | ax + b>, b |,
| a, ax + bjy | and show that
|ах + Ь>>, Ь| = x|a, b|, |a, ах + Ъу\ = у\л,Ь\.
3. Show that |a, b| = Ai|a, ui| + Ьг\л, иг| and that |a, щ| = — аг, |а, иг| =
αι. Check that this gives the correct value for the determinant. This expansion
is called the expansion in minors of the second column.
4. Show that аг \ иг, b | + Ьг | а, иг | = | a, b |. This is the expansion in minors of
the second row.
5. Let a, b be any two vectors. Show that
|дпа + x2b, yi* + угЪ \ = χι\a,>>ia + уг,Ъ\ + хг\b,>>ia + угЪ\;
that
\»,yiA + y2b\ = jr2|a, b|,
Ib, yiA + у2Ъ\ = -yi|a, b|,
and hence that
|*ia +дсаЬ,ла +jrab| = *ijra|a, b| - x2yi\a., b| = |x, y|-|a, b|
where χ = [xu x2], γ - [yi,y2\.
6. Show that if (αϊ, аг), (Αι, Аг), (ci, сг) are vertices of a triangle, then the area is
±i
Ai — αϊ ei — αϊ
ог — аг сг — аг
7. Expand the transpose | a, b |' of | a, b | in minors of the first column and show
that the result is
αι*2 — *1аг = βι | щ, Ъ | + Αι | a, ui |.
This is called the expansion of | a, b | in minors of the first row. Use this method
to obtain the expansion of | a, b | in minors of the second row.
8. Show that if a· b = 0, then | a· b | = ± | a| · | b |. See Problem 7 of Section 12.
15. SIMULTANEOUS LINEAR EQUATIONS
Let a, b, с be any three vectors and let us attempt to find scalars x, у
satisfying the equation
ax + by = с

Sec. 15]
SIMULTANEOUS LINEAR EQUATIONS
55
If x, у do satisfy, then
|c, b| = |a* + by, b| = *|a, Ь| +>|Ь,Ь|
= *|а,Ь| +0,
and similarly
|a,c| = |a,a* + by| = 0 + >|», b|.
Hence if | а, Ь | И 0 and x, у satisfy the equation, then the solution must be
|c, b| |a,c|
* I», ЬГ У |а,Ь|'
If a = [аи а2], Ь = [bu Ь2], с = [с,, с2], then
ад: + by = (^щ + а2м2)х + (^щ + Ь2м2)у
= {αλχ + biy)\ii + (а2х + b2y)u2 = [αλχ + b^, a2x + b2y].
Hence the equation алг + by = с is equivalent to
[alX + biy, a2x + b2y\ = [cu c2]
and in turn equivalent to the simultaneous linear equations
a\x + biy = cb
a2x + b2y = c2.
We shall show presently that the above values of x, у do satisfy the equations,
and this will complete the proof of the following theorem.
T15.1. // a = [aly a2], b = [bu b2], с = [cly c2] are such that |a, Ь| И 0, then
ях + by = с
or equivalently
if and only if
"ix + biy = a
a2x + b2y = c2
|c,b|
|a,b|
Cl
c2
"l
a2
bi
b2
bi
b2
а, с
У =
"ι
a2
"ι
a2
Cl
c2
bi
b2
To remember this solution, note that the elements of the common
denominator determinant are the coefficients of x, у in the linear equations and
that these elements are in the same relative positions as in these equations.

56 PLANE ANALYTIC GEOMETRY (Ch. 2
The numerator determinant in the expression for χ is obtained from the
denominator by replacing the column consisting of the coefficients of χ
(i.e., the first column) by the column formed from the right-hand sides of
the equations. The numerator for у is obtained by replacing the column
consisting of the coefficients of у by the column formed by the right-hand
sides. It is advisable to form the denominator determinants before forming
the numerators.
EXAMPLE 15.1
Solve the equation
(3u! + u2)x + (-4u! + 7u2)y = 2ui
or equivalently the equations
Ъх - Ay = 2
5u2
χ + ly = -5.
The solution is
2
-5
3
1
-4
7
-4
7
25
3
1
3
1
2
-5
-4
7
-17
25
It remains to show that the values for x, у given by T15.1 actually do
satisfy the vector equation. We consider two special cases and use the results
thus obtained to prove T15.1. We first show that
is satisfied if
a*! + byi = щ
l«i,b| b2
X\ =
Ia,b|
Ia,b|
We have
У\
I a, ui|
|a,b|
|a,b|
a*! + byi = (a^i + а2и2)х\ + (ftiU! + b2u2)yi
= (a\Xl + ftl>l)U! + (02*! + ft2>l)«2

Sec. 15] SIMULTANEOUS LINEAR EQUATIONS 57
where
b2 —a2 0i*2 — *ιθ2
^ |a,b| |a,b| |a,b|
α2*2 — b2a2
a2*i + *2> = : —> = °>
I a, b|
and hence
a*! + hyi = щ + 0.
That is, x\,y\ satisfy the equation. Similarly if
|u2, b| -bi
x2 =
У2
|a,b| |a,b|
a, u2| αϊ
|a,b| |a,b|
then
a*2 + by2 = (P\X2 + 1>1у2)щ + (a2x2 + b2y2)u2
= 0 + u2.
Finally, for an arbitrary vector c, we let
|c, b c,|u,, b| + c2|«2, b|
|а,Ь| |а,Ь|
^1*1 + c2*2
I a, c I I a, u,| |a, u2|
У = —-Г7 = *i -г—г— + c2 —— = с1У1 + c2y2.
|a,b| |a,b| |a,b|
Then
αχ + by = a(cixi + c2x2) + b(ciyi + c2y2)
= ci(axi + byi) + с2(лх2 + Ъу2)
= CiUi + C2U2 = C.
This completes the proof of Tl 5.1.
Questions
1. Show that if ax + by - с and |a, b| И 0, then χ = |c, b|/|a, b| and у
| а, с | /1 a, b |. 2. Show that if | a, b | ^ 0 and
, ._ |a, "Η
XI = ι .1 ' У\ =
|ui
la,
|U2
.b|
b|
.b|
|a,b|
I a, uj|
|a,b| |a,b|

58 PLANE ANALYTIC GEOMETRY (Ch. 2
then
a*l + byi = Ul, ΛΧ2 + Ъу2 = U2
3. Show that if * = | c, b | /1 a, b|,> = | а, с | /1 a, b |, then
* = Ctfi + C2*2, У = Ciyi + Сгуг
and
a* + by = с
Problems
1. Write down two scalar equations equivalent to
(Зщ — 5u2)* + (ui + 4u2)>1 = 7щ + Зиг
and find *, у. Check that these values satisfy the vector equation and the two
scalar equations.
2. Leta = [6, -9], b = [-10, 15], с = [1,1].
(a) Compute |a, b|, |c, b|, |а, с|.
(b) Show that if *, у satisfy the vector equation
ax + by = c,
then
*|a,b| - |c,b| =0, у\л,Ъ\ = |а,с| =0.
(c) Can the vector equation have a solution?
3. Let a, b be defined as in Problem 2, but let с = 0. Show that the equation
ax + by = с is satisfied if дг = 5, у = 3. Are these values of дг, у given by the
formulas χ = |с, b|/1а, Ъ\, у = |а, с|/1a, b| ?
4. Let a = [4, -3], b = [-5, 4], с = [3, -2].
(a) Find *i, ylf хг,уг such that
a*i + byi - ui, ахг + Ъуг - u2.
(b) Let χ — 3*i — 2*2, у = Ъу\ — 2уг. Compute a* + Ъу and check that this
vector is equal to с
5. Given a = [01,02], b = [Ai, Д2], | a, b | = 0^2 — 02A1.
(a) Solve the equation | a, b | =0 for Ьг, assuming αϊ τ* 0, and solve this
equation for b\, assuming α2 ^ 0.
(b) Show that if αι j* 0, then
Ьг — a2S, bi — ais, b = as
where s — bi/αχ, and that if аг ^ 0, then
bi — ms, Ьг = a2S, b = as
where s = Ьг/аг.
6. Let a, b be defined as in Problem 2. Find s such that b = as.

Sec. 16]
INTERSECTION OF LINES
59
16. INTERSECTION OF LINES
Consider the lines
Ol*l + 02*2 + 03 = 0, *1*1 + *2*2 + *3 = 0.
If (χι, x2) is a point of intersection of these lines, then it lies on both lines
and its coordinates satisfy both equations. To find these coordinates, we
solve the equivalent equations
αιΧι + a2x2 = —оз,
*i*i + *2*2 = —*з-
If the determinant
αϊ a2
bi b2
a, bl
a2 b2
= |a,b|
is not zero, then these equations are satisfied if and only if (see T14.1)
x\
— оз a2
— *з b2
Ol 02
*i b2
al —a3
bi —b3
x2 =
al a2
b\ b2
Thus, when a, b И 0, there is one and only one point of intersection.
EXAMPLE 16.1
Find the point of intersection of the lines
3*! + 2*2 - 7 = 0,
Xl + 5*2 + 2 = 0.
The coordinates of the intersection are
7 2
-2 5
x\
3 2
1 5
39
*2
3 2
1 5
-13
13
= -1.
Thus (3, —1) is a point of intersection and the only point of intersection.
We shall show that the lines have the same direction when a, b =» 0.
The numbers αλ, a2 are not both zero because the equation αγχγ + аг*2 +

60 PLANE ANALYTIC GEOMETRY [Ch. 2
a3 = 0 is linear in xu x2. If αϊ И 0, we can solve the equation
| a, b = а^2 — a2bi = 0
for b2 and obtain
a2bi aibi
b2 = = a2s, bi = ais =
where s = b^/a^. If a2 И 0, we can solve for bi and obtain
axb2 a2b2
bi = = ais, b2 = a2s =
a2 a2
where s = b2/a2. In either case we have
bi = ais, b2 = a2s.
Moreover, s И 0, since s = 0 implies bi = 0 = b2, and this is impossible.
The vectors
a = k, "2], Ь = [bu b2]
are normals to the lines, and the directions of these lines are given respectively
by the vectors
ν = [a2, -a,], w = [b2, -bx].
The equation | a, b | =0 implies that bi = ais, b2 = a2s and hence implies
that
w = [a2s, -a^s] = [a2, -a^s = vs;
i.e., the two lines have the same direction. The lines are then either identical
or parallel, and we shall show how to determine which is the case. When
bi = ais, b2 = a2s, we have
Mi + b2x2 + b3 = a^xi + a2sx2 + b3
= s(aixi + a2x2 + a3) + *3 — sa3.
Consider first the case b3 — sa3 = 0. Then
*i*i + b2x2 + b3 = s(alxl + a2x2 + a3).
Hence, if the coordinates of (χχ, χ2) satisfy the equation
al*l + 02*2 + 03 = 0,
they also satisfy the equation
bixi + b2x2 + b3 = 0.

Sec. 16]
INTERSECTION OF LINES
61
Conversely (since s И 0), if they satisfy the second equation, they also
satisfy the first. That is, every point of one line lies on the other and the lines
are identical.
Consider next the case b3 — sa3 И 0. If (χγ, x2) is a point of the line
Ol*l + 02*2 + 03 = 0,
then for these values of x\, x2 we have
biXi + b2X2 + b3 = s(aixi + a2x2 + a3) + b3 — sa3
= b3 — sa3 И 0.
That is, there is no point of the first line which is also a point of the second,
and hence the lines do not intersect; hence they are parallel.
Thus the equation | a, b | =0 implies that there exists s such that b = as
and that the lines are identical when b3 = a3s and parallel when b3 И a3s.
The equation | a, b | =0 implies the algebraic condition either that the two
simultaneous linear equations are satisfied by all values of xu x2 which
satisfy one of the equations or that there are no values of χι, x2 which satisfy
both equations.
We have interpreted a line as a set of points. In general the points that
are common to two sets constitute a set called the intersection of the two given
sets. If the two sets are lines having respectively the normals a, b and if
| a, b | И 0, then the intersection is a set consisting of a single point.
If | a, b | =0, the two lines may be identical, and in this case the intersection
is the common line. However, if | a, b | =0, the lines may be parallel but
not identical, and in this case the intersection does not contain any points.
The intersection is then a strange set, but it is still called a set even though it
does not contain any points. It is called the empty set.
EXAMPLE 16.2
Find the intersection of the lines
6*! - 15*2 + 21 = 0, -4*! + 10*2 - 14 = 0.
We have
b = —4u! + 10u2 = sa. = i(6u! — 15u2)
where s = — %, and since
-14 = (-|)21,
the lines are identical. That is, the two equations represent the same line
and the intersection of the two sets is the common line. If the coordinates of a
point (χι, x2) satisfy one equation, they satisfy the other. For example if

62 PLANE ANALYTIC GEOMETRY (Ch. 2
*i = 0, then both equations imply *2 = %. The coordinates of (0, "]/£) satisfy
both equations.
EXAMPLE 16.3
Find the intersection of the lines
6*! - 15*2 + 21 = 0, -4*! + 10*2 - 13 = 0.
Again we have s = — % but —13 И ( — %)21. The lines are parallel and
the intersection is the empty set.
Questions
1. Consider two lines with normals a, b. Show how to find the intersection of the
lines when | a, b | ^ 0. 2. Show that if | a, b | =0 and a^O, then there exists s such
that b = ar. 3. How are the lines related when their normals a, b are such that
| a, b| = 0? 4. Discuss the intersection of the lines when |a, b| =0 and show how
this is related to the simultaneous solution of linear equations. 5. What is meant by
the intersection of two sets and by the empty set? 6. Show how the concepts of
Question 5 are related to intersections of lines.
Problems
1. Find the point of intersection of the lines 4*i — 3*2 + 3 = 0, — 5*i + 4*2 —
2 = 0, and show that the coordinates of this point satisfy both equations.
2. Find the intersection of the lines 6*i — 9*2 + 1 =0, — 10*i + 15*2 + 1 =0.
3. Find the intersection of the lines 6*i — 9*2 + 21 =0, — 10*i + 15*2 — 35 = 0.
4. Let a, b be respectively normals to the lines of Question 2. Find a, b and find
s such that b = ar.
5. Let a, b be nonzero vectors and s, t be scalars such that
&s + Ы = 0
where i^O. Solve the equation for a and state how a, b are related.
6. Show that if a, b are related as in Problem 5, then | a, b | =0.
17. LINEAR DEPENDENCE
We have seen that two lines are parallel when their normals have the
same direction. We shall show that if a, b are nonzero vectors satisfying
the vector equation
ia + /b = 0,

Sec. 17]
LINEAR DEPENDENCE
63
where s, t are not both zero, then a, b have the same direction. We first show
that if s, t are not both zero, then neither is zero. Thus s И 0, / = 0 imply
sa = 0 where s И 0, а И 0, and this is impossible. Similarly s = 0, / И 0 is
impossible. We can therefore solve the above vector equation for a and also
for b. Thus
~b(f> b-.(?)
and hence a, b have the same direction. The vector equation always has the
solution s = t = 0, and this is called the obvious solution. If the obvious solution
is the only one, then a, b are said to be linearly independent, but if there exists a
solution other than the obvious one, then a, b are linearly dependent. We have
now shown that when two nonzero vectors are linearly dependent, they
have the same direction. Conversely, if a, b have the same direction, then
(by definition) there exists a number s such that b = sa; i.e.,
0 = sa + (-l)b = sa + tb,
where / = — 1 И 0, and since this is a nonobvious solution, a, b are linearly
dependent. Thus two lines are parallel if and only if their normals are
linearly dependent. Two vectors a, b are always linearly dependent when one
of them (say, b) is zero, since in this case the vector equation can be satisfied
by setting j = 0, Μ 0. Thus
sa + tb = 0a + t0 = 0.
In Section 39 we show that if a is nonzero, then the set of vectors b such
that a, b are linearly dependent (i.e., such that b has the form sa) is a one-
dimensional vector space. Thus two vectors with a common origin are
linearly dependent if and only if there is a line containing both. Note that
two zero vectors with a common origin are contained in any line containing
the common origin.
We show next that two vectors a, b are linearly dependent if and only if
the determinant | a, b | is zero. If a, b are nonzero vectors such that | a, b | =
0, then we have seen that they have the same direction. Hence they are
linearly dependent. If at least one of these vectors is zero, then | a, b | =0
automatically, and again a, b are linearly dependent. If | a, b | И 0, then by
T15.1 there is only one solution of the equation
sa + tb = 0
and this is of course the obvious one:
|0,b| |a,0|
s = = 0, / = = 0.
|a,b| |a,b|

64 PLANE ANALYTIC GEOMETRY (Ch. 2
We have now proved the following theorem:
T17.1. Two vectors a, b are linearly dependent if a, b =0 and are linearly
independent if | a, b | И 0. // at least one of them is zero, they are linearly dependent,
but if neither is zero they are linearly dependent if and only if they have the same
direction.
EXAMPLE 17.1
The vectors a = 6u! — 15иг, b = —4u! + 10u2 of Example 16.2 have
the same direction and hence are linearly dependent. One can readily check
that | a, b| =0. On the other hand two vectors vb v2 forming an ortho-
normal system are linearly independent, since |vi, v2| = ±1 И 0. See
Problem 8 of this section.
Questions
1. Show that if a, b are nonzero vectors and s, t are scalars such that ar + bi = 0,
where s, t are not both zero, then neither s nor t is zero. 2. Show that if a, b, s, t are
related as in Question 1 and s, t not zero, then a, b have the same direction. 3. What is
meant by the obvious solution of the equation ar + bi = 0? 4. Define linear
dependence and linear independence of two vectors. 5. Show that two nonzero vectors
are linearly dependent if and only if they have the same direction. 6. Show that two
vectors are linearly dependent if at least one of them is zero. 7. Discuss the statement
that two vectors with a common origin are linearly dependent if and only if there is a
line containing both. 8. Show that a, b are linearly dependent if and only if | a, b | =0.
Problems
1. Find s, t not both zero such that
x(6ui - 3u2) + i(-10ui + 5u2) = 0.
2. Show that the vectors 3ui + 4u2, — 4ui + 3u2 are linearly independent.
3. Find x, y, ζ not all zero such that
*(6u, - 3u2) -M-lOu, + 5u2) + z(ui + u2) = 0.
Hint: Let ζ — 0 and use the result of Problem 1.
4. Find x, y, ζ not all zero such that
дг(3щ + 4u2) + y(-4ui + 3u2) + z(ui + u2) = 0.
Hint: Let ζ — — 1 and solve the resulting equation for x, y.
5. Show that щ, u2 are linearly independent.

Sec. 18] PROPERTIES OF LINEAR EQUATIONS 65
6. Show that if *ivi + *2V2 = 0, where vi, V2 form an orthonormal system, then
x\ — *2 = 0. Hint: vr (*ivi + хгУг) — vi-0. Are vi, V2 linearly independent?
7. Let h = b — ai, where t is such that h*a = 0. Show that a, b are linearly
independent if and only if h τ* 0.
8. Let a, b be linearly independent vectors and let a, b, h be related as in Problem
7. Show that a, h are nonzero vectors. Let vi = a/|a|, V2 = h/|h|. Show
that vi-vj = 0 and that vi, V2 is an orthonormal system.
9. Show that if vi, V2 are defined as in Problem 8, then |vi, Уг| = ±1. See
Problem 8 of Section 14.
18. GEOMETRICAL PROPERTIES OF LINEAR EQUATIONS
The normals to the lines
<>1*1 + 02*2 + a3 = 0, *1*1 + *2*2 + *3 = 0
ate respectively the vectors a = [au аг], Ь = [Ь\, ^г], and the directions of
the lines are given respectively by the vectors ν = [a2, — aj, w = [b2, —fti].
We have seen that the lines are parallel or identical when the normals have
the same direction (i.e., when a, b are linearly dependent). When the two
normals are perpendicular, we have a-b = 0 and hence
vw = a2b2 + ( — βι)( — *i) = aibi + a2b2 - a-b = 0,
i.e., ν _L w. In this case the lines are said to be perpendicular. These results
can be seen geometrically from the fact that the angle between two lines is
the same as the angle between the two normals (see Fig. 18.1).
b
о
w
Figure 18.1
We now have a means of finding an equation of the line which is parallel
to a given line
^1*1 + fl2*2 + fl3 = 0

66 PLANE ANALYTIC GEOMETRY (Ch. 2
and which contains a given point. Thus the line
«1*1 + 02*2 + *3 = 0
is parallel to the given line and Ьз can be determined so that the coordinates
of the given point satisfy the equation of this parallel line. We also have a
means of finding an equation of a line which is perpendicular to the given
line and which contains a given point. Thus the line
02*1 — ai*2 + сз = 0
is perpendicular to the given line and C3 can be determined so that this latter
line contains the given point.
EXAMPLE 18.1
Find an equation of a line which is parallel to 3*i — 7*2 + 1=0 and
which contains the point (3, —1). The desired equation has the form
3*i — 7*2 + *з = 0,
and we must determine b^ so that (3, —1) lies on this line, i.e., so that
3-3 -7(-l) + b3 = 16 + ft3 = 0.
Hence Ьз = — 16 and the desired equation is
3*i - 7*2 -16 = 0.
EXAMPLE 18.2
Find an equation of a line which is perpendicular to 3*1 — 7*2 + 1=0
and contains ( — 1, 3). The equation has the form
— 7*i — 3*2 + c3 = 0
and ( — 1, 3) lies on this line if
-7-(-D - 3-3+ сз = -2 + c3 = 0.
Thus сз = 2 and the equation is
-7*i - 3*2 + 2 = 0.
The equation αϊ*! + аг*2 + аз = 0 can be solved for *2 when a2 И 0.
Thus

Sec. 18] PROPERTIES OF LINEAR EQUATIONS 67
where m = —а^/а2, b = —03/02- In order to interpret the numbers b, m,
we locate two points on the line. Thus x2 = b when χλ = 0 and x2 = m + b
when xi = 1, and hence (0, b), (1, m + b) are points of the line. The point
(0, b) also lies on the *2-axis, since x\ = 0. It is the point at which the line
intersects the *2-axis and is called the x2-intercept. A point at which a line
intersects the *i-axis is called the x\-intercept; this intercept will be considered
later. When m is positive, the point (1, m + b) lies one unit to the right
and m units above the point (0, b). A point moving to the right along the
line rises a distance of m units in moving a horizontal distance of one unit.
Thus m is the rate at which the point is rising per unit of horizontal distance,
and this rate is called the slope of the line. The slope of a line is found by
solving an equation of the line for x2. In the resulting equation the
coefficient of xi is the slope. Of course an equation of a line cannot be solved for x2
when the coefficient of x2 is zero, and in this case the slope is not defined.
When m is negative, the point (1, m + b) lies one unit to the right but \m\
units below (0, b). A point moving to the right along such a line falls at the
rate of I m | units per unit of horizontal distance. The point is also said to be
rising at the negative rate m. When m = 0, then x2 = b for all points of the
line and the line is horizontal. The moving point neither rises nor falls and is
said to be rising at rate zero. A line slopes upward to the right when the slope
m is positive, downward to the right when m is negative, and is horizontal
when m is zero. The slope or grade of a road is quoted in per cent. It is the
rise in feet per hundred horizontal feet. A 13 per cent grade is a steep hill for
automobile traffic and a 2 per cent grade is steep for a train track.
If a\X\ + a2x2 + 03 = 0 is a given line, then any parallel line has an
equation of the form a\X\ + 02*2 + с = 0. If 02 И 0, then we can solve each
of these equations for x2 and obtain respectively
x2 = mxi + bi, x2 = τηχγ + b2,
where m = —ai/a2, bx = — а$/а2, b2 = —c/a2. Thus parallel lines have equal
slopes when their slopes exist. Conversely, lines with equal slopes are clearly
parallel. A line perpendicular to the given line (i.e., to the line πιχγ — x2 +
b\ = 0) has an equation of the form χλ + mx2 + d = 0, and when m И 0,
this equation can be written in the form
where — \/m is the slope of the perpendicular line. Thus if two lines are
perpendicular and their slopes exist, then the slope of one line is the negative
reciprocal of the slope of the other.

68 PLANE ANALYTIC GEOMETRY (Ch. 2
The following forms are standard for equations of lines. These forms
enable one to write down equations of lines when certain properties are
specified.
x2 = πιχγ + b (slope-intercept formula).
This equation states that (*i, x2) is on the line with slope m and ^-intercept
(0, b).
*2 — a2 = m(xi — <*i) (point-slope formula).
This equation states that (xi, x2) is on the line through (αϊ, α2) with slope m.
1 = 1 (intercept formula).
a b
This equation states that (xif x2) is on the line with intercepts (a, 0), (0, b).
(bi — ai)(x2 — a2) = (b2 — α2)(χλ — a{) (two-point formula).
This equation states that {x\, x2) is on the line through (ab a2), (b\, b2).
The alternative form
x2 — a2 _ χγ — αλ
b2 — a2 b\ — a\
can be used when neither denominator is zero.
To prove that the point-slope formula has the properties which we have
attributed to it, first note (this part of the reasoning is important and should
not be neglected) that the equation is linear in xx, x2, and hence represents a
line. If we substitute x\ = aif x2 = a2, the equation is obviously satisfied, and
hence the line contains (αϊ, α2). If we solve the equation for x2, we obtain
x2 = πιχγ + (a2 — ma{).
In this resulting equation the coefficient m of x\ is the slope of the line. In the
problems you will be asked to use this method of proof to show that the
remaining forms have the properties attributed to them.
Questions
1. Show that the lines
(21*1 + (22*2 + <J3 = 0, blXl + ^2*2 + A3 = 0
are parallel or identical when their normals [αϊ, аг], [Αι, Ьг] have the same direction
and perpendicular when their normals are perpendicular. 2. How is the slope m of a
line obtained from an equation of the line? 3. What is the geometric interpretation of
the slope? Consider the cases of m positive, zero, and negative. 4. Show that two lines

Sec. 19] THE PRODUCT OF TWO DETERMINANTS 69
are perpendicular if their slopes exist and one slope is the negative reciprocal of
the other. 5. What is meant by the intercepts of a line? 6. Derive the point-slope
formula.
Problems
1. Find an equation of the line through (7, 2) parallel to the line x\ — 3*2 + 5 = 0.
Show that the intersection of the two lines is the empty set.
2. Find an equation of the line through (7, 2) perpendicular to *i — 3*2 + 5 = 0.
Find the slopes of the two lines and show that one slope is the negative reciprocal
of the other.
3. Find an equation of the line with slope 2 and *2-intercept (0, —3). Draw
the line.
4. Find an equation of a line passing through the point (2, 3) and having the
slope — 1. Draw the line.
5. Find the *i-intercept and ^-intercept of the line — 2*i + 5*2 + 10 = 0.
Draw the line.
6. Show that the intercept formula represents the line with intercepts (a, 0),
(0, b).
7. Show that the two-point formula represents a line through the points (αϊ, аг),
(Αι, Αι).
8. Find an equation of the line with intercepts (3, 0), (0, —5).
9. Find an equation of the line through the points (3, 1), (0, —5).
19. THE PRODUCT OF TWO DETERMINANTS
We shall prove the following formula in which the product of two
determinants is expressed as a single determinant and shall then give a
geometrical application of this formula.
|a'i,a'2|-|x,y|
»i -x a,y
a2 · χ a2 · у
where the determinant on the right is made up of the vectors [a! -x, a2-x],
[a! · y, a2 · y] and я\, a'2 are the vectors forming the transpose of | ai, a21. Thus
|a'i,a'2| = |a,, a2|' = |ai, a2|.
The result is expressed in terms of a'i, a'2 instead of ai, a2, since this form
results from the proof and also since it displays a useful analogy with a

70
PLANE ANALYTIC GEOMETRY
ICh. 2
formula which we shall encounter later. The equations
»i = [«ι·»ι, «2·»ι], a2 = [u! · a2, u2 · a2]
suggest the notation
«1*1 = «и, «2»ι = 021, Uja2 = ai2, u2a2 = a22.
In terms of this notation we have
»i = [on, a2\], a2 = [a12, a22],
Oil 012 ' _ «11 021
α21 a22 α12 a22
a'i = [β», a12], a'2 = [a21, a22].
If χ = [*,, x2], у = bi,y2], then
[ai x, a2x] = [αηχι + a2lx2, αι2χλ + a22x2]
= kl, α12]χχ + [θ21, Я22]дг2 = *'\X\ + a'2*2,
l»'l,»'2| =
and similarly
It follows that
[ai-y, a2-y] = i'iyi + л'2у2.
= \&'\X\ + a'2*2, a'i>i + a'2>2|.
arx a,y
a2-x a2y
This determinant can now be expanded as follows:
|a'i*i + a'2*2, a\yi + я'2у2\ = *i |а'ь a\yi + a'2y2\ + *2|a'2, а\У1 + a'2>2|
where
|a'i,a',>, + a'2jy21 = yi\*\, *'i\ + >2|a'i,a'2| = >2|a'b a'2|
I a'2, a'iyi + л'2у2 \ = yx | a'2) a', | + 0 = -yx | а'ь а'21.
Hence
arx aj-y
a2-x a2y
= |a'i*i + a'2*2, a'i>i + a'2j>2|
= Х1У21 a'i, a'21 - x2yi \ a\, a'21
= (*ι>2 — *2.yi)|a'i, a'21
= |ai,a2|'· |x,y|.

Sec. 19] THE PRODUCT OF TWO DETERMINANTS 71
This formula can be given a more symmetric appearance by replacing x, у
respectively by bi, b2. Thus we have T19.1.
T19.1. If a.i, a2, bi, b2 are any four vectors, then
a^bi а^Ьг
|аьа2|'· |bi, b2| =
a2-bi a2*b2
This theorem can be used as follows: Let vi, v2 be an arbitrary ortho-
normal system. Then
Vi-Vi = V2-V2 = 1, V!-V2
and hence (if Vi = ai = bi, v2 = a2 = b2)
V2-Vi
0,
|Vl, V2
|2 =
1 /
vi,v2|
1
vi, v2| =
Vl-V!
V2-V!
vrv2
v2-v2
^
1
0
0
1
= 1.
It follows that |vi,v2| = ±1. If |vbv2| = +1 = |ub u2|, then Vi, v2
and Ui, u2 are said to have the same orientation, whereas if | Vi, v21 = — 1,
they have opposite orientations. An orthonormal system having the same
orientation as Ui, u2 is said to be right-handed, and one having the opposite
orientation is left-handed. A right-handed system is described as follows:
Point the thumb of your right hand away from that side of the paper on which
Vi, v2 are drawn and let the fingers curl. If your fingers indicate a rotation
from Vi toward v2, then vi, v2 is right-handed. The validity of this
description cannot be established mathematically. Try it on the system Ui, u2.
EXAMPLE 19.1
In Problem 1 of Section 7 you showed that if
V! = fu, + fu2, V2 = ~%Щ + fu2,
then Vi, v2 is an orthonormal system. Since
+ 1,
this system is right-handed.
EXAMPLE 19.2
If wl = u2, w2 = ub then
|wi,
w2|
4
5
3
5
=
= |u2
,«1
= -i,
and hence wi, w2 is a left-handed orthonormal system. Further applications of
T19.1 will be given later.

72 PLANE ANALYTIC GEOMETRY (Ch. 2
Questions
1. Show that if |a'i, a'2| = |ai, a2|', then
[arx, a2-x] = λΊχι + а'глгг, [ary, a2-y] = a'i)>i + а'г^г
and that
arx ary
лг'х аг'У
= |a'i*i + aV2, a'iyi + а'ауа|
= |x,y|-|»'i,»'i|
2. What is meant by a right-handed orthonormal system?
Problems
1. Let vi = (ui + m)/\/2, v2 = ( —ui + и2)/л/2. Show that
VpVi = 1 = V2-V2, Vl'V2 = 0
and that |vi, V2I = 1. How is the system vi, V2 described?
2. Let h = b — ai, where t is such that h-a = 0. Show that (using T19.1)
|h|2 = h-(b - ai) = h-b
(a-b)^
= b-b - b-ai = ( (aa)(bb) - {-^l-\
|»,b|*
a-a ba
ab b b
a-a
and hence that
|h|= ±
|а,Ы
3. Show that the square of the area of a parallelogram, two of whose sides are
formed by a, b, is
a-a ba
ab b b
4. Show that the distance from a point X = (*i, дгг) to the line through С =
(ci, сг) with direction a = \a\, аг\ is
\*,CX\
a\ x\ — c\
"2 *2 ~ Ci
VaS +
02
5. Let vi = cui -(- SU2, V2 = —sui + сиг, where c2 + s2 = 1. Show that vrV2 =
0 and that vi, V2 is a right-handed orthonormal system.

Sec. 20]
LOCI
73
6. Let a, b, χ = [*i, x2], у = [jyi, y2] be such that
a*i + byi = ui, ax2 + Ъу2 = u2.
Show that
I a, b | - |x, y| = |a*i + hyh лх2 + Ъу2\ = 1.
20. LOCI
A set of points satisfying a geometrical condition is called a locus. An
equation that is satisfied by the coordinates of any point of the locus and by
no other point is called an equation of the locus. For example, if A is a given
point and с is a given vector, then the set of points X such that AX is
perpendicular to с is a locus. An equation of the locus is
c-AX = Ci(xi — ai) + c2(x2 — a2) = 0
and this equation is equivalent to the sentence: AX is perpendicular to с
This locus is a line through A with direction ν = f2ui — fiu2- A locus is often
visualized as the path of a point that moves so that the geometrical condition
is satisfied. Thus the above locus can be described as the locus of a point X
which moves so that AX is always perpendicular to с
EXAMPLE 20.1
Find the set (locus) of points X, each of which is equidistant from the
line 5*! + 5*2 — 2 = 0 and the line 7*i + x2 — 3 = 0; i.e., the locus of a
point that moves so that its distance from one line is always equal to its
distance from the other. By T13.1 the equation 5*1 + 5*2 — 2 = 0
represents a line, and if a point (xlf x2) is not necessarily on the line, then its
distance to the line is
(5*! + 5*2 ~ 2) _ (5*t + 5*2 ~ 2)
± V52 + 52 ~ ± 5V2
The distance from (*b x2) to 7*i + x2 — 3 = 0 is
(7*t + x2 - 3) _ (7*t + *2_~ 3)
V49 + 1 ~ 5V2
Since the condition of the locus is that these two distances are to be equal,
we equate the expressions for these distances. If the ambiguous signs are both

74 PLANE ANALYTIC GEOMETRY (Ch. 2
positive or both negative, we obtain
5*1 + 5a:2 — 2 7*i + *2 — 3
bVi ~ 5V2
or
0 = 2*1 - 4*2 - 1.
If one sign is positive and the other negative, we obtain
5*i + 5*2 — 2 7*i + *2 — 3
5V2 ~ 5V2
or
12*1 + 6*2 - 5 = 0.
The two lines thus obtained are the bisectors of the angles formed by the
given lines. These bisectors are sets of points. The set of those points that
belong to either or both of two sets is called the union of the two sets. The locus
we are considering here is the union of the angle bisectors.
EXAMPLE 20.2
A point moves so that its distance from the point (3, 0) divided by its
distance from the line *i — 25/3 = 0 is always %. Find the equation of the
locus. The distances from (*i, *2) to (3, 0) and to the line are respectively
V(*i - 3)2 + *22, ± (*1 - ψ)
It is given that the first of these distances divided by the second is %. Hence
V(*l - 3)2 + *22 3 , ;
± ΈΙ = - or ±5V (*i - 3)2 + *22 = 3*, - 25.
χι - -£■ 5
Actually the minus sign is needed in order for *i and *2 to be real, but we
shall square both sides to eliminate the radical and to eliminate the
ambiguous sign. Thus
25 (*i2 - 6*1 + 9 + *22) = 9*i2 - 150*i + 625,
25*i2 - 150*i + 225 + 25*22 = 9*i2 - 150*, + 625,
16*i2 + 25*22 = 400,

Sec. 20]
LOCI
75
EXAMPLE 20.3
A point moves so that its distance from the point (3, 0) plus its distance
from the point ( — 3, 0) is always 10. Find the equation of the locus. The two
distances are
V(Xl - 3)2 + x22 and V(Xl + 3)2 + *22.
The following equation states that the sum of these distances is 10.
V(Xl - з)2 + *22 + V(Xl + з)2 + *22 = ίο.
We can obtain a much simpler equation if we get rid of the radicals. To
accomplish this, we subtract one of the radicals from both sides of the
equation and then square. Thus
V(Xl + з)2 + *22 = ίο - V(Xl - з)2 + *22.
To square the right-hand side, we note that this side is equal to a — ft,
where a = 10 and ft = V(*i - 3)2 + *22 and that (a - ft)2 = a2 - lab +
ft2. Thus
*i2 + 6*! + 9 + *22 = 100 - 20V(Xl - 3)2 + *22 + Xl2 - 6Xl + 9 + *22,
20V(x, - 3)2 + л:22 = 100 - 12^,
5V(Xl - 3)2 + л:22 = 25 - Зх,.
Note that this last equation is the same as an equation obtained in the
previous problem and hence the elimination of the second radical also
produces
2 2
iL + 5-.ι.
25 16
This locus is called an ellipse.
Problems
1. A point moves so that its distance from the line 7xi + *2 — 8 = 0 is always
equal to its distance from the line *i + *2 — 2 = 0. Find the equation of the
locus. Answer:
-χι + 2*2 - 1 = 0, 2Xl + *2 - 3 = 0.

76 PLANE ANALYTIC GEOMETRY (Ch. 2
Check that the point (1,1) lies on all four of the lines, by showing that its
coordinates satisfy all four equations. Find the intercepts of these lines and
draw the lines.
2. A point moves so that its distance from the line 3jri + 4дсг — 12 = 0 is always
2. Find the equation of the locus. Answer:
3*i + 4*2 - 2 = 0, 3*i + 4*2 - 22 = 0.
Draw the three lines.
3. A point moves so that its distance from the point (4, 0) plus its distance from
( — 4, 0) is always 10. Find the equation of the locus. Subtract from both sides
of the equation the radical involving the distance from (4, 0); square and
show that the resulting equation can be converted into
V(*i ~ 4)2 + *2* = 4
¥-*i 5'
Interpret this equation as a new locus.
4. Eliminate the radical appearing in the answer to Problem 3 and show that the
resulting equation can be converted into
2 2
*1 . *2 _ ■
25 9 '
5. Find the equation of the locus of a point that moves so that its distance from
the origin is always 5. The locus is, of course, a circle with radius 5 and center
at the origin.
6. Find the equation of the circle with radius 3 and center at (2, 1).
7. Find the equation of the locus of a point that moves so that its distance from
(— 5, 0) minus its distance from (5, 0) is always 6. Subtract the radical involving
the distance from (5, 0) from both sides and show that the equation can be
converted into
V(*i- 5)' + *22 5
*i-£ 3'
Interpret this equation as a new locus.
8. Show that the answer to Problem 7 can be converted into
2 2
*1 *2 _ ■
9 16
This locus is called a hyperbola.
9. Find the equation of the locus of a point that moves so that its distance from
the point (3, 0) is always equal to its distance from the line *i + 3 = 0.
Answer *22 = 12*i. This locus is called a parabola.

Sec: 21]
THE PARABOLA
77
21. THE PARABOLA
The parabola, ellipse, and hyperbola are called conic sections. At a later
point we shall show that a conic section is the curve of intersection of a plane
with a cone, and that by choosing different relative positions between the
plane and the cone, we can obtain different conic sections. At present we
shall define a conic section as a locus.
D21.1. A conic section is the locus of a point that mooes so that its distance from a
fixed point divided by its distance from a fixed line is a positive constant. The fixed
point is called the focus, the fixed line is the directrix, and the constant is the eccentricity.
The conic section is an ellipse, parabola, or hyperbola, according as the eccentricity is
less than one, equal to one, or greater than one.
Consider first the parabola; i.e., consider the locus of a point that moves
so that its distance from the point (c, 0) divided by its distance from the line
xi + с = 0 is always one (where с > 0). The equation of the locus is
±V(Xl - c)2 + x22 r -r-i ~2
= 1 or ± V (*, - c)2 + x22 = xi + с
χι + с
Actually, the plus sign must be chosen, but we eliminate the ambiguity by
squaring. Thus
x2 — 2cxi + c2 + *22 = *i2 + 2c*i + c2,
or
x2 = 4«i.
Let us see how to draw this curve. Note that if χλ = 0, x2 = 0, the
equation is satisfied and hence the parabola passes through the origin. The origin
is called the vertex of the parabola. Let us see where the line x\ = с intersects
the parabola. If xi = c, then x22 = 4c2 and x2 = ±2c. Thus the points
(c, 2c), (c, —2c) lie on the curve. The segment joining these points is called
the latus rectum. Next note that if (χγ, χ2) satisfies the equation, then (xly — x\)
also satisfies the equation. If (xlf x2) lies above the *i-axis, then (xlf —x2) lies
an equal distance below the *i-axis. That is, the parabola is symmetric with
respect to the *i-axis. Since x2 cannot be negative for real x2, it follows that x\
cannot be negative (if с is positive). That is, the parabola lies to the right of
the x2-axis. As x\ increases, x2 either increases through positive values or
decreases through negative values. That is, the parabola opens up to the
right. It can be shown that this parabola is tangent to the x2-axis at the origin.

78 PLANE ANALYTIC GEOMETRY (Ch. 2
Xi+C
Xi+C
One can now plot the points (0,0), (c, 2c),
(c, — 2c) and draw a free-hand curve through
these points, conforming to the above
description. Sometimes it may be convenient to
replace the extremities of the latus rectum (c, 2c),
(c, —2c) by some other pair of points. For
example, (4c, 4c), (4c, —4c) also lie on the curve.
In drawing the curve, turn the paper so that
your wrist is on the concave side and your
hand can act as a compass. Make a light pencil
sketch because you may have to make
several attempts before obtaining a satisfactory
looking drawing. Select your best attempt,
draw a firm smooth curve over it, and then erase the false attempts (see
Fig. 21.1).
Figure 21.1
Questions
1. Define a conic section and tell what is meant by the focus, directrix, and
eccentricity. 2. State how the value of the eccentricity determines whether the curve
is a parabola, ellipse, or hyperbola. 3. Discuss symmetry and the location of special
points on the parabola.
Problems
1. Find the equation of the parabola with focus (3, 0) and directrix x\ + 3 = 0.
Draw the curve.
2. Find the equation of the parabola with focus ( — 3,0) and directrix x\ — 3 = 0.
Draw the curve.
3. Find the equation of the parabola with focus (0, 4) and directrix дгг + 4 = 0.
Draw the curve.
4. Find the equation of the parabola with focus (0, —4) and directrix хг — 4 = 0.
Draw the curve.
5. Find the equation of the parabola with focus ( — 1, —1) and directrix x\ +
хг — 2 = 0. Simplify the equation and show that it can be transformed into
χι2 - 2xlX2 + X22 + 8*i + 8*2 = 0.
6. Find where the line хг — x\ — 3 = 0 intersects the parabola of Problem 1.
Hint: Solve the equation of the line for хг and substitute this value into the
equation of the parabola. The resulting equation determines x\ and the
corresponding value of хг can be determined from the equation of the line. Note
that this line intersects the parabola in only one point. Add this line to your
figure of Problem 1 and correct the figure if necessary.

Sec. 22]
THE ELLIPSE
79
7. Find the two points where the line хг — x\ = 0 intersects the parabola of
Problem 1. Add this line to your figure of Problem 1 and correct again if
necessary.
22. THE ELLIPSE
In Section 20 we defined an ellipse in terms of two fixed points. Let us
show that this definition agrees with D21.1. Let the two fixed points be (c, 0)
and (— c, 0), where с is positive. The distance between these points is 2c.
Consider the locus of a point (xlf x2) which moves so that its distance from
(c, 0) plus its distance from (— c, 0) is a positive constant 2a. The equation of
the locus is
V(Xl - c)2 + x22 + V(Xl + c)2 + x22 = 2a.
The left-hand side of this equation is the sum of the lengths of two sides of a
triangle, and this sum (i.e., 2a) is greater than or equal to the third side
(i.e., 2c). Hence 2a ^ 2c or с ^ a. The equality holds only for the trivial case
in which the three vertices are collinear. We exclude this case, and hence
с < a. We now obtain
xi2 + 2cxx + c2 + x22 = 4a2 - 4aV(Xl - c)2 + x22
+ Xl2 - 2cXl + c2 + x22,
4aV(Xl - c)2 + x22
V(*i ~ c)2 + x22
a2/c - Xl
This latter equation is the equation of the locus of a point that moves so
that its distance from the point (c, 0), divided by its distance from the line
a21с — x\ = 0, is always equal to с /a. Thus (c, 0) is the focus, a2 /c — χλ = 0
is the directrix, and с/a is the eccentricity e. Moreover e < 1, since с < а.
Hence the conic section is an ellipse. We now remove the second radical.
Thus
aV (χι — c)2 + x22 = a2 — cx\,
a2(Xl2 - 2cXl + c2 + x22) =a* - 2a2cx, + с2х,2,
{a2 - c2)Xl2 + a2x22 = a* - a2c2 = a2(a2 - c2).
4a'
с
a
(a \
— 4cx! = 4c I χι 1.

80 PLANE ANALYTIC GEOMETRY (Ch. 2
Since a < с it follows that a2 — c2 is positive, and we can set this difference
equal to b2. Thus
b2Xl2 + a2x22 = a2b2,
or
*12 *22 ,
—+ —=i·
a
-2 b2
If e is very small, then с is small relative to a, and hence b2 is
approximately equal to a2 and the equation of the ellipse is approximately
x4+x4 = i or Xl2+X22 = «2.
This is the equation of a circle with center at the origin and radius a. Thus
an ellipse is approximately a circle when the eccentricity is approximately
zero. However if e = 0, then с = 0, and the equation a2/c — xi = 0 is
meaningless.
We next consider the problem of drawing an ellipse. If (χχ, x2) is a point
of the ellipse
2 2
«2 + b2 '
then (*!, — x2) is also a point of this curve. Hence this ellipse is symmetric
with respect to the *i-axis. Similar reasoning shows that it is symmetric with
respect to the *2-axis. Moreover, if (xif x2) is on the ellipse, then ( — xu —x2)
is also on it. If (x\, x2) is above the xi-axis and to the right of the *2-axis, then
(— xi, —x2) is an equal distance below the *i-axis and an equal distance to
the left of the *2-axis. That is, the ellipse is symmetric with respect to the
origin. Let us see where the ellipse crosses the axes. If x2 = 0, then
Xl < 2 2
-7=1, *i = a , Χι = ±α·
a1
Thus (a, 0), (—a, 0) are the ^-intercepts of the ellipse. Similarly (0, b),
(0, —b) are the ^-intercepts. Note that χλ2 ^ a2, since otherwise the left-
hand side of the equation of the ellipse would be greater than one, and this is
impossible. Hence xi lies between —a and a. Similarly x2 lies between — b
and b. That is, the ellipse lies inside the rectangle whose sides are x\ = a,
xi = —a, x2 = b, x2 = —b. It touches the sides of the rectangle at the four
points where the axes intersect these sides, and at each of these points it is
tangent to the side it touches (see Fig. 22.1).

Sec. 22]
THE ELLIPSE
81
(0,b)
(-o,0)
,(-c.O)
"J
L
Ο "ι
(c 0)J
(o,0)
(0,-b)
Figure 22.1
f-,= o
Questions
1. Discuss the symmetry of the ellipse. 2. Discuss the enclosure of an ellipse in a
rectangle.
Problems
1. Draw the ellipse of Problem 4 of Section 20.
2. Subtract the radical involving the distance from (*i, дгг) to ( — c, 0) from both
sides of the equation
>/(χι - с)л + x,* + VU, + e)« + „» = 2a;
then square and convert in the manner discussed in the text. Show that the
resulting equation represents an ellipse with focus ( — c, 0) and directrix
*i + a2/c = 0. Show that the equation can be converted into
2 2
a2 b2
3. A point moves so that its distance from (0, c) plus its distance from (0, —c) is
always 2a. Show that we must have с ^ a. Find an equation of the locus.
Remove one of the radicals by squaring and show that the resulting equation
can be interpreted as a new locus involving a focus and directrix. Remove the
other radical by squaring and show that the equation can be transformed into
x\
+
*2
= 1,
4. A point (да, хг) moves so that its distance from (1,1) plus its distance from
( — 1, —1) is always 4. Find an equation of the locus and show that it can be

82 PLANE ANALYTIC GEOMETRY (Ch. 2
transformed into
V(*i- D'+(*2- !)»_ V2
(4 - X1 - xt)/y/l 2
Interpret this latter equation as a new locus.
5. Find the intersection of the line Χϊ — b = 0 with the ellipse
and show that this intersection is a single point. Similarly for the lines хг + b =
0, *i - a = 0, *i + a = 0.
6. What does the equation given in Problem 2 become when с — 0? Interpret
this equation.
23. THE HYPERBOLA
Consider the locus of a point that moves so that its distance from ( — c, 0)
minus its distance from (c, 0) is always 2a. The equation of the locus is
V(Xl + c)2 + *22 - V(Xl - c)2 + X22 = 2a.
The left-hand side is the difference between the lengths of two sides of a
triangle and hence is less than or equal to the length 2c of the third side.
Hence a ^ c, and we exclude the equality. If we subtract the second radical
from both sides of the equation and square, we obtain
V(Xl - c)2 + X22 _ с
X\ — а2 /с а
This is a conic with focus (c, 0), directrix Xi — a2/c = 0, and eccentricity
e = с /a. Moreover e > 1, since с > a. Hence the conic is a hyperbola. The
equation can be converted into
(c2 - a2)Xl2 - aW = a2(c2 - a2).
Since с > a, we can let c2 — a2 = b2 and obtain

Sec. 23]
THE HYPERBOLA
83
Next consider the locus of a point that moves so that its distance from
(c, 0) minus its distance from (— c, 0) is always 2a. We obtain the equation
V(*l " C? + X22 - V(„ + c)2 + X22
2a.
If we subtract the first radical from both sides of the equation and square,
we obtain
V(Xl - c)2 + x22
с
а /с — X\ a
This is also the conic with focus at (c, 0), directrix χλ — а2/с =· 0, and
eccentricity e = с /a. Its equation also reduces to
2 2
X\ X2 _
~a^~~^~ '
The reader can check that this hyperbola is symmetric with respect to
both axes and with respect to the origin. The .«i-intercepts are (a, 0), ( — a, 0).
There are no ^-intercepts, since if χλ were 0, then x2 would be imaginary.
Let us solve the equation for x2. Thus
2 2 2 2 i2
*2 X\ X\ ~ a f. "
1 2 /2 2\
77 = -r - 1 = 2 X2 =-z{xi -a ),
b1 a1 a1 a2
or
x2 = ± - Vx,2 - a2,
a
If Χγ is very large and positive, then a2 is negligible with respect to x2, and
x2 is approximately ,
±-*i
a
To show this, consider the case of the plus sign and let us form the difference
b b , b ,
- xx - - Vx,2 - a2 = - (Xl - Vx,2 - a2)
b(xi ~ У*.2 ~ *2)(*i + V*i2 ~ a2)
Φι + ^V - a2)
b(Xl2 - (xi2 - a2)) ab
a(xi + Vxi2 - a2) xx + Vxi2 - a2
Since both terms in the denominator are large when xi is large, the fraction
is small; i.e., the difference is small. Moreover, the fraction is positive, and

84 PLANE ANALYTIC GEOMETRY (Ch. 2
hence (ί/«)νχι! — α2 is slightly less than (b/a)xl when χλ is large. That is,
the hyperbola lies close to but slightly under the line хг = (b/a)xi when χχ
is large. This line is called an asymptote of the hyperbola. The case of the
negative sign is similarly treated, as also is the case in which Xi is large but
negative. There are two asymptotes whose equations are given by
xS xS b
— ~lfi = ° or *a = ±-*i-
a b a
To sketch the hyperbola, we again draw the rectangle whose sides are the
lines xi = a, xi = —a, *2 = by x2 = —b. The diagonals of this rectangle are
the asymptotes (see Fig. 23.1). A point moving away from the origin along
Figure 23.1
the hyperbola moves continually closer to the asymptote. Don't draw a
curve that turns away from the asymptote.
In Problem 4 you will show that the equation
2 2
*2 X\ _
also represents a hyperbola. If we solve for x\, we have

Sec. 23]
THE HYPERBOLA
85
which for numerically large *2 is approximately
a
That is, the asymptotes of this hyperbola are given by
= 0.
2 2
*2 X\
~ 1^
b
It follows that the hyperbolas
2 2 2 2
Xl X2 _ X2 Xl _
have the same asymptotes. These hyperbolas are said to be conjugate (see
Fig. 23.2).
Figure 23.2
In the preceding discussions, the foci and directrices of the various conies
were chosen so that the resulting equations were in simple forms. These
forms are called standard. Later we shall consider conies whose equations are
not in standard form.
Questions
1. Discuss the symmetry of the hyperbola. 2. Show how to obtain the asymptotes.
3. Show that the hyperbola is close to one of its asymptotes at a point a long way
from the origin. 4. Show how to obtain a rectangle whose diagonals are the
asymptotes. 5. Show that the hyperbola touches two sides of this rectangle.

86 PLANE ANALYTIC GEOMETRY (Ch. 2
Problems
1. Draw the hyperbola of Problem 8 of Section 20.
2. Show that the equation
V(XI - c)2 + x22 - >/(*, +е)* + Ж1« = 2a
can be converted into an equation of a hyperbola with focus at ( — c, 0) and
directrix x\ + a2/c — 0. Show that this latter equation can be converted into
2 2
*1 *2 _
а2 b2
3. Show that the lines x\ — a = 0, *i + a — 0 intersect the curve
а2 Ьг
in one point each.
4. A point moves so that its distance from (0, — c) minus its distance from (0, c) is
always 2b. Show that the equation of the locus can be converted into the
equation
V*!2 + (*2 ~ С)г С
хг-Ь1/с b'
Interpret this equation as a locus and show that the equation can be converted
into
*l _ xj1 - ι
A2 2
where a2 = c2 — b2.
5. Find the *i-coordinates of the two points of intersection of the line хг = mx\
with the curve of Problem 3, where —b/a < m < b/a. What happens if
m = ± b/a?

SOLID
ANALYTIC
GEOMETRY
3
24. SPHERES, LINES, AND PLANES
Solid geometry is concerned with the set of points associated with a
three-dimensional vector space. In solid analytic geometry every point X is
referred to a point 0 chosen as origin. The vector space contains an ortho-
normal system ub u2, из with the property that every vector OX has the form
OX = a^Ui + x2u2 + X3U3, where xu x2, *з are called coordinates (or more
specifically, cartenan coordinates) of X. The point X = (*i, x2, X3) determines
its coordinates uniquely, and conversely, any ordered triplet xly x2, X3 de-

88 SOLID ANALYTIC GEOMETRY (Ch. 3
termines a unique point. The distance between the points X = (*i, x2, *з),
У = (уъУ2,Уз) is \XY\, where
\XY\=\XY-XY = V(yi -Xl)2+ (У2 - x2)2 + (у, - χ3)\
This is called the distance formula. We have Theorem T24.1.
T24.1. The distance between the points (xi, x2, хз), (у1,У2,Уз) is
^(yi ~ *i)2 + (>2 - *2)2 + (уз ~ хз)2·
A sphere with radius r and center at (αϊ, a2, a3) is the set of points
(*i, *2, *з) at a distance r from (au a2, a3). An equation of the sphere is
^(xi ~ αϊ)2 + (X2 ~ a2f + (x3 - a3)2 = r,
and this equation is a statement that (xlf x2, x3) is at a distance r from
(βι, Д2, аз)· The sphere is the set of points that satisfy this equation.
Let ν = [»i, v2, v3] be a nonzero vector with origin A = (βι, α2, a3), and
let X be the point such that AX has the form AX = vt. Since ЛЛ7, ν are
linearly dependent, the vector AX is contained in the line containing ν and
the tip X of AX lies on this line. The set of points X for which AX has the
form vt is a one-dimensional space, i.e., a line. Since
OX = О A + AX = OA + vt,
it follows that
[*i, *2, *з] = [«ι + "it, "2 + M, a3 + 03/],
and since the components of a vector are unique, it follows that this vector
equation is equivalent to the parametric equations
χι = al + υ-it, x2 = a2 + v2t, x3 = a3 + v3t.
By substituting various values for the parameter /, we obtain points of the
line. These points and the line containing them can be depicted by drawing
the appropriate isometric projection.
If A = (βι, α2, α3), Β = (bu b2, b3) are distinct points, then AB is a
nonzero vector and the set of points X = (x\, x2, x3) for which AX has the form

Sec. 24] SPHERES, LINES, AND PLANES 89
ABt is a line. Since
OX = 0A+ {OB - OA)t = 0A{\ - i) + OBt = OAs + OBt
where
it follows that
s = \ — t or s + t = 1,
[*i, *2, *з] = [ais + M, a2s + b2t, a3s + b3t]
and hence that
Χι = a^ + b^, x2 = a2s + b2t, x3 = a3s + b3t.
The scalars s, t are called barycentric coordinates of X with respect to A, B; and X
is denoted by (s, t). The barycentric coordinates are subject to the condition
s + t = 1. Thus, if we substitute s = 1, / = 0 in the above equations, we
obtain *i = βι, x2 = a2, x3 = a3 and hence (1, 0) denotes the point A.
Similarly (0, 1) denotes the point B. Figure 24.1 shows various points of the
(-1,2)
(2,-1)
Figure 24.1
line through A = (2, —2, — 1), В = (1, 3, 4). The points are designated by
their barycentric coordinates. If we substitute 1 — / for the barycentric
coordinate s, we obtain the following parametric equations of the line:
*i
a,(l - /) + b^, x2 = a2{\ - t) + b2t, x3 = a3(l - /) + b3t.
In Section 39 we show that if ν = [vi, v2, v^], w = \w\, w2, w3\ are linearly
independent vectors with common origin A = (a\, a2, a3), then the set of
points X for which AX has the form vs + wt is a two-dimensional space.

90 SOLID ANALYTIC GEOMETRY (Ch. 3
We call this space the plane containing the vectors v, w with common origin
A. In Problem 5 you will be asked to show that this plane has the parametric
equations
x\ = αϊ + vis + wit, x2 = a2 + v2s + w2t, x3 = a3 + V3S + u^t.
If the points A = (αλ, а2, a3), В = (bl, b2, b3), С = (сь с2, c3) are such that
AB, AC are linearly independent, then the set of points X for which AX has
the form ABs + ACt is a plane. Since
OX = О A + (OB - OA)s + (ОС - OA)t
= OA(\ - s - t) + OBs+ OCt
= OAr + OBs + OCt,
where
г = 1 — s — t or r + s + t = 1,
it follows that
[*i, *2, *з] = [air + bis + cit, a2r + b2s + c2t, a3r + b3s + c3t]
and hence that
xi = air + bis + cit, x2 = a2r + b2s + c2t, x3 = a3r + b3s + c3t.
The scalars r, s, t are called barycentric coordinates of X with respect to the points
А, В, С; and X is denoted by (r, s, t). In particular (1, 0, 0), (0, 1, 0),
(0, 0, 1) denote respectively the points А, В, С The barycentric coordinates
locate X on the plane containing А, В, С and are always subject to the
condition г + s + t = 1.
Questions
1. Derive the distance formula. 2. Define a sphere. 3. Show how to obtain an
equation of a sphere. 4. Discuss the one-dimensional space containing a nonzero
vector ν with origin A. 5. Show how to obtain parametric equations of this line.
6. Show how to obtain barycentric coordinates of a point X with respect to two points
A, B. 7. Show how to obtain parametric equations of the line through A, B. 8. Discuss
the two-dimensional space containing two linearly independent vectors with common
origin. 9. Show how to obtain barycentric coordinates of a point X with respect to
three points А, В, С

Sec. 25] THIRD-ORDER DETERMINANTS 91
Problems
1. Show that a point X determines its coordinates uniquely and that an ordered
triplet *i, *2, *з determines a unique point.
2. Find the distance between the points (1, 4, —3), (2, 6, —1).
3. Find an equation of the sphere with radius 3 and center at (1,4, —3). Show
that the points (2, 6, —1), (3, 6, —2), ( — 1,5, —1) lie on this sphere.
4. Find parametric equations of the line containing the vector ν = [ — 2, 4, 3],
with origin (1, —2, —1). Draw the isometric projection of the vector v.
5. Derive the parametric equations of the plane containing the vectors v, w, with
common origin A.
6. Let А, В, С be respectively (1, -2, —1), (2, 5, 0), (0, 3, 4). Draw the isometric
projection of the triangle whose vertices have the barycentric coordinates
(1, 0, 0), (0, 1, 0), (0, 0, 1) with respect to А, В, С. The point whose barycentric
coordinates are (И, И, И) is called the barycenter. Draw the isometric projection
of the barycenter.
7. Show that if X is such that AX has the form ABt then XB - AB{\ — t) =
ABs where s, t are barycentric coordinates of X. Show that | ЛЛГ | / | ЛГ2? | =
i/(l — i) = t/s when s, t are both psoitive. Show that if i/(l — /) = m/n,
then X has the barycentric coordinates s = 1 — t = n/(m + n), t = m/{m +
n), and the cartesian coordinates
na\ -f- fnb\ пач -f- tnbi naz -\- тпЬг
χι = г—' Х2 = r—' *3 = :
m + n m + η m + η
U А, В are respectively (αϊ, аг, аз), (bi, Ьг, Ьз).
8. Let X have barycentric coordinates (r, s, t) with respect to А, В, С. Show that
if r = 0, then X has the barycentric coordinates (s, t) with respect to A, B.
25. THIRD-ORDER DETERMINANTS
We have defined determinants in terms of two vectors in a space of two
dimensions. These determinants are said to be of second order. Third-order
determinants are defined in terms of three vectors in a space of three
dimensions and are denoted as follows:
Ol *1 C\
|a, b, c|= a2 b2 c2
аз *з сз
where
a = ^iii + a2u2 + a3u3, b = iiUi + b2u2 + b3u3
С = CiUl + C2M2 + C3U3.

92 SOLID ANALYTIC GEOMETRY (Ch. 3
To define third-order determinants, we generalize the properties, PI 4.1 to
PI 4.4, which defined the second-order determinants. Thus a third-order
determinant is required to have the properties:
P25.1. |ub u2, u3| = 1.
P25.2. |a,b,c + d| = |a,b,c| + |a,b,d|.
P25.3. |a, b, zc| = z|a, b, c|.
P25.4. | а, с, Ь | = |c,b,a| = |b,a,c| = -|a,b,c|.
Note that P25.4 states that the interchange of any pair of vectors changes
the sign of the determinant. You will be asked to prove the following
additional properties in the problems:
P25.5. |a,b + c,d| = |a,b,d| + |a,c,d|,
|a + b,c,d| = |a,c,d| + |b,c,d|.
P25.6. \я,уЪ, с| =у\я, Ь, с|, |ха,Ь,с| = *|a, Ь, с|.
P25.7. |а,Ь,Ь| = |а,Ь,а| = |а,а,с| =0.
Р25.8. |а, Ь, 0| = |а,0,с| = |0,Ь,с| =0.
We shall show that the above properties determine the value of the third-
order determinant. Thus
| а, Ь, с | = I^U! + a2u2 + a3u3, b, c|
= a, | ub Ь, с| + a2\u2, Ь, с| + a3|u3, Ь, с|,
where | Ui, b, с |, | u2, Ь, с |, | u3, b, с | are called ngned minors of the elements
ai, a2, аз, respectively. These signed minors are evaluated as follows:
|u1; Ь, с| = 0 + i2|ui, u2, с| + i3|ubu3, c|,
b21 ub u2, с | = 0 + 0 + b2c31 u1; u2, u31 = b2c3,
b31 u1; u3, с | = 0 + b3c21 ub u3, u21 + 0 = — b3b2 | ub u2, u31 = — b3c2.
Hence
l«i, Ь, с | = b2c3 — b3c2 =
and similarly
|u2, Ь, с | = -bYc3 + b3a = -
|u3, Ь, с| = btc2 — b2ct
b2 c2
b3 c3
b\ cl
b3 c3
b\ fi
b2 c2

Sec.25] THIRD-ORDER DETERMINANTS 93
Thus the third-order determinant has the value
\a, b,c\
«i|ui, b,
Ol *1
d2 b2
аз *з
с I + a21 u2:
c2
f3
b, c|+ a3|u3, b, c|
= Ol
b2
b3
c2
сг
— a2
br
b3
C\
f3
+ a3
*1
b2
C\
c2
A signed minor is the determinant that results when one of the vectors
a, b, с is replaced by one of the vectors Ui, U2, U3. The signed minor of a
given element is determined as follows: The column in which the element is
located is made up of the components of the vector to be replaced, and this
vector is replaced by that one of the vectors Ui, U2, U3 whose subscript is the
same as that of the given element. Thus the signed minor a2 is obtained by
replacing a = [αϊ, α2, 03] by U2. The signed minor of each element is equal,
except for sign, to a second-order determinant.
The second-order determinant is called the minor of the given element and
is obtained as follows: The element is located in a certain row and certain
column, and if the elements of this row and column are removed, then the
elements that remain form the second order which is the minor of the given
element. Thus the minor of 03 is made up of those elements that do not appear
in the row and column containing 03. A signed minor differs at most in sign
from the corresponding minor.
EXAMPLE 25.1
Let a = [3, -2, 1], b = [5, 0, -7], с = [4, -2, 3]. Then
3 5 4
-2 0 -2
1 -7 3
0-2 5 4
- (-2)
-7 3 -7 3
|a,b,c|
= 3
+ 1
4
-2
3 (-14) + 2-43 + (-10) = 34.
Questions
1. Write down the four properties that define a third-order determinant and the
four additional properties that are consequences of the defining properties. 2. Show
how to obtain the expansion of a determinant in signed minors of the first column.
3. Show how to evaluate |ui, b, c|. 4. Define the signed minor of a given element.
5. Define the minor of a given element. 6. Show how to obtain the expansion in minors
of the first column.

94 SOLID ANALYTIC GEOMETRY (Ch. 3
Problems
1. Prove P25.5.
2. Prove P25.6.
3. Prove P25.7.
4. Prove P25.8.
5. Show how to evaluate |ui, b, c|, and check with the result given in the text.
6. Show how to evaluate |из, b, c|.
7. Compute.
3 4 0
-2 0 -5
7 2 6
8. Let a, b, c, χ = [χι, дг2, дг3], у = [yi,yt,yt\, ζ = [*i, Z2, zj] be any six vectors,
and let
x" = λχι + b*2 + cx3, y" = ayi + b>2 + cy3
ζ" = azi + bz2 + cz3.
Show that
| x", y", z" | = *, | a, y", 2" | + *21 b, y", 2" I + *31 c, y", 2" I
and that
|a,y",2"|= (у#г-у**)\я,Ъ,с\=\у* ** |-|а,Ь,с|.
I Уз zj I
" П|а,Ь,с|
У2 Z2 I
Assume that
b,y",2"|= -
and show that
\П Zl |-|a,b,c|, |cy", ж"
1 >з zj 1
|«",y",2"| = |«,y,2|-|a,b
26. FURTHER PROPERTIES OF DETERMINANTS
The problem of expanding a determinant in minors of the second column
is reduced to that of expanding in minors of the first column as follows:
|a, b, c| = -|b, a, c|
= — *i|ub а, с| - b2\u2, a, c| - b3\u3, a, c|
-bi
02 C2
03 C3
+ b2
"1
оз
C\
f3
- b3
"l Cl
Д2 ^2

Sec. 26] FURTHER PROPERTIES OF DETERMINANTS 95
This is the expansion of the determinant in minors of the second column.
Since
I a, b, c| = 6, |a, ui, c| + i2|a, u2, c| + i3|a, u3, c|,
the signed minor of b2 is equal to the corresponding minor, whereas the signed
minors of b\, 63 are the negatives of the corresponding minors. Similarly, the
expansion in minors of the third column is
I а, Ь, с I = - I а, с, ЬI = I с, а, Ь I
a2 b2
03 *3
— c2
αλ bi
03 b3
+ c3
02 b2
= с 11 a, b, U! I + c21 a, b, u21 + c3 | a, b, u3 |.
The signed minor of an element is equal to the corresponding minor or to
the negative of this minor, according as the element occupies the position of a
plus or minus sign in the following checkerboard diagram:
+ - +
- + -
+ - +
Thus the signs of the terms in the expansion in minors of the second column
are the same as the signs of the second column of the checkerboard, namely,
— ,+,—. In the minor expansion in the third column the signs are the same
as the third column of the checkerboard; similarly for the first column.
We shall show that the expansion
I a, b, c| =o1|u1,b, c| +*i|a, u,, c| +c,|a, b, u,|
is also valid. This is called the expansion in signed minors of the first row.
We have
I а, Ь, с I =a1|u1, b, c| + |a2u2 + o3u3, b, c|
= a-ι I ub Ь, с | + bY I a2u2 + агмг, ub с |
+ I a2\i2 + a3u3, i2u2 + b3u3, с \
= β, I ub Ь, с I + *! I a2u2 + a3u3, ub с |
+ fi I a2u2 + a3u3, i2u2 + b3u3, щ \
+ 102U2 + a3u3, i2u2 + b3u3, c2u2 + C3U3 |

96 SOLID ANALYTIC GEOMETRY (Ch. 3
where the last of these determinants is zero, since
I o2u2 + o3u3, b2\i2 + b3u3, c2u2 + c3u3 \
0 0 0
0 0 0 0
a2 b2 c2 = 0 — a2 + a3 = 0.
b3 сз b2 c2
a3 b3 c3
= β! | ub Ь, с | + h I a2n2 + a3u3, щ, с |
+ fi I o2u2 + a3u3, b2u2 + b3u3, щ \
Hence
I a, b, с I
where
-|ui, a, c| = -
a2u2 + a3u3, ui, c| = a — (цщ, ub c|
= la. ub cl - «il«i, «i. c| = I a, uu с
a2 c2
a3 c3
а2м2 + a3u3, Ь2м2 + b3u3, и1\ = \я — a,^, Ь — biub U! |
= | а, Ь - o,U!, U! | - 0 = | a, b, U! |
a2 b2
a3 b3
= |ui, а, Ь
It follows that
| а, Ь, с | = β! | ui, Ь, с | + 0! | а, иь с | + C! | a, b, U!
= a\
b2
b3
c2
сг
-Ьх
a2
оз
c2
f3
+ Cl
a2
оз
b2
b3
where the first line in the above equation is the expansion in signed minors of
the first row and the second line is the expansion in minors of the first row.
We shall use this result together with the fact that a second-order
determinant is equal to its transpose, to show that a third-order determinant
is also equal to its transpose. The transpose |a, b, c|' of |a, b, c| is the
determinant that results from interchange of the rows of |a, b, c| with the
columns. The transpose can be expanded in minors of the first column as
follows :
Ol 02
bi b2
оз
*з
Cl C2 C3
= a\
= Ol
b2
c2
b2
b3
b3
сг
c2
сг
- h
- bl
a2
c2
02
03
03
сг
c2
сг
+ Ci
+ fi
02 a3
b2 b3
a2 b2
03 b3
= |a,b,c|.

Sec. 26] FURTHER PROPERTIES OF DETERMINANTS 97
That is, the expansion of the transpose | a, b, c|' in minors of the first
column is equal to the expansion of | a, b, с | in minors of the first row. Thus
we have T26.1.
T26.1. A third-order determinant | a, b, c| is equal to its transpose | а, Ь, с |'.
As a corollary to T26.1 note that a determinant with two equal rows is
zero, since the transpose of this determinant has two equal columns and
is therefore made up of two equal vectors (see P25.7). Thus we have T26.2.
T26.2. A third-order determinant with two equal rows or two equal columns is zero.
As a second corollary note that a determinant can be expanded in minors
of any row since this expansion is equal to the expansion of the transpose in
minors of the corresponding column. The signs follow the checkerboard
scheme, since the checkerboard is unchanged if its rows are interchanged
with its columns.
Questions
1. Write down the expansion of a determinant in minors of the second column
and show how this expansion is obtained. 2. Write down the expansion in signed
minors of the second column. 3. Show how to obtain the expansion in minors of the
third column. 4. Write down the expansion in signed minors of the first row and show
how to obtain it. Also show how to obtain the expansion in minors of the first row.
5. Prove that a third-order determinant is equal to its transpose. 6. Prove that a third-
order determinant is zero if two rows or two columns are equal. 7. Show how to
obtain the expansion in minors of any given row. 8. Describe a scheme for keeping
track of the signs in the various expansions.
Problems
1. Check the result of Problem 7 of Section 25 by expanding in minors of the
second row.
2. Let (*ι, Χ2), (αϊ, аг), (*ь *г) be three points (plane analytic geometry). Consider
the equation
ДС1 *2 1
01 02 1 = 0.
bx b2 1
Expand the determinant in minors of the first row and show that the resulting
equation represents a line (unless certain minors are zero). Substitute a\, аг
for χι, *2 in the determinant and show that the equation is satisfied. Similarly
show that (ii, Ьг) is a point of the line.

98 SOLID ANALYTIC GEOMETRY (Ch. 3
3. Use the result of Problem 2 to find the equation of the line through (4, —2),
(7, 1).
4. Show that the equation
χι хг 1
a 0 1
0 b 1
0
is equivalent to the intercept formula.
5. Show that if ах + Ъу + cz = d, then |d, b, c| = *|a, b, c|; and that if
| a, b, с | τ* 0, then
|d,b,c|
|a,b,dl
|a, b, c| |a, b, c| |a, b, c|
6. (a) Show that if a = [αϊ, ач, аз], b = \b\, Ьч, *з]> с = [а, сч, с»], then
|a,b, c| |ui, Ь, с| |a, m, c| |a, b, m|
_ 01 -: ; [- + 01 -. ; [- + С\ -
|а,Ь, с
(b) Show that if
|ш, b, cl
|a,b,c|
a,b,c|
|a,b,c|
X\
У1 =
|a,ui,c|
|a, b, c|
Zl =
|a, b, c|
la, b, m|
|a,b,c|
then αι*ι + b\yi + C1Z1 = 1.
(c) Show that
Д2 Ьч Сч
ач Ьч Сч
03 bz Cz
|a,b,c|
= αηΧ\ + *2>1 + C4Z1 = 0.
(d) Show that а&\ + *з>1 + C3Z1 = 0.
(e) Show that a*i + \>y\ + czi = щ.
7. (a) Show that |a, b — xa, c| = |a, b, c|.
(b) Show that |a — хЪ,Ъ,с — yb\ = a, b, c|.
8. Show that
a\ — хач b\ — xbi c\ — хсг а\ b\ c\
ач Ьч сч — ач Ьч сч
аг — уач Ьг — уЬч сг — усг "г Ьг сг

Sec.27] THREE EQUATIONS IN THREE UNKNOWNS 99
27. THREE EQUATIONS IN THREE UNKNOWNS
Let a, b, c, d be four vectors such that |a, b, c| τ* 0, and let us solve the
vector equation
ax + by + cz = d
for x, y, z. If x, y, ζ satisfy the equation, then
| d, b, с | = | ax + by + cz, b, с |
= ж|а, Ь, с| +>|b, b, с| + z|c, b, c|
= *|», b, c|
|d, b, c|
and hence
χ =
Similarly
and hence
la. b, c|
I a, d, c| =у\я,Ъ,с\, | а, Ь, d | = z\d, b, c|,
| a, d, с | | a, b, d |
у = ;—;—7' z
a, b, c| |a, b, c|
We have shown that if x, y, ζ satisfy the equation, their values must be the
above ratios of determinants. It remains to show that these values do satisfy.
First we consider three special cases.
We shall show that the equation
is satisfied if
a*i
lUl
la,
la>
+ ъУ1
., b, c|
b, с
b, uj
+ CZ! =
• У\
■■ U!
Ia.
|a,
Ul,
b,
c|
c|
*1 =
I a, b, с
We have
a*i + byi + czi = (aiUi + a2u2 + азиз)*1
+ (iiUi + i2u2 + *зиз)>1 + (fiUi + c2u2 + c3u3)zi
= (ai*i + biyi + ciz,)ui
+ (a2xi + b2yx + c22i)u2 + (α3·«ι + *3>i + c3zi)u3,

100 SOLID ANALYTIC GEOMETRY (Ch. 3
where each of the latter three parentheses produces the expansion of a
determinant in minors of the first row, namely,
(alxl + biyi + c-lz-l) =
ai | ub Ь, с | + &i | a, ub с | + ci | a, b, ut |
|a,b,c|
Similarly,
(a3xi + b3yi + c3zi)
|a,b,c| i
|a,b,c|
У\ + C2Zl) =
... _L ,.,Λ —
=
02
02
Ol *1
Cl
a2 b2 c2
03 b3 c3
|a, b, c|
b2 C2
b2 C2
оз *з сз
|a,b,c|
03
02
03
*3 C3
Ь2 C2
*3 £3
- t
I a, b, с I
The second and third parentheses are both equal to zero, since in each case
the corresponding numerator determinant has two rows equal. Since the
first parenthesis is one, we have
a*l + ty, + CZ! = Щ,
i. е., these values of xu у\, ζλ satisfy the vector equation.
Similarly it can be shown that the equation
a*2 + Ьуг + CZ2 = U2
is satisfied if
and
*2 =
z2 =
|u2, b, c|
|a,b,c|
1 a, b, u21
|a,b,c|
а*з + Ъу3
У2 =
+ cz3
1 a, u2, с |
|a, Ь, с|
= u3

Sec.27] THREE EQUATIONS IN THREE UNKNOWNS 101
is satisfied if
I из, Ь, с I I a, u3, с |
'з = -—: r< Уз =
•гз
I a, b, c| | a, b, c|
I a, b, u3|
Now let
|a,b,c|
|d, b, c| |u,, b, c| |u2, b, c| |u3, Ь, с
* = -—: г = «ι -—: г + «2 -—: г + «з
а, Ь, с| |а, Ь, с| |а, Ь, с| |а, Ь, с1
d\x\ + d2x2 + d3x3,
|a,d,c|
|a, Ь, с|
|a,b,d|
ζ =
|a, b, c|
Then
= diyi + d2y2 + d3y3,
= άγΖγ + d2Z2 + <^3Z3·
ад: + by + cz = Λ{άλχλ + d2x2 + d3x3) + b(diyi + d2y2 + d3y3)
+ c^i*! + d2z2 + d3z3)
= difai + Ъух + czj + d2(ax2 + Ъу2 + cz2)
+ d3(*x3 + by3 + cz3)
= diUi + d2u2 + d3u3 = d,
and hence the vector equation is indeed satisfied by the above values of x, y, z.
This vector equation is equivalent to three scalar equations. Thus
a* + by + cz = [αλχ + biy + c,z, a2x + b2y + c2z, a3x + b3y + c3z]
= [du d2, d3] = d,
and hence
a\x + biy + cxz = du
a2x + b2y + c2z = d2,
a3x + b3y + c3z = d3.
We now have T27.1.
T27.1. // a, b, c, d are four vectors such that |a, b, c| τ* 0, then the vector
equation
a* + by + cz = d

102 SOLID ANALYTIC GEOMETRY (Ch. 3
is satisfied, or equivalently, the scalar equations
a\x + bly + cxz = du
are satisfied if and only if
a2x + b2y + c2z = d2,
a3x + b3y + c3z = d3
|d, Ь, с |
la. b, c|
la, d,
I а, Ь, с |
b,d|
la> b, c|
di
d2
d3
a\
a2
a3
a\
a2
a3
a\
a2
a3
a\
a2
a3
a\
a2
a3
*1
b2
b3
*1
b2
b3
di
d2
d3
*1
b2
b3
*1
b2
b3
*1
b2
b3
C\
c2
f3
C\
f2
c3
C\
c2
c3
C\
c2
c3
di
d2
d3
C\
c2
f3
Note that the elements of the common denominator determinant are
the coefficients of x, y, ζ and that these elements appear in the same relative
positions as in the three equations. The numerator in the formula for χ is
obtained from the denominator by replacing the column formed from the
coefficients of χ (first column) by the column formed by the right-hand sides
of the equations; similarly for y, z.

Sec.27] THREE EQUATIONS IN THREE UNKNOWNS 103
Questions
1. Show that if a* + by + cz — d and | a, b, с | ^ 0, then
|d,b,c|_ |a,d,c| __ |a, b,d|
|a, b, c| ' У |a, b, c| ' * |a, b, c|
2. Show that a*i + byi + cz\ = ui if
= |m, b, d = |a, m, cl = |a, b, ut|
I a, b, c| ' >l |a, b, c| ' *' |a, b, c|
3. Show that if *i, >i, z\, хг, уг, ζϊ, x3, у3, z3 are appropriately defined, if a*i + byi +
czi - ub алг2 + b>2 + cz2 = u2, ax3 + Ъу3 + cz3 = u3, d = [d\, аг, d3] and if x, y, ζ
are the three ratios of Question 1, then χ = d\x\ + dixi + d&3, у — d\y\ + diyi +
d3y3, ζ = dizi + d2Z2 + «/3^3 and that these values do satisfy the equation ал +
by + cz = d. 4. Write down the three scalar equations equivalent to ax + Ъу +
cz = d.
Problems
1. Solve the equation
[3, 1, -5}x + [2, -4, Ъ\у + [5, -3, \\z = [0, 3, 0].
Write down the corresponding system of scalar equations.
2. Find x, y, z, t not all zero such that
[3, 1, -5}x + [2, -4, Ъ\у + [5, -3, \\z + [0, 3, 0]i = 0.
Hint: Set ί = — 1 and use the result of Problem 1.
3. Let a = [3, 1, -5], b = [2, -4, 13], с = [5, -3, 8], d = [0, 3, 0].
(a) Compute |a, b,c|, |d, b, c|, |a, d, c|, |a, b, d|.
(b) Show that if a* + by + cz = d, then
*|a,b,c| = |d,b,c| =0, |a,d,c| =0, |a,b,d| =0.
(c) Does the equation of (b) have a solution?
4. Show that if a, b, с are defined as in Problem 3, then
a* + by + cz = 0 if χ = 1,> = 1, ζ = —1.
5. Let a, b, c, d be defined as in Problem 3. Find x, y, z, t not all zero such that
ax + by + cz + dt — 0. Hint: Let t — 0 and use the result of Problem 4.
6. Let a, b, с be defined as in Problem 3. (a) Show that ax + by + cz — 0 if
у - χ, ζ - —χ. (b) Find χ, у, ζ such that ax + by + cz = 0 and x2 + y2 +
z2= 1.

104 SOLID ANALYTIC GEOMETRY (Ch. 3
28. THE PRODUCT OF TWO DETERMINANTS
Some geometric results will be obtained by means of the following formula
for the product of two determinants:
I a ι, a 2, а з|
a^x a^y ai*z
a2 · χ a2 · у a2 · 2
a3-x a3-y a3-z
where the determinant on the right is made up of the vectors [ai* x, a2-x, a3-x],
[ару, a2-y, a3-y], [ai-2, a2-2, а3-г], and a\, a'2) a'3 are the vectors forming
the transpose of | at, a2, a3|. Thus
Let
|ai,a2, a3| = |ab a2, a3| = |ab a2, a3|.
ai = [ni-ai, u2-ab u3-aj = [an, a2i, «3il,
a2 = [u!-a2, u2-a2, u3-a2] = [αγ2, o22, o32],
a3 = [ura3, u2-a3, u3-a3] = [α13, a23, 033].
Then
[a 1, a 2, a 3J =
Oil Ol2 Ol3
021 022 023
«31 a32 033
Oil 021 031
012 o22 a32
θι3 ο23 α33
Let
a'i = [оц, o12, a13], a'2 = [α21, α22, a23], a'3 = [α31, a32, 033].
x = [*i, *2, хз\, У = 1у1,У2,уз\, ζ = [zu z2, z3],
x" = [arx, a2-x, a3-x], у = [ary, a2-y, a3-y],
2" = [a!-2, a2-2, а3-г].
Then
x" = [anxl + a21*2 + a3lx3, al2X\ + o22*2 + a^xz, αιζχλ + a23x2 + 033*3]
= [оц, о12, α13]*! + [α21, a22, a23]*2 + [а31, а32, азз]д:3
= a'!*! + a'2*2 + а'3*з,
and similarly,
у" = л\ ух + а'2 у2 + а'зуз, 2" = aVi + a'2z2 + а'3г3.

Sec.28] THE PRODUCT OF TWO DETERMINANTS 105
Moreover,
ai'X ai-y ai*z
a2 · χ a2 · у а2 · 2
a3 · χ a3 · у а3 · ζ
l*",y",«"l
= |aVi + a'2*2 + а'3д:з, у", ζ'
where
= Xl | a'b y", 2" | + x2 | a'2, y", 2" | + ,81 a'3, y", 2" |,
Ι «Ί, У", «" I = >i I a'i, «Ί, 2" | + y2 | a'b a'2, 2" | + >81 «Ί, a'8, «" I
= >a|a'i,a'a, 2" | + y3|a'i, a'3, ж"|
= 0 + 0 + >2г31 a',, a'2, a'31 + 0 + ji3z2 | a\, a'3, a'21
and similarly,
Hence
|*",y",«"| =
=
>2
>3
|a'2,y",
|а'з,у",
/ I >2
1*1
\ 1
УЗ
*2
*3
■ |a'i,a'2, a'3|,
."|= -
У1
Уз
*"\ =
У\ ζ,
>2 *2
*2
4
— *2
У1
Уз
Zl
1я' ϋ' ϋ' I
la 1, a 2, а з|,
Z3
• |a'i, a'2, a'3|.
•Zl
*3
+ *3
У\ zi
y2 *2
|)|a'i,a'2,a'3
= |ai,a2, a3| ·|χ, у, z|
and this establishes the formula for the product of two determinants. A more
symmetric formula is obtained if we replace x, y, 2 respectively by bi, b2, b3.
Thus we have T28.1.
T28.1. If ai, a2, a3, bi, b2, b3 are any six vectors, then
|ai, a2, a3|'· |bb b2, b3| =
ai-bi ai-b2 ai-b3
a2-bi a2-b2 a2-b3
a3*bi a3-b2 a3-b3

106 SOLID ANALYTIC GEOMETRY (Ch. 3
As an application of T28.1 consider an orthonormal system Vi, V2, V3.
Then
I vi, v2, v312 = I vb v2, v31' · I vb v2, v31
vrvi
v2-vi
V3-Vi
vrv2
ν2·ν2
ν3·ν2
vrv3
ν2·ν3
ν3·ν3
1 0 О
0 1 0
0 0 1
= |ub u2, u3| = 1
and |vi, v2, v3| = ±1 И0. Thus, if χ is any vector, then by T27.1 there
exist scalars y\, y2, уз such that
Moreover,
x = vi>i + v2y2 + УзУз-
x-Vi=>i, x-v2=j»2, x-v3 = y3.
That is,
T28.2. If Vi, v2, v3 is any orthonormal system and χ is any vector, then
χ = vi(x-vi) + ν2(χ·ν2) + ν3(χ·ν3).
The orthonormal systems vi, v2, v3 and Ui, u2, u3 are said to have the
same orientation if |vi, v2, v3| = |ui,u2, u3| = +1 and opposite
orientations if |vi, v2, v3| = —1. A system |vi, v2, v3| is said to be right-handed
or left-handed according as its orientation is the same or opposite to that of
ui> u2, u3· A right-handed system is described as follows: Point the thumb
of your right hand in the direction and sense of v3 and let your fingers curl.
A right-handed system is such that your fingers indicate a rotation about v3
from vi toward v2. This result cannot be established mathematically. Try it
on the system Ui, u2, u3.
We have defined a determinant in such a manner that the value assigned
to it depends on the choice of orthonormal system in that P25.1 states that
|ui, u2, u3 I = +1. However, we have seen that |vi,v2, v3| is also +1 if
Vi, v2, v3 is a right-handed orthonormal system. By the same reasoning as
that used in Section 25, we can obtain an expression for the value of a
determinant I a, b, с | in terms of the components of a, b, с with respect to the
system Vi, v2, v3. This value must be that implied by P25.1 to P25.4 and hence
equal to that given in terms of the components of a, b, с with respect to
Ui, u2, u3. Thus, if we replace P25.1 by the condition | vi, v2, v3| = +1, we
shall not thereby alter the value assigned to the determinant | a, b, с |. That
is, the value of a determinant depends only on the orientation of the ortho-
normal system in terms of which it is defined, and two right-handed systems
assign the same value.

Sec. 28] THE PRODUCT OF TWO DETERMINANTS 107
Questions
1. Show that if χ = [*i, дг2, дгз], у = [yi, уг, уз], ζ = [ζχ, ζ2, ζ3] and | аь а2, а31'
|a'i, a'2, a'3|, then
[&ι·χ,Λ2·χ, a3-x] = a'i*i + a'« + а'з*з = ж"
[агу, а2-у, а3-у] = a'i>i + a'2>>2+ а'з>>з = у"
[arz, a2-z, a3-z] = a'izi + a'2z2 + а'з*з = z"
and
|ж",у",ж"|
|аь a2, a3| -|x, y, z|
2. Discuss the orientation of orthonormal systems.
Problems
1. Show that
2. Given
|a,b,c|2 =
a-a a-b a-c
b-a b-b b-c
c-a c-b c-c
xa-a + ia-b = a-c
sb-a + ib-b = b-c
sc-a + ic-b + r = c-c
These are three equations in the unknowns s, t, r. Solve for r by determinants
(assuming the denominator is not zero), expand the denominator determinant
in minors of the last column, and show that the answer can be expressed in
the form
|a, b, c|2
I a-a a-b I
I b-a b-b I
3. Given χ = [*ι, Χ2, дез], b = [bi, b2, b3], |ab a2, a3|' = |a'i, a'2, a'3|, ai*i +
аг*2 + аз*з = b. Show that
a'i-ж = b\, a'2-x = Ьг, aV* = Ьз.
4. Given that |a, b, c| = 5. Show that
13a - 2b - 3c, 5a + 4b + 7c, 2b| = -360.
5. Given that v, = [\i, %, %\, v2 = \%, Ц, -%]. v3 = {-%, %, -Ц] is an
orthonormal system. Show that it is right-handed.

108 SOLID ANALYTIC GEOMETRY (Ch. 3
29. DISTANCES FROM POINTS TO LINES AND POINTS TO PLANES
As in the two-dimensional case the distance from a point X to the line
containing the vector a with origin С is equal to the distance | g | from Υ to X
where g = YX and Υ is that point of the line which makes | g | a minimum.
Since Υ is on the line, there exists a scalar r such that
g = CX - CY = CX - at = Ь - ar
where b = CX. To make | g | a minimum, we must have
ba
g-a = b-a — ra-a = 0, r =
8 aa
The square of the distance | g | is then
g.g = g.(b - ar) = g-b - 0
(a-b)2
= b-b — ra-b = b-b
a-a
The vectors a, b form two edges of a parallelogram whose base and altitude
are formed by the vectors a, g. The area of the parallelogram is |a| · |g|
and the square of the area is
(a-a)(g-g) = (a-a)(b-b) - (a-b)2 = *"* *"
ba b-b
Since (a-a)(g-g) cannot be negative, we have
(a-a)(b-b) - (a-b)2 = |a|2|b|2 - |a-b|2 ^ 0,
(where |a-b| denotes the absolute value of a-b) and hence
|а|2|Ь|2^ |а-Ь|2.
By extracting the positive square roots of both members of this inequality,
we obtain
|а|-|Ь| Ш |a-b|
and this is called the Schwarz inequality.
We shall use the Schwarz inequality to obtain an inequality relating the
sides of a triangle А, В, С Let
ВС = a, CA = b, BA = ВС + CA = a + b = с

Sec. 29] POINTS TO LINES, POINTS TO PLANES 109
Then the sides have the lengths | a |, | b |, | с | and
|c|2 = (a + b)-(a + b) = aa + 2ab + bb
= |a|2 + 2a-b + |b|2
(|a| + |b|)2= |a|2 + 2|a|-|b| + |b|2.
Let us compare the middle terms 2a · b, 21 a | · | b | of the right-hand members
of these equations. Since a-b < |a-b| when a-b is negative, and a-b =
| a · b | when a · b is positive or zero, we have a-b ^ | a · b | in all cases and
in turn |a.·Ь| ^ |a| - |b| by the Schwarz inequality. Thus
a-b й a-b й |а| -|b|.
From this it follows that
|c|2±S(|a| + |b|)2,
and by extracting the positive square roots of both members of the latter
inequality, we obtain
|c| й |a| + |b|.
That is, the length of a side of a triangle is less than or equal to the sum of the
lengths of the remaining two sides. This is called the triangle inequality.
The distance from a point X to the plane containing the linearly
independent vectors a, b with common origin D is defined to be the distance
| h | from Υ to X, where h = YX and Υ is that point of the plane which makes
|h| a minimum. Since Υ is on the plane, there exist scalars s, t such that
DY = as + bt,
h = DX - DY = с - as - bt,
where с = DX. We suspect that h is a minimum if s, t are chosen so that
h-a = 0 = h-b. Let us assume for the moment that s, t can be so chosen,
and let us show that the corresponding | h | is a minimum. Let Y' be any
other point of the plane and let h' = Y'X. Then there exist s', t' such that
h' = с - a/ - bt' = с - ai - Ы + a(i - s') + b(/ - /')
= h + e
where e = a(j - Л + b(/ — /')· Then
|h'|2 = (h + e)-(h + e) = h-h + 2h-e + e-e
= |h|2+ lei2,

110 SOLID ANALYTIC GEOMETRY (Ch. 3
since
he = h-a(j - /) + h-b(/ - /') = 0.
If |e|2 = 0, then
e = a(j- s') + b(/- /') = 0
and since a, b are linearly independent, this implies s — s' = 0, t — t' = 0;
i.e., s' = s, t' = t, Y' = Y. Thus |h| is a minimum when h-a = 0 = h-b,
since for any other point Y' of the plane, we have | h'|2 > |h|2 and hence
|h'| > |h|.
We wish to find s, t, h | such that
a-h = a-c — sa-a — /a-b = 0,
b-h = b-c — Л-а — /b-b = 0,
|h|2 = h-h = h(c - aj - Ы) = h-c - 0 - 0
= (c — as — b/)-c = c-c — sa-c — /b-c.
These equations can be rearranged as follows:
ia-a + /a-b = a-c
Л-а + /b-b = b-c
я;-а + /c-b + |h|2 = c-c.
Thus we have three linear equations in the unknowns s, t, |h|2. By T27.1
these equations have a unique solution, and the value of |h|2 is
a-a a-b a-c
b-a bb b-c
c-a c-b c-c
Ihl2 =
a-a a-b 0
b-a bb 0
c-a c-b 1
if the denominator determinant in this expression for | h |2 is not zero. We
evaluate the denominator by expanding in minors of the third column. Thus
a-a a-b 0
b-a b b 0
c-a c-b 1
0 - 0 +
a-a a-b
b-a bb
(a-a)(g-g),

Sec.29] POINTS TO LINES, POINTS TO PLANES 111
where a, g form a base and altitude of the parallelogram, two of whose
sides are a, b. The denominator cannot be zero unless
a = 0 or
g
ar = 0.
Since a, b are given linearly independent and each of these latter equations
implies that a, b are linearly dependent, the denominator cannot be zero.
We conclude that s, t can be chosen so that a-h = 0 = b-h. Moreover, the
numerator in the expression for | h |
hence
is equal to |a, b, c|2 by T28.1, and
Ihl2 =
and
|a|2|g|2
|h|= ±
|a,b,
Igl
The vectors a, b, с form three edges of a parallelepiped (see Fig. 29.1)
whose base is a parallelogram, two of whose edges are formed by a, b, and
whose altitude is formed by h. We
assume without proof that the
volume is the area of the base a | · | g |
times the length of the altitude | h |.
Thus the volume of the
parallelepiped is
|a|-|g|-|h| = ±|a,b,c|.
We also assume without proof that
the tetrahedron, three of whose edges Figure 29.1
are formed by a, b, c, has one-sixth
the volume of the parallelepiped. Thus the volume of the tetrahedron is
±|a, b,c|/6.
Questions
1. Show that if a, b form two edges of a parallelogram, then the square of the area
of the parallelogram is
I a-a a-b I
I b-a bb I
2. Prove the Schwarz inequality and the triangle inequality. 3. Let h = YX, b/ =
Y'X, с = DX, where Υ, Υ' are points of the plane containing the linearly independent
vectors a, b with common origin 0. Show that there exist scalars s, t, s', t' such that
h = с - as - bi, h' = h + a(s - s') + b(i - i')·

112 SOLID ANALYTIC GEOMETRY (Ch. 3
4. Show that if a-h = 0 = b-h and Y' * Y, then
|h'|2 = |h|2 + |a(x-/)+b(i-i')l2> |h|2.
5. Show that there exist s, t such that a-h = 0 = b-h and
a-a
b-a
c-a
a-a
b-a
c-a
|a, b, с
M2|g
a-b
bb
cb
a-b
bb
cb
I2
I2
a-c
be
c-c
0
0
1
where |a| · |g| is the area of the parallelogram, two of whose edges are formed by
a, b. 6. Discuss the volume of the parallelepiped, three of whose edges are formed
a, b, с
Problems
1. Find the area of a parallelogram, two of whose edges are formed by the vectors
a = [1,2,2], b = [2, 1, -2].
2. Let a = [2, 4, 4], b = [3, 3, 0], с = [1, 5, -1], g=b-ar, h = с - as -
bi, where r, s, t are such that a-g = 0, a-h = 0 = b-h. Find r, s, t, g, h,
| g |, | h |. Check your results by the formulas
l2|g|2 =
a-a a-b
b-a bb
|h|= ±
|a,b,c|
|a|-|g|
3. Let g = [3, -12, 15] - [-5, 20, -25]r.
(a) Find r such that g-g is a minimum, and find the corresponding value of g.
(b) Find x, у not both zero such that
[3, -12, 15]* + [-5, 20, -2S]y = 0.
Hint: Set χ = — 1 and use the result of (a).
4. Let h = [5, -9, 14] - [3, -1, 2}s - [2, 3, -4]i.
(a) Find s, t such that h- h is a minimum, and find the corresponding value of h.
(b) Find x, y, ζ not all zero such that
[5, -9, 14]* + [3, -1, 2]у + [2, 3, -4]z
Hint: Let χ = —\ and use the result of (a).
0.

Sec. 30]
VECTOR PRODUCTS
113
5. Find x, y, z, t not all zero such that
[5, -9, 14]* + [3, -1, 2\y + [2, 3, -4]z + [7, -5, 6]i = 0.
Hint: Let t = 0 and use the result of Problem 4.
6. Let a, b, с be such that а·с = 0 = b-c and
Show that
|a,b,c|2 =
a-a a-b
b-a b-b
a-a a-b 0
b-a b-b 0
0 0 c-c
= (c-c)2.
7. By means of the Schwarz inequality show that
|Ь|2^ 1а + Ь1-1а-Ь|
for any two vectors a, b.
30. VECTOR PRODUCTS
If we expand the determinant | a, b, с | in signed minors of the third
column and if с = [c\, c2, £3], we obtain
| а, Ь, с | = | a, b, U! | C! + | a, b, u21 c2 + | a, b, u31 c3.
This sum of three products can be regarded as the scalar product of two
vectors, namely, of the vectors
ν = | a, b, ui | ui + | a, b, u21 u2 + | a, b, u31 u3,
С = C!U! + C2U2 + C3U3.
That is,
|a,b, c| = vc.
In particular, if с = a, we have
I a, b, a I = 0 = va,
and hence ν _L a. Similarly, ν _L b. We now have a method of finding a
vector ν perpendicular to two given vectors a, b. This vector is called the
vector product (or cross-product) of a, b and is denoted by ν = a X b. We shall
see that this product has properties similar to those of the product of two
numbers. The above formula, which gives the components of v, is analogous
to the expansion of a determinant in minors of the third column. To
remember this formula, we write down an array having the form of a determinant

114 SOLID ANALYTIC GEOMETRY (Ch. 3
and expand it as though it were a determinant. That is, if a = [a1; a2, o3],
Ь = [bu b2, b3], then
a χ b = ν =
01 *1 Ul
02 b2 U2
03 *з u3
= I a, b, ui IU! + I a, b, u21 u2 + I a, b, u31 u3
a2 b2 ax bx
Ui —
оз *з o3 b3
EXAMPLE 30.1
If a = Ui + 2u2 + 2из, b = щ — u2 — 4из, then
a X b =
u2 +
01 *1
02 b2
U3.
1
2
2
2
2
1
-1
-4
-1
-4
Ul
u2
U3
Ul -
1 1
2 -4
u2 +
1
-1
из
= — 6ui + 6u2 — 3u3.
We can obtain an additional property of the vector product by
substituting с = ν in the equation |a, b, c| = ν-с and then squaring the determinant
by means of T28.1. Thus, since va = 0 = vb, we have
I a, b, v| = vv = |a X b|2,
|a,b,v|2=(vv)2
a-a a-b a-v
b*a b'b b*v
va vb vv
We expand the determinant on the right in minors of the third column. Thus
a-a a-b 0
(vv)2 = b-a b-b 0 = 0 - 0
-ΙΟ 0 vv
a-a a-b 0
b-a ЬЬ О
0 0 vv
a-a a-b
b-a b-b
(vv)

Sec. 30]
VECTOR PRODUCTS
115
If ν И 0, we can cancel ν · ν from the extreme members of this equation
and obtain
a-a a-b
vv =|a X Ы =
ba bb
where the determinant on the right has been interpreted as the square of the
area of the parallelogram, two of whose sides are formed by the vectors
a, b. Hence
la X bl
a-a a-b
ba bb
is interpreted as the area of this parallelogram.
Let us show that this formula for the magnitude | a X b | of a X b holds
even when
a X b = | a, b, ui | ui + | a, b, u21 u2 + | a, b, u31 u3 = 0;
i.e., when
| a, b, ui | = 0 = a2b3 — a3b2,
|a, b, u2| = 0 = — aib3 + a3bu
| a, b, u3| = 0 = aib2 — α2*ι·
If one of the components of a (say, a3) is different from zero, we can solve the
first of these equations for b2 and the second for ^ and obtain
h Ьз
b2 = — a2,
b = —a =
оз
и Ьз
bi = —βι,
оз
га
оз
where r = —. Hence
оз
a-a a-b
ba bb
and
a-a r(a-a)
r(a-a) r2(a-a)
ι a-a
r(a-a)
r(a-a)
= 0
la X bl=
a-a a-b
ba bb
since both members of this equation are zero, and similarly, when one of
the other components of a is different from zero. The equation holds also
when all components of a are zero, since again both members are zero.

116 SOLID ANALYTIC GEOMETRY (Ch. 3
The vector product ν = a X b depends on the vectors a, b, but does it
also depend on the orthonormal system щ, u2, U3? Our definition requires
that it be such that
vc = I a, b, с I
for every vector c, and we have seen that a determinant is assigned the same
value by all right-handed orthonormal systems. Hence, if some right-handed
orthonormal system assigns the value w to this vector product, the w must
be such that
wc = |a, b, c| = vc
for every vector с In particular
W-Ui = VUi, WU2 = VU2, W'U3 = VU3.
It follows that
w = [w-u1; wu2, w-u3] = [vui, vu2, vu3] = v.
That is, the vector product is assigned the same value by every right-handed
orthonormal system.
The product
(a X b)-c = vc = I a, b, c|
is called the scalar triple product of a, b, с It is customary to omit the
parentheses and write a X b-c in place of (a X b)-c and also с-a X b in place of
с-(a X b). The symbol a X (b-c) is never used and is not even assigned a
meaning.
Questions
1. Define the vector product. 2. Show how it can be evaluated as a symbolic
expansion of a determinant. 3. Show that a X b-c = |a, b, c|. 4. Show that
a X b _L a, a X b _L b. 5. Show that if с = a X b, then
(c-c)2 = |a,b,c|2= I *"a l'l l(c-c),
I b-a b'b
and that if с ^ 0, then
'ахьНь."а ba:bbH|aHsi>2.
where |a| · |g is the area of the parallelogram, two of whose edges are formed by
a, b. 6. Show that |aXb|=\/
» b-a b-b
even when a X b

Sec. 31] PROPERTIES OF VECTOR PRODUCTS 117
Problems
1. Show that if a-b = 0, then |a X b|2 = |a|2-|b|2·
2. Let a, b be linearly independent; let h = b — ar, where r is such that h-a = 0.
Show that a, b. are nonzero vectors. Let vi = a/1 a |, V2 = h/1 h |, V3 = vi X V2.
Compute vrv2, |vi|, | V21, | V31. Show that vi, V2, V3 is an orthonormal
system.
3. Show that the orthonormal system of Problem 2 is right-handed; i.e.,
Vl X V2"V3 = 1.
4. Let vi be a unit vector such that vi, ui are linearly independent, and let
V2 = vi X ui/1 vi X ui I, V3 = vi X V2. Show that vi, V2, V3 is a right-handed
orthonormal system.
5. Let vi be a unit vector such that vi, ui are linearly dependent. Show that
vi, U2 are linearly independent. Find V2, V3 such that vi, V2, V3 is a right-handed
orthonormal system.
6. Let vi, V2, V3 be an orthonormal system and w be any vector. Show that there
exist x, y, ζ such that w = x\\ + yv2 + ZV3. Show that if | w | = 1, w _L V2,
w _L V3, then у = ζ = 0 and χ = ±1.
7. (a) Show that b X a = [|b, a, ui|, |b, a, u2|, |b, a, u3|] = —a X b.
(b) Show that a X a = 0 for every vector a.
8. Show that a X (b + c) = [ I a, b + c, m I, | a, b + c, u21, | a, b + с, из | ]
= aXb+aXc,
and that (a + b)Xc = aXc + bXc.
9. Show that a X (уЪ) = y(a X b) and (*a) X b = *(a X b).
10. Let h = b — ar, where r is such that a-h = 0. Show that a X b = a X h,
alaXb,hlaXband|aXb| = | a | · | h |. See Problem 1.
11. Let vi, V2, V3 be defined as in Problems 2, 3.
(a) Show that V2 X V3*vi = |v2, V3, vi| = νι·νι, V2 X V3*V2 = νι·ν2>
V2 X V3-V3 = vi*V3, and hence that V2 X vj = vi.
(b) Show that vj X vi = V2.
(c) Compute V3 X V2, vi X V3 and V2 X vi.
31. PROPERTIES OF VECTOR PRODUCTS
We shall develop some of the algebraic properties of vector products.
T31.1. Ь X a = -a X b.
Proof:
b X a = I b, a, ui |ui + I b, a, u2 |u2 + | b, a,u3 |u3
= — I a, b, ui |ui — I a, b, u2 |u2 — | a, b, u3 |u3 = —a X b.

118
SOLID ANALYTIC GEOMETRY
(Ch.3
T31.2. a X a = 0 for every vector a.
Proof: It follows from T31.1 that a X a = -a X a and hence that a X a = 0.
T31.3. aX(b + c)=aXb+aXc.
Proof:
a X (b + c) = | а, Ь + c, u! |u! + | а, Ь + c, u2 |u2 + | а, Ь + c, u3 |u3
= | a, b, ui |ui + | a, b, u2 |u2 + | a, b, из |из
+ I а, с, ui |ui + | a, c, u2 |u2 + | a, c, u3 |u3
= aXb + aXc.
T31.4. (a + b)Xc = aXc + bXc.
Proof: By T31.1 and T31.3 we have
(a + b) X с = -с X (a + Ь)
= -сХа-сХЬ = аХс + ЬХс.
T31.5. (*a) X b = x(a X b).
Proof:
(xa) X b = \xa, b, ujui + |xa, b, u2|u2 + |xa, b, из|из
= x\a, b, ui |ui + x\a, b, u2|u2 + x\a, b, u3|u3 = x(a X b).
T31.6. a X (yb) = y(a X b).
Proof:
a X (yb) = - (yb) X a = -у(Ъ X a) = y(a X Ь).
Consider certain special vector products involving the vectors ui, u2) из.
Thus
1 0 u,
ui X u2 =
= 0 - 0 +
1 0
0 1
из = из.
0 1 u2
0 0 u3
Of course by T31.2 we have щ X ui = 0. Similarly, we obtain the following
multiplication table:
ui X ui = u2 X u2 = из X из = 0,
ui X u2 = из, и2 X ui = —u3,
u2 X u3 = ui, и3 X и2 = — ui,
U3 X U[ = U2, Ui X U3 = — U2.

Sec.31] PROPERTIES OF VECTOR PRODUCTS 119
This table can be remembered as follows: Any three consecutive numbers in
the sequence 1, 2, 3, 1, 2 are said to be in "cyclic order." If three numbers
are in cyclic order and the first two are subscripts of two of the above vectors,
forming a product, then the third number is the subscript of the vector that
is equal to the product. The reversal of order of multiplication reverses the
sign of the products by T31.1. These considerations make it easy to remember
the above table.
The concept of linear dependence is extended as follows:
D31.1. Vectors vi, · · · v„ are said to be linearly dependent if there exist scalars
*i, '" ' *n not all zero such that
vi*i -\ v„*„ = 0.
Otherwise they are linearly independent.
T31.7. Two vectors a, b in three-dimensional space are linearly dependent if and
only if a X Ь = 0.
Proof: We have seen that |aXb| = | a | · | g | is the area of the
parallelogram, two of whose edges are formed by a, b, and that this area is zero if
and only if a, b are linearly dependent.
T31.8. Three vectors a, b, с in three-dimensional space are linearly dependent if
and only if | a, b, с | =0.
Proof: The volume of the parallelepiped, three of whose edges are formed by
a, b, c, is
|a| · |g| · |h| = ± |a, b, c|.
Hence | a, b, с | = 0 if and only if at least one of the equations
h = с - as - b/ = 0,
g = b - ar + cO = 0,
a = a + b0 + cO = 0,
holds. Since each of these equations expresses the fact that a, b, с are linearly
dependent, it follows that they are linearly dependent if and only if
|a, b, c| = 0.
T31.9. Four vectors a, b, c, d in three-dimensional space are always linearly
dependent.
Proof: If | a, b, с | =0, then by T31.8 there exist x, y, ζ not all zero such that
ax + by + cz + dO = 0.

120 SOLID ANALYTIC GEOMETRY (Ch. 3
If I a, b, с | И 0, then by T27.1 there exist x, y, ζ such that
ax + by + cz + d(-l) = 0.
In either case a, b, c, d are linearly dependent.
As in two-dimensional space, two vectors with a common origin are
linearly dependent if and only if there is a line containing both of them.
Consider three linearly dependent vectors a, b, с with a common origin and
suppose that a, b are linearly independent. Then there exist x, y, ζ not all
zero such that ax + by + cz = 0. In particular ζ И 0, since otherwise we
would have ax + by = 0 with x,у not both zero; i.e., a, b linearly dependent.
Hence
с = (т)а + (т)ь = 'а + /ь
where s = —x/z, t = —y/z. Thus с lies in the plane of a, b. If no two of
the three vectors are linearly independent, then there is a line containing all
three.
Questions
1. Prove that aXb=-bXa, aXa = 0,aX(b + c)=aXb + aXc,
(a + b)Xc = aXc + bXc, (*a) Xb=»(aXb), a X (уЪ) = y(a X b). 2.
Write down the multiplication table for ui, иг, из, and show how it is obtained.
3. Define the linear dependence of η vectors. 4. State and prove the condition for the
linear dependence of two vectors. 5. State and prove the condition for the linear
dependence of three vectors. 6. Show that four vectors in three-dimensional space are
always linearly dependent. 7. Discuss the statement that three vectors with a common
origin are linearly dependent if and only if there is a plane containing them.
Problems
1. Let a = [3, 2, -1], b = [1, -3, 5], с = [4, -1, 1]. Find a X b and a X b-c,
and check that a X b*c = |a, b, c|.
2. Find the volume of the parallelepiped, three of whose edges are formed by the
vectors a, b, c, of Problem 1. Find the area of the base containing a, b, and the
length of the corresponding altitude.
3. Let a = [2, -14, 8], b = [-3, 21, -12], с = [0, 1, -2], d = [1, -1,1].
(a) Compute a X b. (b) Find x, у not both zero such that ax + by = 0.
(c) Find дг, y, ζ not all zero such that ax + Ъу + cz = 0. (d) Find x, y, z, t not
all zero such that ax + Ъу + cz + di = 0.

Sec.31] PROPERTIES OF VECTOR PRODUCTS 121
4. Let a, b, с be defined as in Problem 1 and let d = [1, —1,1]. Find x, y, z, t not
all zero such that ax + by + cz + di = 0.
5. Let a = [3, 2, -1], b = [1, -3, 5], с = [5, -4,9]. Compute |a, b, c|. Find
x, y, ζ not all zero such that лх -\- by -\- cz = 0. Hint: The vector equation is
equivalent to
3* -\-у = — 5z, 2x — Ъу = Az, —x + by = —9z.
Solve two of these equations for x, у in terms of ζ and show that these values
satisfy the third equation.
6. Find x, y, ζ satisfying the equation лх + by + cz = 0 of Problem 5 and the
additional condition x2 + y2 + z2 = 1.
7. Let vi = cui + Л12, V2 = —sui + сиг, where c2 + s2 = 1. Show that vi*V2 =
0, vi X V2 = из, and that vi, V2> из is a right-handed orthonormal system.
8. Show that if a X b ^ 0, then |a, b, a X b| ^0 and that a, b, a X b are
linearly independent.
9. Let a = [a2, 0], a* = [ — a2, au 0]. Show that
a* = u3 X a.
10. Let a, b, vi, V2, V3 be defined as in Problems 2, 3, 11 of Section 30 and let с be
any vector.
(a) Show that
a = aivi, b = iivi + *2V2,
where a\ = \ a |, bi = r \ a |, bz = \ b. |. Show that there exist scalars
си с2, сз such that
С = ClVl + C2V2 + C3V3.
(b) Compute a X b and (a X b) X с and show that
(a X b) X с = aibifiV2 — а\Ь#г\\.
(c) Compute (a*c)b and (b'c)a and show that (a'c)b — (b«c)a = aicii2V2 —
aitovi = (a X b) X с
11. (a) Show that χ X (у Χ ζ) = -(у X ζ) Χ χ = (i-z)y - (i-y)z. You can
use the result of Problem 10.
(b) Show that a X (b X c) = (a'c)b — (a-b)c. You can use the result of
part (a). The products (a X b) X c, a X (b X c) are called vector triple
products. These two products are not equal, and hence the vector product is
not associative.

122 SOLID ANALYTIC GEOMETRY (Ch. 3
32. LINEAR EQUATIONS
We have seen that the square of the distance from a point X to the plane
containing the linearly independent vectors a, b, with common origin D, is
a
•a b-a c·
a
a«b b-b c-b
a«c b-c cc
a-a b*a
a-b b*b
|a, Ь, с |2
|aXb|2
= |h|
where с = DX and where the denominator is not zero, since a, b are linearly
independent. The distance from X to the plane is thus
la, b, cl a X b-c
I hi = ± = ±
|a X b| |a X b|
This distance is zero if and only if
h = с - aj - b/ = DX - as - b/ = 0,
and hence if and only if X lies on the plane. Let a X b = η = [n\, η-ι, яз].
Then
|a X b| = |n| = Vni2 + n22 +
"3
a X b-c = n-DX = n-OX - n-OD
= ЩХ1 + П2Х2 + "3*3 + "4,
where
nA = -Ώ.-ΟΏ.
Hence the distance from X to the plane is
n-c nixi + n2x2 + "3*3 + "4
|h| = ± = ±
I n |
V„,2 + „22 + „32
The point X lies on the plane if and only if | h I =0, and this is the case if
and only if the numerator is zero; i.e.,
"1*1 + "2*2 + "3*3 + "4=0.

Sec. 32]
LINEAR EQUATIONS
123
This equation is called an equation of the plane. This equation implies that
n-c = n-DX = 0;
i.e., that η is perpendicular to any vector that connects D to some point X
of the plane. The vector η is called a perpendicular, or normal to the plane.
An equation of the form
il*l + i2*2 + '3*3 + '4 = 0,
in which f1( e2, e3 are not all zero, is of first degree in χλ, x2, x3, and such an
equation is said to be linear. We shall show that every linear equation in
*i> *2> *з represents a plane (i.e., is an equation of a plane). We select three
points A, B, D, which have coordinates satisfying the linear equation and
for which DA and DB are linearly independent. We then show that the
linear equation represents the plane of the vectors DA, DB. Consider the case
e3 И 0. If x\ = x2 = 0, then x3 = —e^/e^ and the coordinates of D =
(0,0, — *4Аз) satisfy the equation. If^ = l,x2 = 0, then *з = —(е\+ e^/e^,
and the coordinates of A = (1, 0, — (fi + e^/e^) satisfy the equation.
Similarly the coordinates of В = (0, 1, — (e2 + й)/*з) satisfy the equation.
ОЛ- Γΐ,Ο, — pi.
Бг-[м.-5].
DX = xu x2, *3 + - -
DA X DB =
|n|
n-DX
1 0 ui
0 1 u2
— il/'3 —'2 Аз "3
uifi u2e2 ui
+ + u3 -
'3 '3
Vei2 + e22 + i32
1*з|
elxl + e2x2 + e3x3 -
?i + u2e2 + u3i8
<?3
Ь ί4
*3

124 SOLID ANALYTIC GEOMETRY (Ch. 3
The distance from X to the plane of DA, DB is therefore
n-DX Qfi*i + e2x2 + f3*3 + ^)Аз
|n| V^2 + ί22 + ί32/|ί3|
l?l*l + f2*2 + f3 + '4
= ± V,,2 + ,22 + i32
and the equation of the plane is
i!*! + i2X2 + '3*3 + *4 = 0.
The cases fi И 0, f2 И 0 are similarly treated. Thus we have T32.1.
T32.1. Every linear equation
il*l + i2*2 + '3*3 + i4 = 0
represents a plane. If (χχ, x2, *з) is a point not necessarily on the plane, then its
distance from the plane is
e-ίΧ-ί + e2x2 + *з*з + e*
± V(l* + e2* + i32
Note that an equation such as Ъх\ + 4x2 — 7 = 0, in which the coefficient
of *з is zero, is linear in xlt x2, x$ and hence represents a plane (not a line) in
solid analytic geometry. If x1( x2 are any two numbers satisfying the equation,
then the coordinates of X = (χχ, χ2, xz) satisfy, and X is a point of the plane
no matter what value is assigned to x$. For example (1, 1, 0), (1, 1, 9) are
points of the given plane.
Questions
1. Show that the distance from the point X = (*i, *2, *з) to the plane containing
the linearly independent vectors a, b, with common origin D, is
II, I — nvDZ _ «1*1 + «2*2 + «3*3 + «1
1 l_± |n| _± Vni* + nS + «з2 '
where a X b = η = [«ι, «2, «з] and in = —n-OD. 2. Show that X lies on the plane
if and only if
«1*1 + «2*2 + «3*3 + «4 = 0.
3. Show that if сз y* 0 and D, А, В are respectively the points (0, 0, — C4A3),
(1, 0, — («i + ei)/ei), (0, 1, — (c2 + ei)/ez), then the distance from (*ь x2, *з) to the

Sec. 32]
LINEAR EQUATIONS
125
plane containing the vectors DA, DB is
ei*i + егхг + ез*з + «
vV + e22 + e32
and that the equation of the plane is
«1*1 + «2*2 + «3*3 + et = 0.
Problems
1. Given the plane x\ — 2*2 + 2*з — 14 = 0,
(a) Find three points D, А, В of the plane such that η = DA X Di? ^ 0.
(b) Show that if η is defined as in (a) and X = (*i, *2, *з), then
n-DX _ χι - 2*2 + 2*з ~ Η
|n| _± 3
(c) Find the distance from (3, 1,2) to the plane.
(d) Show that if m = AD X AB, where D, A, B are defined as in (a), then
ια·ΑΧ _ *i — 2*2 + 2*з — 14
Ί^Γ = ± 3
2. Find the distance from (3, 1,2) to the plane
2*! + 3*з - 7 = 0.
3. Points Υ = (уьУ2,уз), Ζ = (zi, z2, z3), lie on the plane «i*i + «2*2 + «3*3 +
m = 0. Express this fact by equations, subtract one equation from the Other,
and show that YZ _L n, where η = [щ, щ, из].
4. Given that D = (d\, «/2, 03) lies On the plane a\X\ + 02*2 + 03*3 + "i = 0,
show that X = (*!, *2, *з) lies on this plane if and only if DX· a = 0, where
a = [αϊ, α2, α3].
5. Given that D lies On the intersection of the planes a\X\ + 02*2 + 03*3 + 04 =
0, Mi + *2*2 + *з*з + bi = 0, show that if DX = (a X b)i, where a =
[αϊ, 02, a3], b = [ii, *2, *з], and a X b 7s 0, then Ζ is also a point of the
intersection.
6. Show that three planes αι*ι + 02*2 + Яз*з + "t = 0, *i*i + *2*2 + *з*з +
bt = 0, ci*i + f2*2 + сз*з + Ci = 0, with normals a = [αϊ, аг, аз], b =
[61, *2, Ьз], с = [с], С2, сз], intersect in one and only one point if |a, b, c| r* 0.

126 SOLID ANALYTIC GEOMETRY (Ch. 3
33. THE INTERSECTION OF TWO PLANES
We shall show that if two distinct planes
<?!*! + a2x2 + а3хз + Д4 = 0, fti*i + b2x2 + b3x3 + ft4 = 0,
have a point D = (rf1( d2, d3) in common, then their intersection is a line.
The normals to these planes are respectively a = [a1( a2, a3], b = [b\, b2, b3].
If X = (χι, x2, x3) is a point of the first plane, then (since D is also a point of
the plane)
oi*i + 02*2 + «3*3 + 04 = 0, αλά\ + a2d2 + аз<4 + &ι = 0.
Subtracting the second equation from the first, we obtain
ai(*i — di) = a2(x2 — d2) + a3(x3 — d3) = л-DX = 0.
Similarly, X is a point of the second plane if and only if b'DX = 0. That is,
DX has a direction such that DX J_ a, DX J_ b. Let us consider the case
a X b И 0. Then a X Ы a, a X Ы b, and this suggests that the
intersection of the planes is probably the line through D with direction a X b.
If X is a point of this line, then DX has the form (a X b)/, and hence
л-DX = a-a X Ы = 0, Ъ-DX = 0.
That is, every point of the line is a point of the intersection, and it remains to
show that every point of the intersection is a point of the line. It follows from
T27.1 that there exist scalars r, s, t such that for any X,
DX = яг + bs + a X Ы,
since
|a, b, a X b| = a X b-a X Ь И 0.
We wish to show that r = s = 0 and therefore DX = a X b/ when X is a
point of the intersection of the planes; i.e., when я-DX = Ъ-DX = 0. The
above vector equation implies the scalar equations
л-DX = 0 = a-ar + a-bi,
Ъ-DX = 0 = b-ar + Ъ-bs,

Sec.33] THE INTERSECTION OF TWO PLANES 127
and by T15.1 these equations have a unique solution, namely, the obvious
one r = s = 0, since
a*a a-b
ba bb
= la X Ь|2 И 0.
Therefore the intersection of the planes is the line through D with direction
a X b.
Consider next the case a X b = 0. Then a, b are linearly dependent,
and since neither vector is zero, there exists a scalar к distinct from zero such
that b = ka, or equivalently,
*i = kau b2 = ka2, b3 = ka3.
It follows that if X is a point of the first plane, then
a\X\ + a2x2 + azxz + a4 = 0,
biXi + b2x2 + bzxz + b4 = k(aiXi + a2x2 + a3x3) + b4
= k(aiXi + a2x2 + a3x3 + 04) + b4 — ka^
= b4 — ka4.
Hence X is a point of the second plane if and only if b4 — ka4 = 0. Thus,
when b4 — ka4 = 0, every point of the first plane is also a point of the second,
and similarly (since t^0), every point of the second plane is also a point of
the first; i.e., the planes are identical. When b\ — ka4 И 0, no point X of the
first plane is also a point of the second, and hence the intersection of the two
planes is the empty set. In this case the planes are said to be parallel. It
follows that the two planes are parallel or identical if and only if their
normals have the same direction; i.e., if and only if a X b = 0. It also
follows that if the two planes are distinct but have a point in common, then
a X b И 0, and hence their intersection is a line.
Suppose that the two planes are such that a-b = 0. Let Χ, Υ be such
that a = DX, b = DY, where D is a point of the intersection; i.e., suppose
that the normals have the common origin at D. Then
а-/>Г = а-Ь=0 and Ъ-DX = 0,
and hence the tip of the normal to one of the plane lies in the other plane.
That is, each plane contains the normal of the other plane. In this case the
two planes are said to be perpendicular to one another. Thus two planes are
perpendicular when their normals are perpendicular.

128 SOLID ANALYTIC GEOMETRY (Ch. 3
EXAMPLE 33.1
Find parametric equations of the line of intersection of the planes
2*i - *2 + 9*3 + 4 = 0, -4*! + 3*2 - 13*з -7=0.
To locate one of the many points of the line, we consider the third plane
*з = О and find where the three planes intersect. It is easy to show that the
intersection is the point ( —%, —1,0) = D. The intersection of the two
given planes is thus the line through D with direction a X b, where a =
[2, -1,9], b = [-4,3, -13], and a X b = [-14, -10, 2]. The vector
equation DX = a X b/ is equivalent to the parametric equations
*i = —| - 14/, *2 = -1 - 10/, *3 = 2/.
As another example note that ui, u2, из are so related that each is a
normal to the plane containing the other two vectors. These three planes
are called coordinate planes. Thus the plane containing ui, u2 with common
origin О has the normal щ X иг = из, and the equation ОЛ'-из = *з = 0.
It can be interpreted as the plane of plane analytic geometry. The plane
containing из, ui has the normal u2 and equation x2 = 0. The intersection of
the two coordinate planes is the line through О with direction u2 X из = ui;
i.e., the line containing the axis щ. It is the set of points X for which there
exists a scalar r such that OX = гщ. The parametric equations are *i =
r, *2 = *з = 0. The line containing the axis иг is the intersection of the
coordinate planes *з = 0, *i = 0 and has the parametric equations χλ = 0,
*2 = s, *з = 0. The line containing the axis из is the intersection of *i = 0,
*2 = 0 and has the parametric equations *i = *2 = 0, *з = /. In drawing a
three-dimensional figure representing a plane, it is customary to draw the
lines in which that plane intersects the
coordinate planes.
EXAMPLE 33.2
Draw a figure indicating the plane
2*i + 4*2 + 3*з -12 = 0.
To draw the intersection of this plane
with *з = 0, we locate the point of
intersection of this line with *2 = 0;
i.e., the point (6, 0, 0) and the point
of intersection of the line with *i = 0,
(0,0,4)
(0,3,0)
(6,0,0)
Figure 33.1

Sec.33] THE INTERSECTION OF TWO PLANES 129
i.e., the point (0, 3, 0). We draw the segment connecting these points. The
intersection of the given plane with x2 = 0 is the line through (6, 0, 0),
(0, 0, 4), and the intersection with χλ = 0 is the line through (0, 3, 0),
(0, 0, 4). We draw the corresponding segments of these lines (see Fig. 33.1).
Questions
1. Given that two planes with normals a, b have a point D in common, show that
X lies on the intersection of these planes if and only if a·DX = 0 and Ъ-DX = 0.
2. Show that if a X b ^ 0, then X lies on the intersection of the planes if and only if
there exists a scalar t such that DX = a X bi. 3. Show that if two planes have normals
a, b, such that a X b = 0, then either they are identical or their intersection is the
empty set. 4. Discuss perpendicular planes. 5. Describe the drawing of an isometric
projection of a plane.
Problems
1. Find parametric equations of the line of intersection of the planes 3*i — 2*2 —
5*з - 1 = 0, 2*! + 7*2 + 5*з - 9 = 0.
2. Find where the line of Problem 1 intersects the plane 5*i + 5*2 — *з — 9 = 0,
by substituting in the equation of this plane the values of *i, *2, *з given by the
parametric equations, solving the resulting equation for the parameter, and
then finding the corresponding point. Check the result by finding the
intersection of the three planes.
3. It is given that the planes 2*i + *2 + *з — 3 = 0, 2*i + *2 + 4*з — 6 = 0,
2*i + *2 — 2 = 0, intersect in a line.
(a) Find parametric equations of the line.
(b) Draw an isometric projection showing the three planes and their line of
intersection.
4. (a) Find the intersection of the planes 4*i + 6*2 + 3*3 —12 = 0, 4*i +
6*2 + 3*з - 24 = 0, *i + 2*2 - 2 = 0.
(b) Draw an isometric projection showing the three planes, the intersection of
the first and third, and the intersection of the second and third.
5. Draw an isometric projection showing the planes — *i + 3*з —3 = 0, 3*i +
*з — 11 =0, their line of intersection, and their normals with common origin
at the point (3, 0, 2) on their line of intersection. How are the planes related?

130 SOLID ANALYTIC GEOMETRY (Ch. 3
34. THE INTERSECTION OF THREE PLANES
The three equations
Ol*l + 02*2 + 03*3 + 04 = 0,
bixi + b2x2 + *з*з + *4 = 0,
fi*i + c2x2 + c3x3 + c4 = 0,
represent three planes with normal vectors a = [a1( a2, a$], b = [bi, b2, b^],
с = [ci, c2, C3] if these vectors are not zero. A point X lies on all three planes
if and only if its coordinates satisfy all three equations. According to T27.1,
equations have a solution and only one solution if
Ol
a2 a3
b2 b3
c2 C3
4\ *i fi
a2 b2 c2
03 *з сз
I a, b, c| И 0.
The point X thus obtained is called the intersection of the three planes.
EXAMPLE 34.1
Find the intersection of
2*! — 3*2 + *з — 9 = 0 2xx — 3*2 + *з = 9
*i + 4*2 — 3*з + 10 = 0 or *! + 4*2 — 3*з = —10
The solution is
*! + 5*2 + 9 = 0
*! + 5*2 = —9
*1 =
*2
9 -3
-10 4
-9 5
2 -3
1 4
1 5
2 9
1 -10 -
1 -9
2 -3
1 4 -
1 5
1
-3
0
1
-3
0
1
-3
0
1
-3
0
40
40
= 1
80
40
= -2

Sec.34] THE INTERSECTION OF THREE PLANES 131
2
1
1
2
1
1
-3
4
5
-3
4
5
9
-10
-9
1
-3
0
40
1
40
*3
and (1, —2, 1) is the point of intersection of the three planes.
Consider three planes with normals a, b, с such that | a, b, с | =
a X b-c = 0. If a X b = 0, then the planes with normals a, b are either
parallel or identical. If they are parallel (but not identical), then the intersection
of the three planes is the empty set. If they are identical and this common
plane is neither parallel to nor identical with the third plane, then the
intersection of the three planes is a line. If they are identical and the common
plane is parallel to (but not identical with) the third plane, then the
intersection of the three planes is again empty. If all three planes are identical,
then the common plane is the intersection of the three planes.
Next consider the case |a, b, c| =0 but a X b И 0. By T31.8 there
exist x, y, ζ not all zero such that
ax + by + cz = 0,
and T31.7 implies that ζ cannot be zero when a X b И 0. Hence
= (— Ja + (—\ Ь = Ла + kb,
where h = —x/z, к = —у/ζ. Thus
Ci = fuii + kbi, c2 = ha2 + kb2, c3 = ha3 + kb3.
Let X be a point of the line of intersection of the planes with normals a, b.
Then _ι_ _ι_ _ι_ η
αι*ι + a2x2 + вз*з + 04 = 0,
bixi + b2x2 + b3x3 + b4 = 0.
The point X also lies on the plane with normal с if and only if
CiXi + C2X2 + C3*3 + C4 = Λ(α1*1 + fl2*2 + Я3*з)
+ k(bixi + b2x2 + *3*3) + c4
= h(alxl + a2x2 + a3x3 + a4)
+ *(*!*! + Ь2Х2 + Ь3Х3 + Ь4)
+ C4 — hat — kbi
= C4 — Л^4 — kbt = 0.

132 SOLID ANALYTIC GEOMETRY (Ch. 3
If C4 — ha\ — kbi = 0, then the intersection of the three planes is the line of
intersection of the planes with normals a, b; i.e., every point of this line is
also a point of the plane with normal с If c4 — ha4 — kb^ И 0, then no point
of the line of intersection of the two planes is also a point of the third plane,
and hence the intersection of the three planes is the empty set. We have now
shown that if three planes have normals a, b, с such that | a, b, с | =0, then
the intersection of the planes is the empty set, or a line, or an entire plane.
EXAMPLE 34.2
Find the intersection of the planes
x\ + 7*2 + 4*3 — 5 = 0,
4x, - 10л:2 + 6*3 + 4 = 0,
-6*! + 15л:2 - 9*з + 3 = 0,
with normals a = [1, 7, 4], b = [4, —10, 6], с = [ — 6, 15, —9], respectively.
Since с = ( — %)Ъ, the last two planes are either parallel or identical,
and since 3 — ( — J^)4 = 9 И 0, they are not identical. Hence the
intersection is the empty set.
EXAMPLE 34.3
Find the intersection of the planes
*i — 2*2 + 4*з + 4 = 0,
2*! + 3*2 — 5*3 —13 = 0,
7*! + 2*3 - 14 = 0,
with normals a = [1, — 2, 4], b = [2, 3, —5], с = [7, 0, 2]. Since
a X b = [-2, 13,7] И 0, a X b-c = |a, b, c| = 0,
there must exist h, к satisfying the vector equation /za + кЪ = с, and hence
the equivalent scalar equations
h + 2k = 7, - 2h + Ък = 0, 4h - 5k = 2.
When we solve the first two scalar equations, we see that they imply h = 3,
к = 2. Hence h = 3, к = 2, must be the values that satisfy the vector
equation. It can easily be checked that they do satisfy. Since —14 — Ah —
( —13)A; = —14 — 12 + 26 = 0, the third plane contains the line of
intersection of the other two. This line is the intersection of the three planes. One

Sec. 35] PROPERTIES OF INTERSECTING PLANES 133
point of the line is (2, 3, 0), and the parametric equations of the line are
xi=2-2/, *2 = 3 + 13/, *3 = 7/.
The parametric equations enable us to obtain various solutions of the three
linear equations representing the three planes.
Questions
1. Show that three planes with normals a, b, с intersect in a point if |a, b, c| r* 0.
2. Discuss the intersection of three planes such that a X b = 0. Consider all cases.
3. Show that if three planes have normals a, b, с such that a X Ь ^ 0, but | a, b, с | =
0, then the intersection of the planes is either a line or the empty set.
Problems
1. Find the intersection of the planes 3*i + 4*2 — 7 = 0, 4*i + 5*2 — 3 = 0,
7*! + 2*2 + *з - 4 = 0.
2. Find the intersection of the planes 3*i — 2*2 — 5*з — 1=0, 2*i + 7*2 +
5*з - 9 = 0, *i + *2 - 2 = 0.
3. Find the intersection of the planes 3*i — 2*2 — 5*3 — 1 = 0, 2*i + 7*2 +
5*з - 9 = 0, *i + 2*2 - 3 = 0.
4. Find the intersection of the planes 6*i — 15*2 — 12*3 + 21 =0, — 4*i +
10*2 + 8*3 - 8 = 0, *i + *2 + *з - 1 =0.
5. Find the intersection of the planes 6*i — 15*2 — 12*з + 21 =0, — 4*i +
10*2 + 8*3 - 14 = 0, *i + *2 + *з - 1 = 0.
6. Find the intersection of the planes 6*i — 15*2 — 12*3 + 21 =0, — 4*i +
10*2 + 8*3 - 14 = 0, 14*i - 21*2 - 28*3 + 1=0.
35. FURTHER PROPERTIES OF INTERSECTING PLANES
Suppose that the planes
Ol*l + ^2*2 + 03*3 + 04 = 0, *i*i + ft2*2 + *3*3 + *4 = 0,
intersect in a line and that X is a point of the line. Then the coordinates of X
satisfy the two equations and hence satisfy the equation
Λ(αι*ι + α2χ2 + 03*3 + 04) + A:(*i*i + b2x2 + b3x3 + ft4)
= (Aa, + A*,)x, + (ha2 + kb2)x2 + (haz + kb3)x3 + (ha4 + kb4) = 0.
where h, к are not both zero. This third equation represents a plane con-

134 SOLID ANALYTIC GEOMETRY (Ch. 3
taining every point X of the line of intersection of the two given planes.
The scalars h, A can be chosen so as to satisfy some additional condition.
EXAMPLE 35.1
Find an equation of the plane that passes through the intersection of
the planes
2дс, + *2 + 2*з + 5 = 0,
*i — *2 + *з — 3 = 0,
and contains the point (1, 0, —2). The desired equation has the form
A(2«j + *2 + 2*з + 5) + k(Xl - x2 + *3 - 3) = 0.
If we substitute the coordinates of (1, 0, —2) in this latter equation, we
obtain
Μ - 4k = 0,
and this additional condition is satisfied if h = 4, к = 3. Thus the desired
plane has the equation
4(2*! + x2 + 2*з + 5) + 3(*! - *2 + *3 - 3)
= 11*! +*2 + 11*3 + И = 0.
EXAMPLE 35.2
Find an equation of the plane that passes through the intersection of the
two given planes of Example 35.1 and is perpendicular to the first of these
planes. The desired equation has the form
(2Л + k)x1 + (A - k)x2 + (2A + A)*3 + (5A - ЗА) = 0.
The normal to this latter plane is [2Л + k, h — k,2h + A], and the normal to
the first of the given planes is [2, 1, 2]. The two planes will be perpendicular if
the scalar product of these normals is zero; i.e.,
(4A + 2A) + (A - A) + (4A + 2A) = 9Л + ЗА = 0.
We may thus choose h = 1, A = — 3. The desired equation is
(2*i + *2 + 2*з + 5) — 3(*! — *2 + *з — 3)
= -*! + 4*2 - *3 + 14 = 0.
We have seen how to find parametric equations of the line of intersection
of two planes. Let us see how to find planes intersecting in a line whose
parametric equations are given. There are many planes containing the given line,

Sec. 35] PROPERTIES OF INTERSECTING PLANES 135
but certain special planes can readily be found. Suppose that the parametric
equations are
*i = di + ait, x2 = d2 + (fyt, x3 = d3 + a3t.
Let us consider first the case in which none of the numbers a\, a2, a3 is zero.
We can then solve each of the parametric equations for /. Thus
_ *i — d\ _ x2 — d2 _ *з — d3
al a2 a3
By equating various members of this system of equalities, we obtain the
following linear equations:
*i — di _ x2 — d2 X! — dx _ x3 - d3
αϊ a2 ai a3
X2 — d2 _ x3 — d3
αϊ a3
If the coordinates of a point X satisfy the parametric equations, they must
also satisfy all three of the linear equations. Hence all three planes contain
the given line. Next consider the case a\ = 0, but neither a2 nor a3 is zero.
We then have the planes
x2 — d2 x3 — d3
x\ = d\, =
a2 a3
The remaining cases are similarly treated.
Questions
1. Discuss the problem of finding an equation of the plane that contains the line of
intersection of two given planes and which satisfies an additional condition. 2. Discuss
the problem of finding equations of planes containing a line whose parametric
equations are given.
Problems
1. Find an equation of the plane that passes through the intersection of x\ +
5*2 + *з — 5 = 0, x\ — 3*2 — 2*з = 0, and contains the point (1,1, 0).
2. Find an equation of the plane that passes through the intersection of the planes
of Problem 1 and is perpendicular to 2*i — 3*з + 14 = 0.
3. Find equations of three planes containing the line x\ = 1 — 2i, *2 = i, *з =
-3 + 3ί.

136 SOLID ANALYTIC GEOMETRY (Ch. 3
4. Find equations of two planes containing the line x\ = 1 — 2/, д* = 1, xj =
-3 + 3/.
5. Find an equation of the plane containing the line of Problem 3 and the point
0,0,1).
6. Find an equation of the plane containing the line of Problem 4 which is
perpendicular to a plane with normal [1, 2, 0].
36. LOCI IN THREE-DIMENSIONAL SPACE
PROBLEMS
1. A point moves so that its distance from the origin is always 5. Show
that the equation of the locus is
*i2 + *22 + *з2 = 25.
Note that the locus is a sphere with center at the origin and radius 5.
2. Find the equation of the sphere with center at (2, —1, 3) and
radius 2.
3. A point moves so that its distance from (2, —1, 3) is always equal to
its distance from (3, 1, 4). Show that the equation of the locus is a plane.
Show that the vector from (2, — 1, 3) to (3, 1, 4) is normal to the plane and
that the midpoint of this segment lies on the plane. Note that the plane is
the perpendicular bisector of the segment.
4. A point moves so that its distance from the point (3, 0, 0) is always
equal to its distance from the plane X\ + 3 = 0. Show that the equation of
the locus is „
*22 + *з2 = 12xi.
This is a surface called a paraboloid.
5. A point moves so that its distance from (4, 0, 0) plus its distance from
( — 4, 0, 0) is always 10. Show that the equation of the locus can be
converted into
V{xi _ 4)2 + χ2 + χ32 ^ 4
¥ - *i 5'
Interpret this equation as a new locus. Then show that this last equation
can be converted into
2 2 2
χι x2 *з
— + — + — = 1.
25 9 9
This surface is called an ellipsoid.

Sec.37] DRAWING THREE-DIMENSIONAL FIGURES 137
6. A point moves so that its distance from ( — 5, 0, 0) minus its distance
from (5, 0, 0) is always 6. Show that the equation of the locus can be con-
2 2 2
*1 *2 '3 _
9 16 16 ~
This surface is called a hyperboloid.
37. DRAWING THREE-DIMENSIONAL FIGURES
In three dimensions a set of points whose coordinates satisfy a quadratic
equation is called a quadric surface. We shall show how to make isometric
drawings of various quadric surfaces. This is accomplished by drawing the
curves in which the surfaces intersect appropriately chosen planes.
Figure 37.1 shows the ellipsoid
2 2 2
*1 *2 *3
— + — + — = i.
9 16 4
16 4
Figure 37.1
This surface intersects the plane *3 = 0 in the ellipse whose equations are
2 2
— + — = 1, *з = 0.
9 16
This ellipse is contained in the rectangle whose sides are the four lines
*i = 3, *3 = 0; χι = —Ъ,х3 = 0; x2 = 4, *3 = 0; x2 = —4, *3 = 0; it is
tangent to the sides of this rectangle at the midpoints. The isometric pro-

138 SOLID ANALYTIC GEOMETRY (Ch. 3
jection of this rectangle is a parallelogram. We first draw the parallelogram
and then draw the ellipse tangent to the sides at the midpoints. In a similar
manner we draw the isometric projection of the ellipses
2 2 2 2
*1 *3 , „ _, *2 *3
1 =1, x2 = 0 and 1
9 4 16 4
1, X!
0.
These ellipses are the intersections of the ellipsoid with the planes x2 = 0,
χι = 0. All construction lines are dotted.
9 + 4 9
Figure 37.2
Figure 37.2 shows the hyperboloid of one sheet whose equation is
2 2 2
*1 *2 *3 _
9 4 9'
This surface intersects the plane x\ = 0 in the hyperbola
2 2
— = 1, χι = 0.
The asymptotes of this hyperbola are the diagonals of the rectangle whose
sides are the lines x2 = 2, X\ = 0; x2 = — 2, x\ = 0; x$ = 3, xx = 0;

Sec.37] DRAWING THREE-DIMENSIONAL FIGURES 139
*з = — 3, Xi = 0; and the hyperbola is tangent to two of the sides at the
midpoints. We draw the isometric projection of the rectangle and the
diagonals, and then draw the hyperbola. In a similar manner we draw the
hyperbola
2 2
= 1» *2 = 0.
The hyperboloid intersects the plane *з = 4 in the ellipse
*!2 *22 16
— + — = 1 + — · x3 = 4,
9 4 9
and this ellipse is tangent to the sides of an appropriate rectangle at the
midpoints. Two of these midpoints lie on that one of the above hyperbolas
which is contained in the plane X\ = 0. These points are the intersections of
this hyperbola with the line X\ = 0, x$ = 4. We draw this line and locate
these two midpoints. Similarly, the remaining two midpoints are the
intersections of the remaining hyperbola with the line x-i = 0, *з = 4. We can
now draw the rectangle, since its sides are parallel to the xi-axis and дгг-axis
and we have located the midpoints of these sides. After drawing the rectangle,
we draw the ellipse. In a similar manner we draw the ellipse
V x22 16
— + — = 1 +— · *3 = -4.
9 4 9
We erase the portions of the hyperbolas that lie above the plane *з = 4 and
below the plane *з = —4, to simplify the figure.
Figure 37.3 shows the hyperboloid of two sheets whose equation is
2 2 2
X\ Xi *з _
14 4'
The two sheets refer to the two separate pieces making up this surface. Note
also the two minus signs in the equation. The hyperboloid intersects the
planes x\ = 0, *2 = 0 in the hyperbolas
2 2 2 2
x2 X3 „ , *i *з ч
1 · Xi = 0 and 1 =1, *2 = 0.
4 4 14
We draw the isometric projections of the appropriate rectangles and their
diagonals and then draw these hyperbolas. The intersections of the hyper-

140 SOLID ANALYTIC GEOMETRY (Ch. 3
1 4 4 "'
Figure 37.3
boloid with the planes *з = 4, *з = —4, are the ellipses
16 л2 *22 16
1= 1 · *з = 4 and —
4 14 4
2 2
1=T + _, ,3=_4.
These ellipses are contained in rectangles, the midpoints of whose sides are
located in the manner described above. We draw the isometric projections of
these rectangles and then draw the
ellipses.
Figure 37.4 shows the elliptic
paraboloid
2 2
«1 *2 _ *3
4 1 ~ 1
The intersection of this surface with
the plane *з = 4 is the ellipse
2 2
χι дг2
77 + — = 1, *з = 4.
16 4
We draw the isometric projection of
Figure 37.4

Sec. 37] DRAWING THREE-DIMENSIONAL FIGURES 141
the appropriate rectangle containing this ellipse and then draw the ellipse.
The intersections of the paraboloid with the planes x\ = 0, x2 = 0 are the
parabolas
*2 — *3i
x\ = 0 and
2 _
4*я
*2
0.
The first of these parabolas is tangent to the *i-axis at the origin and passes
through the midpoints of two of the sides of the rectangle containing the
ellipse. The second parabola is tangent to the xi-axis at the origin and
passes through the remaining two midpoints. We draw the parabolas so as
to satisfy these conditions.
Figure 37.5 shows the hyperbolic paraboloid
2 2
—*i *2 'з
The intersections of this surface with the planes *з =
hyperbolas
-1, *з = 1 are the
1
*2
4
= 1, *з = — 1 and
1
+
*2
=1, *з = 1,
and the intersections with x2 = 4 and x2 = — 4 are the parabolas
4 — x3 = *i2, x2 = 4 and 4 — x3 = *i2, x2 = —4.
We draw the hyperbola contained in the plane *з = — 1 after drawing the
appropriate rectangle and its diagonals. Each of the lines x2 = 4, *з = — 1,
and x2 = —4, *3 = —1, intersects
this hyperbola in two points. We
draw the lines and locate the
intersections. Each of the parabolas
passes through two of these four
intersections. One of the parabolas is
tangent to the line x2 = 4, *з = 4
at the point (0, 4, 4), and the other
is tangent to x2 = — 4, *з = 4 at
(0, —4, 4). We draw the two
tangents and then draw the parabolas
satisfying the above conditions. Each
of the lines x2 = 4, *з = 1, and x2 =
— 4, x3 = 1, intersects one of the
parabolas in two points. We draw Figure 37.5

142 SOLID ANALYTIC GEOMETRY (Ch. 3
these lines and locate the intersections. The remaining hyperbola passes
through these four intersections, and we now draw this hyperbola. The
paraboloid intersects the planes x\ = 0, x2 = 0 in the parabolas
*22 = 4*з, xi = 0 and — χι2 = x3, x2 = 0.
One of these parabolas is tangent to the xi-axis and passes through the
midpoints of two of the sides of the rectangle used in constructing the
hyperbola contained in the plane *з = — 1. The other parabola is tangent to the
*2-axis and passes through the points (0, 4, 4), (0, —4, 4). We draw these
parabolas.
Problems
Draw isometric projections depicting the following surfaces by showing the curves
in which these surfaces intersect the planes indicated.
*· ΊΓ + 77 + ¥= 1,*ι = 0,*» = 0,*, = 0.
9 16 4
2. - ^ + ^ + X-4 = 1, X2 = 0,x3 = 0, *, - 4, *, = -4.
9 9 4
3. — ; = 1, *2 = 0, x3 = 0, χι = 4, *i = —4.
4 14
. xi2 .хз2Х2 n n
4. —- + — = —, Χ2 = 4, ДГ1 = 0, x3 = 0.
4 11
5. — — = —, ДГ2 = — 1, xi = 4, *i = —4, x2 = 1, x\ = 0, x3 = 0.
4 11
, xi2 , *22 *32 . Ann
6. —- + —- = —r, x3 = 4, x3 = -4, ДП = 0, X2 = 0.
14 16
38. RULED SURFACES
Figure 38.1 shows the cone
2 2 2
X\ *2 _ '3
4 1 ~ 9
Its intersections with the planes x3 = 3, *з = — 3 are the ellipses
2 2 2 2
X\ *2 X\ X2
1 =1, *3 = 3 and 1 = 1, *3 = -3.
4 1 4 1

Sec. 38]
RULED SURFACES
143
Since *22 = *з2/9 or ±3*2 = *з when χλ = 0, the intersection of the cone
with the plane x\ = 0 consists of the lines χλ = 0, x$ = 3*2, and χι = 0,
*з = — 3x2. Similarly, the intersection with *2 = 0 consists of the lines x2 = 0,
Figure 38.1
3*1 = 2*3, and *2 = 0, 3*i = — 2*3. These curves and lines can be drawn by
the methods described above.
Figure 38.2 shows the parabolic cylinder
X2 = -2χλ.
It intersects the planes *з = 3, *з = — 3 in the parabolas
*22 = — 2*i, *з = 3 and *22 = "2*!, *3 = —3.

144
SOLID ANALYTIC GEOMETRY
(Ch. 3
It intersects the plane xi = —2 in the lines xi = — 2,
X2 = 2, and xi = —2, x2 = —2. This cylinder is a
ruled surface. A ruling is a line contained entirely in a
surface, and a ruled surface is a surface that is entirely
made up of rulings. A cylinder is a surface all of whose
rulings are parallel, and a cone is a ruled surface whose
rulings all pass through a point called the vertex of the
cone. In Problem 2 you will be asked to find parametric
equations of the rulings of the cylinder.
The vertex of the cone of Fig. 38.1 is the origin.
Parametric equations of a line through the origin have the
form
Xl = Vit, X2 = V2t, *3 = Ы
If »з = 3 and »i, v2 are such that
2 2
—+ — = 1,
4 1
then for every point X of the line we have
2 2 2 2.2 2.2 гц2
X\ *2 *3 »1 < »2 < 9i
4 19 4 19
/V »22 \ ,
That is, every point of the line lies on the cone. Such lines are the rulings.
Let Υ = (>i,>2, Уз) be a point of the intersection of the cylinder of Fig. 38.2
with the plane *3 = 0. Then
У2 + 2Л = 0, y3 = 0.
Parametric equations of a line through Υ have the form
xi = У\ + »i/, *2 = >2 + Ы ^з = >з + °з'-
If we choose υ\ = »2 = 0, v3 = 1, then every point Л7 of the corresponding
line is such that
x22 + 2дп = y22 + 2yi = 0.
Hence every point of the line lies on the cylinder and the line is a ruling.
These rulings have the common direction [0, 0, 1] and are therefore parallel.

Sec. 38]
RULED SURFACES
145
We consider next the rulings of the hyperboloid
2 2 2
χι x2 *з _
9 4 9
of Fig. 37.2. Let У be a point of the intersection of this hyperboloid with the
plane X3 = 0. Then
9 4
A point X of the line through Υ with direction [»i, v2, »з] lies on the
hyperboloid if
2 2 2
9 4 9
=^+^_1 + 2^+^v+f- + ---V2
V 9 4/ V 9 4 9 /
„ />1»1 >2»2\ /»12 »22 »32\
= 21 + )/ + (_ + Ь
V 9 4/ V 9 4 9/
We make the coefficient of / equal to zero by choosing »i = 9y2, »2 = — 4>i.
Thus
>1»1 >2»2 _ 9>1>2 4>i>2 _
9 4 9 4'
We make the coefficient of t2 equal to zero by choosing &3 so that
222 2
»1 »2 »3 „ , , »3
V 4 9/9 9
i.e., by choosing
»32 = 9-36 or »3 = ±18.

146 SOLID ANALYTIC GEOMETRY (Ch. 3
Then X is a point of the hyperboloid if it is a point of the line through Υ
with direction [9y2, — 4yb 18] or of the line through Υ with direction
[9.У2, — 4>i, —18]. Thus both lines are rulings. It can be proved that every
point of the surface lies on a pair of rulings, but we omit the proof. Thus the
1 + 1
9
x3 = 3,x3=-3
Figure 38.3
Figure 38.4
hyperboloid is a ruled surface. Figure 38.3 shows some of the rulings of the
hyperboloid
2 2 2
X\ *2 *3
— + = 1.
1 1 9
Figure 38.4 shows some rulings of the paraboloid
2 2
ι +T = 7;
i.e., of the paraboloid of Fig. 37.5. Let us find parametric equations of a
pair of rulings through a point Υ of the intersection of this paraboloid with
the plane x-i = 0. We have
~У\ ~ Уз = 0, y2 = 0.

Sec.39] MODELS OF THE POSTULATES 147
■- — (-?♦$'■
Equations of a line through Υ have the form
*i = У ι + »,/, x2 = v2t, x3 = уз + v3t,
and this line lies in the surface if, for all values of /,
-*i2 *22 *з -(yi+vit)2 v22t2 y3 + v3t
0 = + = +
14 1 1 4 1
= (~>i2 - Уз) ~ (2>i»i + »з)< +
The coefficient of t2 will be zero if », = 1 and v2 = ±2; the coefficient of /
will be zero if »3 = — 2y,t>, = — 2y, and the term — y2 — y3 has already been
chosen equal to zero. Hence the line through Υ with direction (1, 2, —2yi)
and the line through Υ with direction (1, —2, —2yi) are both rulings.
Problems
1. Draw the isometric projections of the intersection of the surface xi2 = 2*2 with
the planes x\ = 3, x\ = — 3, X3 = 0, дгг = 2.
2. Find parametric equations of the rulings of the cylinder of Problem 1.
39. MODELS OF THE POSTULATES
A two-dimensional model of the postulates is constructed as follows. Each
vector a is considered as simply an ordered pair of real numbers [ab a2], and
each point X is considered as an ordered pair (*ι, χ2). The zero vector 0 if
defined to be [0,0] and the negative — [αϊ, α2] of [01,02] is the vector
[ — βι, — a2]. The vector operations and the relations between points and
vectors are defined in terms of algebraic relations between such ordered
pairs. When we have studied this model, it will then be clear how to construct
higher dimensional models that can be used to define fourth- and higher-
order determinants. The two-dimensional model is defined as follows:
D39.1. [βι, αο] = [δι, b2] if and only if αϊ = b\ and a2 = b2.
D39.2. [a,, a2\ + [bu b2] = [a, + bu a2 + b2].
D39.3. χ[αχ, a2] = \xax, xa2] = [ab a2]x.
D39.4. [βι, a2]-[bi, b2] = β,ft, + a2b2.
D39.5. (*,, x2) = {y\,y2) if and only if x, = >, and x2 = y2.
D39.6. If X = (*,, χ2), Υ = (yuy2), then XY = [yi - *„ y2 - x2].

148 SOLID ANALYTIC GEOMETRY (Ch. 3
Our first task is to show that the model satisfies all the postulates of
Section 9. Since [β,, a2] + [ft,, b2] = [β, + ft,, a2 + b2] is an ordered pair of
numbers enclosed in square brackets, it is a vector by definition. Clearly,
k, a2] + [bu b2] = [bu b2] + [au a2],
([βι, a2] + [bu b2]) + [cu c2] = [β,, a2] + ([ft,, b2] + [cu c2]),
[βι, a2\ + [0, 0] = [a„ a2],
k,a2] + [-«i, ~a2\ = [0,0].
Hence the vectors thus defined form a commutative group with respect to
the operation +. The product *k, a2] = [xai, xa2] is a vector and has the
properties
(* + >)k, a2] = [χαλ + yd!, x^ + yb^
= x["i, a2] + y[au a2],
x(y[ai, a2]) = (xy)[au a2],
^([oi, a2] + [bu b2]) = [χαλ + xbu xa2 + xb2]
= x[oi, a2] + x[bu b2],
x[au a2] = [au a2] if л: = 1.
Hence our model is a vector space. The scalar product [01,02] -k> ^2] =
aibi + a2b2 is a number, and [βι, ο2]·[θι, α2] = a\ + a2 is positive except
when [βι, α2] = [0, 0]. This product has the properties
[oi, o2] · [fti, b2] = [ft,, ft2] · [a,, a2],
[α,, α2]·([*ι, b2] + [c,, c2]) = [a,, e2]-[ft,, b2] + [a,, e2]-[c,, c2],
k, a2] ■ (y[bu ft2]) = >(k, a2] ■ [ft,, ft2]).
If X = (*,, x2), Υ = (yi,y2), Ζ = (ζ,, z2) are points, then XY = [y - *,,
>2 — ^2] is 3 vector. Moreover,
XY + Υ Ζ = [>, - xuy2 - x2] + [z, - yu z2 - y2] = [z, - *,, z2 - x2] = XZ.
If [β,, a2\ is a given vector and X = (*,, x2) is a given point, then the point
Υ = (x\ + β,, χ2 + a2) is such that
XY = [χι + α, - *,, л:2 + 02 - a:2] = [a,, a2].
Finally there exists at least one point; for example, the point (1, 3). Thus all
the postulates are satisfied by this model.

Sec. 39] MODELS OF THE POSTULATES 149
To show that this vector space is two-dimensional, we note that the
orthonormal system ui = [1, 0], u2 = [0, 1] has the property that if [xu x2]
is any vector, then x\, x2 are the scalars such that
*i[l, 0] + *2[0, 1] = k, 0] + [0, x2] = [xu x2].
Now suppose that there is an orthonormal system consisting of three vectors
[дь аг]> [^i, b2], [ci, c2]. We have seen that there exist numbers x, y, ζ not all
zero satisfying the scalar equations
i\x + biy + ciz = 0, a2x + b2y + c2z = 0,
and hence satisfying the vector equation
[au a2]x + [bu b2]y + [cu c2]z = [0, 0].
However, if the above system is orthonormal then
[au a2] ■ ([au a2]x + [bu b2]y + [cu c2]z) = χ = [au a2] ■ [0, 0] = 0,
and similarly, у = ζ = 0. Since this is a contradiction, there is no such ortho-
normal system. Hence the vector space is two-dimensional. If О is the point
(0, 0), then to every vector [xi, x2] there corresponds the point X = (*i, x2)
such that OX = [*i, x2], and to every point there corresponds the vector.
The set of points X is a two-dimensional space.
Suppose that the postulates of Section 9 were inconsistent; i.e., that they
implied two mutually contradictory statements. Then these contradictory
statements would be statements about the model, since the model satisfies
the postulates. That is, the model would be inconsistent, and in turn ordinary
algebra would be inconsistent, since the model is defined in terms of the
operations of algebra. Hence the postulates are consistent if ordinary algebra
is consistent.
Let us construct a four-dimensional model in which each vector a is an
ordered quadruple [αϊ, a2, a^, a^] and each point Xis a quadruple (*i, x2, хз, х^).
The zero vector is (0, 0, 0, 0), and the negative of [αϊ, α2, аз, 04] is [ — αϊ, — аг,
— аз, — а4]. We make no attempt to visualize the model geometrically, but
instead we define its properties algebraically as follows:
D39.7. [аь а2, а3, 04] = [bu b2, b3, b4] if and only if αϊ = bu a2 = b2, a3 = b3,
a4 = b4.
D39.8. [au a2, a3, 04] + [bu b2, b3, b4] = [^ + bu a2 + b2, a3 + ft3, 04 + *4]·
D39.9. x[ai, a2, a3, a4] = [xau xa2, xa3, xa4] = [au a2, a3, a4]x.
D39.10. [β,, a2, a3, a4]-[bu b2, b3, b4] = a^ + a2b2 + a3b3 + a4ft4.

150 SOLID ANALYTIC GEOMETRY (Ch. 3
D39.ll. (xb x2, x3, x4) = (y,, y2, y3, y4) if and only if xY = yu x2 = y2, *з = Уз,
Ч = У4-
D39.12. If Χ = (χι, χ2, χ3, χ»), Υ = (yi, Уь Уз, У*), then
XY = Ь\ - х\, Уг - χ2, Уз - *з, У4 - ч]·
The task of showing that this model satisfies the postulates is almost
identical with that of showing that the two-dimensional model satisfies the
postulates. In order to use this model to define fourth-order determinants,
it is not necessary to prove that it is four-dimensional. It is, however, necessary
to define the orthonormal system щ = [1, 0, 0, 0], иг = [0, 1, 0, 0], u3 =
[0, 0, 1, 0], U4 = [0, 0, 0, 1]. This system has the property that every vector
[*i, x2, X3, x4] can be expressed in the form
[*1, x2, X3, Ч} = Ul*! + U2X2 + U3X3 + U4X4.
With the aid of fourth-order determinants it can be proved that the model is
four-dimensional.
We can define a one-dimensional model in which every vector a has the
form a = [a] and every point X has the form X = (x), where a, x are numbers.
The zero vector is [0] and the negative of [a] is [ — a]. The properties of this
model are analogous to those of the two-dimensional and four-dimensional
models.
Let us show that if ν is a nonzero vector and A any point in two-
dimensional or three-dimensional space, then the set of points X for which
AX has the form vt is a one-dimensional space. The unit vector u = v/|v|
is an orthonormal system, and if χ is a vector of the form vt, then χ = vt =
u Ι ν I / = их, where л: = | ν | /. Conversely for any real number x, их =
u| v|/, where / = x/\ v\. We call χ the vector [x]. If X is such that AX has
the form vt, then AX = их, and we call X the point (x). This set of vectors
[x] is a one-dimensional vector space, and this set of points (x) is a one-
dimensional space.
Let us show that if a, b are linearly independent vectors and С is any point
in three-dimensional space, then the set of points X for which AX has the
form as + Ы is a two-dimensional space. Let h = b — ar, where r is such
that h-a = 0. Then г = a-b/a-a and a, h are nonzero vectors. If vi =
a/1 a I, v2 = h/1 h I, then | vi | = | v21 = 1 and Vi · v2 = 0; i.e., vi, v2 is an
orthonormal system. We have a = vi|a|, b = ar + h = vi|a|r + V2|h|,

Sec. 39] MODELS OF THE POSTULATES 151
and hence if χ has the form a.r + b/, then
χ = v,|a|j + (v,|a|r + v2|h|)/
= vi | a | (s + ri) + v21 h J / = vixi + v2x2,
where xY = |a| (s + rt), x2 = \h\t, t = x2/|h|, s = xl/\a\-rt = *i/1 a |
— гдг2/1 hI. We call χ the vector [xlt x2]. If X is such that CX has the form
as + b/ then CX = Vi^i + V2*2, and we call X the point (*i, x2). This set
of vectors [χι, x2] is a two-dimensional vector space, and this set of points
(*i, x2) is a two-dimensional space. We also have
a b ar
Vi = · V2 = — ·
|a| |h| |h|
and hence for any real numbers xu x2,
bx2
/ xi rx2 \
\~\a~\~ Jh\)
where
Via:i + v2x2 = a [ -r-r - γ^ ] +
= as + Ы,
Xl ГХ2 X2
^ _ TaT _ ThT' ~ ThT'
Questions
1. Define a two-dimensional model of the postulates of Section 9 and show that it
satisfies all these postulates. 2. Define a four-dimensional model of the postulates.
3. Show that if A is any point and ν any vector in two-dimensional or three-
dimensional space, then the set of points X for which AX has the form vi is a
one-dimensional space. 4. Show that if a, b are linearly independent vectors and С is
any point in three-dimensional space, then the set of points X for which CX has the
form as + bi is a two-dimensional space.
Problems
1. Show that the distance between the points X = (*i, *2, хг, Χι), У = (у\,уг, уз,
У*) is
^(У1 ~ *l)! + (у* - "О* + 0-1 - *3)2 + 0-4 - Xi)2,
where this distance is defined to be
\~XY\= \XY-XY.

152 SOLID ANALYTIC GEOMETRY (Ch. 3
2. Let v, = [%, H, 0, 0], v2 = [-%, H, 0, 0], v3 = [0, 0, 1/V2, 1/V2], v4 =
[0,0, -1/V2, 1/V2]. Show that vi, v2, v3, V4 is an orthonormal system. To
establish this result, there are ten scalar products to compute.
3. Let 0= (0,0,0,0), X = (*,,*,,*„*«), a = [1,1,2,4], b = [1,-1, -4,6].
The set of points X for which OX has the form as + bi is the plane containing
the vectors a, b with common origin 0. Write down the scalar equations
equivalent to the vector equation OX = as + bi, solve the first two for s, t in
terms of x\, *2, substitute these values in the last two, and obtain the equations
— χι + 3*2 — xz = 0, 5*i — X2 — χκ = 0. These are equations of the plane.
4. The set of points X = (*i, *2, xz, xt) for which OX has the form />щ + уиг is
the plane containing щ, иг with common origin 0. Parametric equations of
this plane are x\ — p, X2 = q, xz = 0, дг4 = 0. Find parametric equations of
the plane containing из, U4, with common origin 0. Let the parameters be s, t.
Show that the two planes have only the point О in common.
5. Show that if AX = ABs + ACt, where A = (αχ, at, a3, at), В = (bi, b^, bz, bt),
С = (ci, C2, с3, ct), X = (*i, *2, xz, Xi), then
x\ — a\r + bis + c\t, *2 = сцг + biS + c2i,
xz = a3r + bzs + c3t, χι = ащг + bis + c4i,
where r = 1 — s — t.
40. DETERMINANTS OF HIGHER ORDER
Fourth-order determinants are defined in terms of four four-dimensional
vectors a = [au a2, a3, 04], b = [bu b2, b3, b4], с = [cu c2, c3, c4], d = [du d2,
аз, di\ as follows:
a\ bi c-i
I a, b, c, d|
02
03
b2
bz
c2
dr
d2
dz
a\ b^ c\ d\
We again generalize the properties of second-order determinants. Thus:
P40.1. |ub u2, u3, U4I = 1.
P40.2. I a, b, c, d + e I = | a, b, c, d | + | а, Ь, с, е |.
P40.3. I a, b, c, t d I = 11 a, b, c, d |
P40.4. la, b, d.
dl = lb.
b| = Id, b
dl =
a, b, c, d
= I a, c, b, d I

Sec. 40] DETERMINANTS OF HIGHER ORDER 153
On the basis of these properties we can prove the following additional
properties by methods closely analogous to those used in Sections 14 and 25:
P40.5. |a, b, c + d, e| = |a, b, c, e| + |a, b, d, e|,
|а, Ь + с, d, e| = |a, b, d, e| + |a, c, d, e|,
ja + Ь, c, d, e| = |a, c, d, e| + | Ь, с, d, e|.
P40.6. |a, b, zc, d| = z|a, Ь, с, d|,
| а, уЪ, с, d J = у | а, Ь, с, d |,
| *a, b, c, d | = * | а, Ь, с, d |.
P40.7. | a, b, c, c| = |a, Ь, с, Ь| = |а, Ь, с, а = |a, b, b, d| =
|a, b, a, d| = |a, a, c, d| = 0.
P40.8. |a, b, c, 0| = |a, b, 0, d| = |a, 0, c, d| = |0, Ь, с, d| =0.
We now show that these properties uniquely determine the value of a
fourth-order determinant. Thus
| a, b, c, d| = lu^i + u2a2 + u3a3 + ща^ Ь, с, d|
= αϊ |ui, b, c, d| + a2|u2, b, c, d|
+ a31 u3, b, c, d I + a41 u4, b, c, d |,
where | ui, b, c, d |, | u2, b, c, d |, | u3, b, c, d |, | U4, b, c, d | are respectively
signed minors of ab a2, a3, a4. Moreover,
|ui, b, c, d| = 0 + ft2|ub u2, c, d| + ft3|u!, u3, c, d| + b4\uu щ, с, d|,
where
I ui, u2, c, d I = c31 ub u2, u3, d I + c41 ub u2, u4, d |
= c3d41 ui, u2, u3, u41 + c4d31 ui, u2, u4, u31
= c3d4 — c4d3 =
сз d3
c4 d4
'
and similarly
|ubu3, c, d| = -
c2 d2
c4 d4
- |u!,u4, c, d| =
Hence
|u!, Ь, с, d| = b2
сз d3
c4 d4
- *з
c2 d2
c4 d
4
+ b4
c2 d2
сз d3
c2 d2
сз d3
b2 c2 d2
Ьз с3 d3
b4 c4 d4

154 SOLID ANALYTIC GEOMETRY (Ch. 3
In a like manner it can be shown that
bi ci di
I "2, b, c,d\= - b3 c3 d3
bi Ci di
I "4, b, c, d\ =
b\ ci di
Ь2 C2 d2
Ь^ Ci di
l"3, b,c, d\ =
bi ci dx
Ь2 C2 d2
*з c3 d3
i.e., the signed minors are equal except for signs to the corresponding minors.
The value of the determinant is thus
Ol
b2
b3
bi
C2
C3
Ci
d2
dz
di
- a2
*1
*3
bi
+ оз
C\
C3
Ci
*1
b2
bi
dy
dz
di
C\
c2
Ci
d,
d2
di
- a4
br
b2
b3
C\
c2
cz
dr
d2
d3
This is called the expansion in minors of the first column. The determinant is
evaluated by computing each minor and then computing the above sum.
The problem of expanding in minors of some other column is reduced to
that of expanding in minors of the first column as follows:
|a, b, c, d| = - |Ь, а, с, d|,
| a, b, c, d | = - | a, c, b, d | = | c, a, b, d |,
|a, b, c, d| = -|a, b, d, c| = |a, d, b, c|
= - |d, a, b, c|.
For example, the expansion in minors of the fourth column is the following:
| a, b, c, d | = rfi |a, b, c, ui | + rf21 a, b, c, u21
+ d31 a, b, c, u31 + di \ a, b, c, u41
= - I d, а, Ь, с I
= ~d1
a2 b2 c2
a3 b3 c3
di bi Ci
+ d2
Ol *1 C\
a3 b3 c3
Qi bi Ci

Sec. 40] DETERMINANTS OF HIGHER ORDER 155
"1
a2
a4
*1
*2
*4
Cl
C2
C4
+ rf4
Ol
02
03
*1
*2
*8
Cl
C2
C3
The signed minors differ from the corresponding minors only in sign, and
the signs are determined by the checkerboard scheme
+ - + -
- + - +
+ - + -
- + - +
The expansion in minors of the first row is obtained as follows:
| a, b, c, d | = a, | Ul b, c, d | + | a2u2 + a3u3 + a4^4> b, c, d |
= αλ | щ, b, c, d | + bi\ a2u2 + a3u3 + a4U4, щ, с, d |
+ | a2u2 + a3u3 + 04Щ, *2u2 + ft3u3 + ft4u4, c, d |
= αλ | ub b, c, d | + bi\ a2u2 + a3u3 + 04114, щ, с, d |
+ ci I a2u2 + a3u3 + 04x14, b2u2 + ft3u3 + *4щ, Ui, d |
+ d\ I 02U2 + a3U3 + 04X14, ft2U2 + ft3U3 + b4Xl4, C2U2 + C3U3 + C4XI4, Ui I
+ I 02U2 +03X13 + 04X14, b2U2 + ft3U3 + ^Щ, C2U2 + C3ll3 + C4Xl4, d2U2 + d3U3 + d4Xl4 I ,
where
|a2u2 + a3u3 + 04x14, щ, c, d| = Iа - aiuu ub c, d| = |a, Ui, c, d|,
1a2u2 + a3u3 + 04x14, Ь2м2 + b3u3 + b4xi4, ub d | = | a, b, ub d |,
I a2u2 + а3из + 04x14, b2u2 + b3u3 + b4xi4, c2u2 + c3u3 + C4U4, Ui | = | a, b, c, Ui |
and
102U2+a3u3 + a4u4,62u2 + ft3u3 + ft4u4,c2u2 + c3u3 + c4u4,rf2u2 + rf3u3 + rf4u4 |
0 0 0 0
a2 b2 c2 d2
03 b3 c3 d3
04 Ь\ С4 d.4
0
»3
»4
0
C3
c4
0
d%
d4
+ a3
0
b2
b4
0
c2
C4
0
d2
d4
- 04
0
b2
bz
0
c2
cz
0
d2
d3

156 SOLID ANALYTIC GEOMETRY (Ch. 3
Hence the expansion in minors of the first row is the following:
| а, Ь, с, d|
= a, | Ul b, c, d | + *i | a, u,, c, d | + cx | a, b, ub d | + rf, | a, b, c, Ui |
Ь2 C2 d2
= αχ ba cz dz — b\
b$ C4 di
a2 b2 d2
+ cl a3 b3 d3
a4 ^4 d\
Since a third-order determinant is equal to its transpose, we have the
additional result
02 "3 i4
02
03
04
dr
C2
C3
C4
02
03
04
d2
dz
dt
b2
b3
*4
C2
C3
C4
I а, Ь, с, d|
Ol
b2 Ьз i4
c2 сз С4
d2 d3 d4
*ι
C2 C3 C4
d2 d3 ά^
+ cl
a2 a3 α4
b2 Ьз bi
d2 аз d4
-d1
a2 a3 α4
b2 Ьз Ь4
C2 C3 C4
Ol 02 03 04
*1 *2 *3 *4
Ci C2 C3 C4
rfl rf2 <^3 <^4
where the last member of these equalities is the transpose \a,b,c,d\' of
I a, b, c, d\, and the middle member of the equalities is the expansion of this
transpose in minors of the first column. Hence a fourth-order determinant is
also equal to its transpose. If the transpose of a fourth-order determinant
is expanded in minors of a given column, this expansion is equal to the
expansion of the original determinant in minors of the corresponding row.
Hence a fourth-order determinant can be expanded in minors of any
row. The signs follow the checkerboard scheme. Note also that if any two rows
or any two columns of a fourth-order determinant are equal, the determinant
is equal to zero by P40.7. Fifth- and higher-order determinants are treated in
a similar manner.

Sec. 40] DETERMINANTS OF HIGHER ORDER 157
Questions
1. Write down the four properties that define a fourth-order determinant. 2. These
properties imply four additional properties. Write down the additional properties.
3. Show how to obtain the expansion in signed minors of the first column. 4. Show
how the signed minors are evaluated in terms of the corresponding minors. 5.
Show how to obtain the expansion in minors of the first row. 6. Show that a fourth-
order determinant is equal to its transpose. 7. Show how to expand in minors of any
row. 8. Show that the determinant is zero if two rows or two columns are equal.
Problems
1. Compute
2
0
1
4
3
-4
0
-2
1
1
3
1
-2
3
0
2
by expanding in minors of the first column. Check by expanding in minors of
the third row.
2. Show that
3. Show that
4. Show that
|a, b, с - хл, d| = |a, b, c,d|.
| a, b — хл, с — ул, d — ζλ | = | a, b, c, d |.
a\ b\ c\ d\
02 — xa\ bi — xb\ сч. — xc\ dz — xd\
аг—уа\ Ьз — ybi с3 — yc\ </3 — yd\
at — zai bt — zb\ сз — zc\ di — zd\
a\ b\ c\ d\
02 Ьг сг di
03 Ьз сз di
Qi bi ci di
5. Given
Xl X2 *3 1
a\ 02 аз 1
b\ Ьг Ьз 1
c\ сг сз 1
О,
expand in minors of the first row and show that the equation represents a plane
(if certain minors are not zero). Substitute a\, a<i, аз for x\, *2, хз in the
determinant and show that the equation is satisfied. Similarly show that (b\, bi, Ьз),
(c\, C2, сз) also lie on the plane.

158 SOLID ANALYTIC GEOMETRY (Ch. 3
6. (a) Show that
Ol 02 03 1
b\ Ьг bz 1
Cl Ci Сз 1
d\ di d3 1
01 02 03 1
b\ — 01 ^2 — 02 Аз — 03 0
ci — αϊ сг — 02 сз — аз О
</ι — αϊ a"2 — 02 аз — аз О
= -|Λθ, ЛС, Л0|,
where Л = (аь а2, аз), Я = (*ь *2, *з), С = (ch c2, c3), D = (βΊ, </2, </3).
//wi: Use the result of Problem 4 and expand the second determinant in
minors of the fourth column,
(b) Show that the volume of a parallelepiped, three of whose edges are
formed by the vectors AB, AC, AD, is
Ol 02 03 1
b\ Ьг Ьз 1
Cl Ci Сз 1
οΊ </г </з 1
41. FOUR EQUATIONS IN FOUR UNKNOWNS
Let the vectors a, b, c, d, e be such that | a, b, c, d | И 0, and let us
attempt to solve the equation
a* + by + cz + d/ = e
for x, y, z, t. If x, y, z, t satisfy the equation, then
| e, b, c, d | = | ax + by + cz + dt, b, c, d |
= *|a, b, c, d| +0 + 0 + 0,
I a, e. c, d| = y\a, b, c, d|, |a, b, e, d| = z|a, Ь, с, d|,
I a, b, c, e | = /1 a, b, c, d |,
and hence the values of x, y, z, t must be
|e, Ь, с, d|
I a, b, c, d|
I a, e, c, d | | a, b, e, d |
1 ζ =
I a, b, c, d| | a, b, c, d|
I a, b, c, e|
I a, b, c, d |

Sec. 41] FOUR EQUATIONS IN FOUR UNKNOWNS 159
It remains to show that these values do satisfy. We consider first the case
e = ui and denote the corresponding values of x, y, z, t by x\, y\, zb t\. We
wish to show that
a*i + by ι + czl + dti = ui,
or equivalently that
Ol*l + hyi + CiZi + dit! = 1,
02*1 + b2yi + c2zi + d2h = 0,
оз*1 + *3>i + с&\ + dzh = 0,
α4*ι + *4>i + cazx + dih = 0.
We have
Ol*l + *1>1 + CiZl + d-i^
_ β! I Ui, b, c, d I + *! I a, ui, c, d I + c! I a, b, U!, d I + rfi I а, Ь, с, ui I
| а, Ь, с, d|
= | а, Ь, с, d | =
| а, Ь, с, d |
02*1 + *2>1 + ^2^1 + d2t\
a2|ui, b, c, d| + Ь2\л, ub c, d| + c2|a, b, ub d| + d2\a, b, c, Ui |
I a, b, c, d I
a2 b2 c2 d2
a2 b2 c2 d2
03 *з cz d$
a± Ьа са d±
= = 0,
I a, b, c, d I
since two rows of the numerator determinant are equal. Similarly,
03*1 + *3>i + c3zi + d3ti = 0, a4xi + b4yi + cazx + d^ = 0
and hence лх\ + Ъух + czi + d/i = ui. In a like manner it can be shown
that if *2, y2, z2, t2 are the values of *, y, z, t corresponding to the substitution
e = U2, then ax2 + by2 + cz2 + dt2 = u2; similarly, for the substitutions
e = U3 and e = U4.

160 SOLID ANALYTIC GEOMETRY (Ch. 3
For an arbitrary e we have
| e, Ь, с, d |
χ =
| а, Ь, с, d|
_ gi | ui, b, c, d | + g21 "2, b, c, d | + g31 u3, b, c, d | + e4 | u4, b, c, d |
| а, Ь, с, d|
= gl^l + g2*2 + e3X3 + e4X4,
У = '\У\ + e2y2 + e3y3 + e4y4,
•ζ = gi^i + g2^2 + e3z3 + e4z4,
/ = gi*i + e2t2 + e3t3 + e4t4,
and hence
ax + Ъу + cz + dt
= afo*! + e2x2 + e3x3 + e4x4) + b(tiyi + e2y2 + e3y3 + e4y4)
+ Φιζι + e2z2 + e3z3 + e4z4) + dfo/i + e2t2 + e3t3 + e4t4)
= (a*! + byi + czi + dti)ei + (ax2 + Ъу2 + cz2 + dt2)e2
+ (ax3 + Ъу3 + cz3 + dt3)e3 + (ax4 + Ъу4 + cz4 + d/4)g4
= uigi + u2g2 + u3g3 + u4g4 = e.
Thus we have T41.1.
T41.1. // a, b, c, d, e are such that a, b, c, d| И 0, then the vector equation
ax + by + cz + dt = e
is satisfied or the equivalent scalar equations
a\x + biy + cxz + dxt = t\,
a2x + b2y + c2z + d2t = g2,
a3x + b3y + c3z + d3t = g3,
a4x + b4y + c4z + d4t = g4,
are satisfied if and only if
| e, b, c, d | | a, e, c, d | | a, b, e, d | | a, b, c, e |
л: = 1 у = 1 ζ = > / = -·
| a, b, c, d| | a, b, c, d| |a, b, c, d| |a, Ь, с, d|
The solution of η equations in η unknowns is treated in a similar manner.

Sec. 41] FOUR EQUATIONS IN FOUR UNKNOWNS 161
Questions
1. Show that if a* + by + cz + di = e and |a, b, c, d| 9^ 0, then
[e, b, c, d|
la, b, c, dl
У =
1 a, e, с, d |
| a, b, c, d|
la, b,
la, b, c,d|
I a, b, с, е |
Ia,b,c,d|
2. Show that if e = ui and x\, y\, z\, t\ denote the corresponding values of x, y, z, t,
then the equation a*i + byi + cz\ + dii = ui is satisfied. 3. Assume that similar
results can be obtained when e = иг, e = uj or e = щ show that the values of
x, y, z, t given in Question 1 do satisfy the equation a* + by + cz + di = e. 4. Write
out the equivalent scalar equations.
Pro blems
1. Solve the following equation for x, y, z, t:
[2, 0, 1, 0]x + [0, 0, 3, -\]y + [2, 0, 4, 2]z + [4, 1, 8, l]i = [10, 1, 14, 6].
Write out the equivalent scalar equation and check that your values of x, y, z, t
satisfy them.
2. Let a = [αϊ, α2, a3], b = [bi, Ьг, b3], с = [η, a, c3], d = [d\, аг, d%\. Show that if
|a, b, c| τ* 0, then
a\ b\ c\ d\
a\ b\ c\ d\
Д2 ^2 C2 di
оз bi сз di
|d»b, c| |a, d, c| |a, b, d|
= αϊ ; + bi -— + ci -.— г — di = 0.
I a, b, c| | a, b, c| |a, b, с
That is, show that αϊ* + biy + c\z = d\ if
|d, b, c| |a, d, c|
la, b, с
|a,b,c|
|a,b,c|
la, b, d]
I a, b, с |
Replace the first row of the numerator determinant by аг, Ьг, сг, di and show
by a similar procedure that a^x + biy + c2z = aV Show that aye + bzy +
сзг = di.
3. Use the method of Problem 2 to show that two equations in two unknowns
have a solution when the denominator determinant is not zero.
4. Let a = [αϊ, α2, α3], b = [b\, i2, b3], с = [ch a, c3], d = [d\, d2, d3]. Suppose that
| a, b, с | τ* 0, but that
αϊ ii ci </i
02 Ьг сг di
az bz cz dz
04 bi a di
= 0.

162 SOLID ANALYTIC GEOMETRY (Ch. 3
Show that a* + by + cz = — di if
— i|d, b, c| -i|a, d, c|
χ =
and that
I a, b, c|
У =
Ia,b,c|
-<|a,b,d|
I a. b, cl
a\ b\ c\ d\
02 02 C2 d"i
b3
сг оз
c\ dt
-i|d,b,c|
\a,b,c\
04"
+ 04
| a, b, с | | a, b, с
= 04ДГ + bty + az + dtt = 0.
i|a,d,c| -f|a,b, d| , w
+ c* —ι—: 1 h tdi
| a, b, с |
Thus show that if t ^ 0, then x, y, z, t are not all zero and that they satisfy
the four equations:
a\x + b\y + c\z + d\t = 0, 02* + biy + czz + dtf = 0,
03* + b3y + c-fi + </j/ = 0, 04* + biy + ο^ζ + dtt = 0.
Suppose that the fourth-order determinant of Problem 4 is such that every
third-order minor is zero but that
01 01
02 02
jiO.
Show that αϊ* + biy
-ciz, aye + Ьгу = — c$z if
and that
Cl
C2
Ol
02
bi
*2
*1
*2
J = -Z-
Ol
02
Ol
02
Cl
C2
*1
*2
*1
02
*3
01 01
02 02
= 03
Cl 01
C2 02
+ *3
01 Cl
02 C2
01 01
02 Ьг
= 03* + 03>> + C3Z = 0.
01 01
02 02
+ C3Z
Similarly show that 04* + bty + ctz = 0. Thus show that there exist *, y, z, t
not all zero, but t = 0, satisfying the four equations given at the end of Question
4. Note that we can take ζ ^ 0.
6. Suppose that the fourth-order determinant of Problem 4 is such that every
second-order minor is zero, but αϊ τ* 0. Show that αϊ* + b\y = 0 if * =
—ybi/ai, and that for this * we have 02* + Ьгу = 0 = аз* + Ьгу = a& + bty.
Thus show that there exist *, y, z, t not all zero satisfying the four equations
at the end of Question 4.

Sec. 42] APPLICATIONS OF DETERMINANTS 163
42. GEOMETRICAL APPLICATIONS OF DETERMINANTS
Let us find an equation of the plane through the points A = (α1; a2, a3),
В = (b\, b2, b3), С = (ci, c2, £3)· The equation has the form
(a) dixi + d2x2 + d3x3 + <4 = 0.
The points А, В, С lie on the plane if and only if
(b) diai + d2a2 + d3a3 + dt = 0
(c) rfifti + d2b2 + d3b3 + d4 = 0
(d) dici + d2c2 + d3c3 + d4 = 0.
The unknown numbers d\, d2, d$, d4 must therefore satisfy equations (b),
(c), (d). Moreover, if X is any point of the plane, then the unknowns must
also satisfy equation (a). The coefficients of these unknowns are χι, χ2, Χ3, 1,
αϊ, a2, etc. The obvious solution d\ = d2 = d$ = d\ = 0 is such that (a) does
not represent a plane, but by T41.1 this is the only solution unless the
determinant of the coefficients is zero. Hence we must have
0 =
= *i
л:
1 X2
*3
Οι a2 a3
bi b2 b3
c\ c2 c3
ι
a2 03
b2 b3
c2 c3
+ *3
Ol
*1
C\
1
1
1
1
1
1
1
- X2
a2 1
b2 1
c2
1
a-ί a3 1
fti *з 1
c\ c3 1
-
a\ a2
bi b2
C\ C2
03
»3
C3
Since this equation is linear xu x2, x3, it represents a plane. Moreover, if we
substitute eb a2, a3 for xu x2, x3, the first two rows of the fourth-order
determinant become equal, the determinant is zero, and the equation is satisfied;
i.e., A is a point of the plane. If we substitute b\, b2, b3 for χι, x2, x3, the first
and third rows become equal, and hence В belongs to the plane; similarly,
for С We have thus found the desired equation unless the coefficients of
*i> x2, хз are all zero, and it can be shown that this is the case only when
А, В, С lie on a line.

164 SOLID ANALYTIC GEOMETRY (Ch. 3
In plane analytic geometry the equation of a line through two points
A, B can be obtained by analogous reasoning. The equation is
X\ X2 1
α, a2 1
ft, ft2 1
Let us find an equation (plane analytic geometry) of the circle through
three points А, В, С The equation has the form
= *1
<*2
ft2
1
1
- *2
"l
*1
1
1
+
Ol
ft.
02
ft2
(a)
diixi2 + x22) + d2xi + d3x2 + <k = 0.
Note that if d\ И 0, we can divide by rf, and obtain an equation in the more
familiar form in which the coefficient of x2 + x22 is one. The points А, В, С
lie on the circle if and only if
(b)
(c)
(d)
rf,(a,2 + a2) + d2ax + d%a2 + d4 = 0,
rf,(ft,2 + ft22) + d2bx + d3b2 + d4 = 0,
di(ci2 + c22) + d2Cl + d3c2 + dA = 0.
The unknowns rf,, d2, d$, <U must satisfy equations (a), (b), (c), (d) and must
not all be zero. Hence we must have
2 ι 2
Xl + *2 Xl X2
"ι2 + α22 αϊ a2
ft, + ft2 fti ft2
r 2 4- r 2
C\ Τ C2
{xi2 + X22)
C\
a\
ft.
C\
C2
a2
b2
c2
- χ\
+ X2
t\2 + a22 a, 1
ft,2 + ft22 ft, 1
c,2 + c2 c, 1
a, + a2 a2 1
ft,2 + ft22 ft2 1
c,2 + c22 c2 1
"i2 + a22 a, a2
ft,2 + *22 ft, b2
2 2
C\ + C2 C\ C2

Sec. 42] APPLICATIONS OF DETERMINANTS 165
When the coefficient of x2 + x22 is not zero, this equation has the appropriate
form of an equation of a circle. By reasoning similar to the above, we see
that А, В, С lie on the circle, since the coordinates of these points satisfy the
equation. The coefficient of χι2 + x2 has the same form as the middle term
in the above equation of a line through two points. Thus this coefficient is
zero if and only if A lies on the line through В, С When the coefficient is
zero, we obtain a linear equation in *,, x2 representing the line through
the three points А, В, С
EXAMPLE 42.1
Find an equation of the circle through (4, 4), (4, —4), ( — 4, 0). The
equation is
*i2 +
X22
32
32
16
(*12 + X22)
+ X2
3
3
1
*1
4
4 -
-4
4
4
-4
2 4
2 4
5 -4
*2
4
-4
0
4
-4
0
1
1
1
1
1
1
1
1
1
1
-
- *i
32
32
16
32
32
16
4
4
-4
4
-4
0
4
-4
0
1
1
1
64(x,2 + x22) + 128*, + 1536.
Dividing by —64, we obtain
X!2 + x22 ~ 2x, - 24 = 0,
V - 2xv+ χ2Δ = 24,
x,2 - 2*, + 1 + x-i = 25,
On - I)2 + xi = 25.
This is the circle with radius 5 and center at (1, 0).

166 SOLID ANALYTIC GEOMETRY (Ch. 3
0 =
*12 + *22
20
5
0
EXAMPLE 42.2
Find an equation of the circle through ( — 2, 4), ( — 1, 2), (0,0). The
equation is
*1 X2 1
-2 4 1
-1 2 1
0 0 1
-2 4
(*i2 + *22) -1 2
0 0
-2 1
-1 1
0 1
+ *2
= -20*,
20
5
0
10x2,
1
1
1
-
- xi
20
5
0
20 4
1
5 2 1
0 0 1
-2 4
-1 2
0 0
2x, + x2 = 0.
It is easy to check that the three points do lie on this line.
Questions
1. Show how to find an equation of a plane through three given points. 2. Show
how to find an equation (plane analytic geometry) of a line through two points and a
circle through three points. What does the equation of the circle become when the
three points lie on a line?
Problems
1. Find an equation of the line through (2, —1), (5, 3).
2. Find an equation of the plane through (2, 3, — 1), (3, 1, 4), (5, 0, 0).
3. Find an equation of the circle through (0, 1), (1, 0), (3, 0).
4. Find an equation of the circle through (0, 1), (1, 0), (2, —1).
5. Show how to find an equation of the sphere through (a,, ai, аз), (6,, bi, bz),
(ci, C2, cz), (d\, di, άΐ). Discuss the case in which the four points lie in a plane.
6. Given the equation
x\ *2 1
a\ аг 1
bi Ьг 1
0,

Sec. 42] APPLICATIONS OF DETERMINANTS 167
show that if the minors of x\ and *2 are both zero, then (αϊ, 02) and (b\, bz) are
the same point.
7. Given the equation
= 0,
X\ *2 *3
a\ 02 az
b\ Ьг bz
c\ сг cz
show that the minor of x\ can be written in the form
02 O3 1
b% — 02 bz — 03 0
C2 — 02 Cz — Oz 0
Show that if the minors of x\, *2, xz are all zero, then AB X AC = 0 and hence
*2
bz
02 C2 — 02
аз cz — a3
that AB and AC are contained in a line.

TRANSFORMATIONS
OF
COORDINATES
4
43. TRANSLATIONS
In plane analytic geometry we studied various conic sections. The points
and lines in terms of which these conic sections were defined were chosen so
that the equations of these curves were in simple forms. In general an equation
of a conic section is somewhat more complicated, and when such an
equation is given, we are faced with the problem of discovering what curve the
equation represents. For example, the equation
4*!2 + 24x^2 + Π*22 - 40*! - 20x2 - 20 = 0
represents the curve shown in Fig. 43.1. The first three terms of this equation
are of degree 2 in xu x2; the next two terms are of degree 1; and the last term
(called the constant term) is of degree 0. The degree of a term is obtained by

Sec. 43]
TRANSLATIONS
169
adding the exponents of χι, χ2· Thus the exponents in the term 24x^2 are 1
and 1, and the degree is 2. The exponents in Ax-f = Ax-^x^ are 2 and 0, and
the degree is again 2. Such an equation is said to be quadratic in x\, x2. In
general the degree of an equation is the degree of a term of highest degree.
Figure 43.1
The curve looks like a hyperbola whose asymptotes are the lines shown in
the figure. We might also guess that if we were to choose a new origin at the
point Η of intersection of the asymptotes and a new orthonormal system
Vi, v2 such that vi, V2 He in the angle bisectors of the asymptotes, then the
equation of the curve referred to this new axis system would be in one of
the simple standard forms. These guesses are correct. The moving of the origin
to a new point Η is called a translation, and the turning of the axes into the
new positions vi, V2 is called a rotation. In this section we shall learn how a
quadratic equation can be simplified by a translation, and in the section 46
we shall learn how the equation can be further simplified by a rotation.
A translation is defined by the vector equation
OX = Ui*i + U2*2 = OH + HX = Ui/Z! + U2h2 + \1\У\ + U2>2
where hi, h2 are the old coordinates of the new origin H, and yit y2 are the
new coordinates of the point X referred to the origin H. The vector equation
is equivalent to the scalar equations
χι = hi + yu x2 = h2 + У2-

170 TRANSFORMATIONS OF COORDINATES (Ch. 4
These scalar equations also define the translation. To apply a translation to an
equation, we substitute the values given by the translation equation for the
old coordinates in terms of the new ones and the coordinates of the new origin.
EXAMPLE 43.1
Apply a translation to the above equation of the curve of Fig. 43.1.
We obtain
4(>i + *ι)2 + 24(Λ + Ai)(>2 + A2) + 110-2 + h2)2
- 40(>! + hx) - 20(j>2 + h2) - 20
= 4j-!2 + Shiyi + 4fn2 + 24у1У2 + 24h2yi + 24hxy2 + 24A,A2
+ lly22 + 22h2y2 + 11A22 - 40yl - 40/z! - 20>2 - 20A2 - 20
= 4>!2 + 24vi>2 + 11>22 + 2(4Ai + 12A2 - 20)^ + 2(12Aj + 11A2 - 10b
+ (4Л!2 + 24A,A2 + ПЛ22 - 40^! - 20A2 - 20) = 0.
The first-degree terms in y\, y2 can be removed by choosing hi, h2 so as to
satisfy the scalar equations
4A, + 12A2 - 20 = 0,
12A, + 11A2 - 10 = 0,
or the equivalent vector equation
[4, 12]*! + [12, 11]A2 + [-20, -10] = a^ + a2A2 + a3 = 0,
where
a, = [4, 12], a2 = [12, 11], a3 = [-20, -10].
The solution is
I ~аз, a2| I a,, -a3|
hi = = — 1, h2 = = 2.
I»i, »г| |аь а2|
If we substitute these values of Ai, A2 in the expression for the constant term,
we obtain
4Л!2 + 24A,A2 + 1U22 - 40*! - 20Л2 - 20 = -20.
Hence when we translate to the new origin Η = ( — 1,2), the equation
becomes
4yi2 + 24yiy2 + Uy22 - 20 = 0.

Sec. 43]
TRANSLATIONS
171
In order to draw the curve represented by this equation in^i, j% we shall
have to produce a further simplification by means of a rotation. The vectors
ai, аг will play an important role in this simplification.
Questions
1. What is meant by a quadratic equation in *ι, *2? 1. What is meant by the
second-degree terms, the first-degree terms, and the constant term. 3. Show how to
obtain the translation equations.
Problems
1. Given the equation
17*i2 - 12*i*2 + 8*22 + 18*1 - 4*2 = 0,
perform an arbitrary translation and show that the first-degree terms of the
resulting equation are
2(17Ai - 6A2 + 9)yi, 2(-6Ai + 8A2 - 2)y2.
Show that the first-degree terms can be removed if Αι, Λ2 are such that aiAi +
агЛг + аз = 0, where
ai = [17, -6], a2 = [-6, 8], a3 = [9, -2].
Solve for Αι, Λ2 and obtain the new equation in which the first-degree terms
are missing.
2. Let ai, аг, аз be defined as in Problem 1 and let χ = [хь *2]. Show that
ari = 17*i — 6*2, Л2'х = —6*i + 8*2, аз·» = 9*ι — 2*2,
and that the equation can be written in the form
(17*i - 6x2)*i + (-6*1 + 8*2)*2 + 2(9*i - 2*2)
= (arx)*i + (a2-»)*2 + 2аз·» = О.
Let h = [Αι, Λ2]. Show that the new constant term can be written in the form
(17Ai - 6A2)Ai + (-6A1 + 8A2)A2 + 2(9A, - 2A2)
= (ai-h)Ai + (a2-h)A2 + 2a3-h
= (aiAi + агАг + аз)'Ь + аз'Ь
and that when aiAi + агЛг + аз = 0, the constant term becomes аз'Ь.

172 TRANSFORMATIONS OF COORDINATES (Ch. 4
3. Let ai, a2, аз be defined as in Example 43.1, and let χ = [*i, *2], h = [Αι, A2],
У = ЬиУг\-
(a) Compute ami, 2aru2, a2-u2, 2аз-и1, 2аз-и2 and check that these numbers
are respectively the coefficients in the equation of this example, exclusive of
the constant term.
(b) Compute an, a2-x, аз·» and show that the equation of this example can
be written in the form
(4*i + 12*2)*i + (12*i + 11*2)*2 + 2(-20*i - 10*2) - 20
= (ai*i + a2*2)-x + 2a3-« - 20 = 0.
Show that after the translation has been performed, the equation becomes
(snyi + а2>>2)-у + 2(aiAi + a2A2 + a3)-y
+ (aiAi + a2A2 + a3)-h + a3h - 20 = 0.
(c) Show that if Ai, A2 are such that aiAi + а2Л2 + аз = 0, the equation
becomes
(aiji + a2^2)-y + (аз-h - 20) = 0.
4. Apply an arbitrary translation to the equation 9*i2 + 16*22 + 18*i — 96x2 +
9 = 0; choose Ai, A2 so that the first-degree terms are eliminated, and show that
the equation becomes 9yi2 + 16ya2 — 144 = 0, or
16 9
5. (a) Apply an arbitrary translation to the equation *i2 + *22 + 2*i — 6дг2 +
9 = 0; choose Ai, A2 so as to eliminate the first-degree terms, and show that
the equation becomes
У12+У22 = 1.
(b) Solve the translation equations xi = yi -\- hi, *2 = yi + A2 for yu y2, where
Ai, A2 are the numbers obtained in part (a); substitute the values in the
quadratic equation of (a) and show that the equation becomes
(*, + l)2 + (*2 - 3)2 = 1.
(c) Find the center and radius of this circle.
6. (a) Substitute χ = у + A in the equation
хг - 6x + 4 = 0;
determine A so that the first-degree term in у is eliminated, and write down
the resulting equation in y.
(b) Solve the quadratic in у of part (a) and show that у = ±λ/5, and hence
x = 3 ±Vs.

Sec. 44] EQUATIONS IN TWO VARIABLES 173
7. Apply an arbitrary translation to the equation дс22 — 4*i — 6*2 + 5 = 0, and
show that the equation becomes
угг ~ 4ji + 2(A2 - Ъ)уг + (A22 - 4Ai - 6A2 + 5) = 0.
Note that the first-degree term in yi can be eliminated by the translation but
that the first-degree term in y\ cannot. Find A2 so as to eliminate the first-degree
term in уг, and show that for this choice of A2 the constant term can be
eliminated by a proper choice of Ai. Find this Ai and show that the equation becomes
Угг = 4>ί.
44. QUADRATIC EQUATIONS IN TWO VARIABLES
The computation of Example 43.1 is laborious, and we shall describe a
method of reducing the labor. The description of the method can be simplified
by choosing an appropriate notation for the coefficients in the quadratic.
The notation that we have selected will seem more reasonable if we write x\X\
in place of x2 and x2x2 in place of χ·?. Some of the terms contain the factor
2. These terms are written in this form to simplify subsequent computations.
The general quadratic has the form
Oii*i*i + 2ax2xxx2 + a22x2x2 + 2αλ3χλ + 2a23x2 + Озз = 0.
If we apply an arbitrary translation to this equation, we obtain
«ιιϋΊ + hi)2 + 2a12(y! + h1)(y2 + h2) + a22(y2 + h2)2
+ 2a13(yi + hi) + 2а2з(>2 + h2) + 033
= auy2 + 2auyihi + auh2 + 2ax2yxy2 + 2ax2yxh2 + 2ax2y2hx + 2ax2hxh2
+ a22y2 + 2a22y2h2 + a22h2 + 2al3yl + 2al3ki + 2агз>2 + 2α23^2 + 033
= any2 + 2a12yly2 + a22>22 + 2(anhx + a12h2 + ai3)yi
+ 2{ax2h\ + a22h2 + агз)>2 + On^i2 + 2ax2hxh2 + a22h22 + 2ахфх
+ 2a23h2 + дзз = О.
The first-degree terms in yi and y2 can be removed from the latter equation if
we can choose Ab h2 so as to satisfy the equations
a\\h\ + d\2h2 + Ol3 = 0, a\ih\ + 022^2 + ^23 = 0.
If
»i = [an, ai2], a2 = [ai2, a22], a3 = [a13, агз],

174 TRANSFORMATIONS OF COORDINATES (Ch. 4
then the equations in hi, h2 are equivalent to the vector equation
iihi + a2/z2 + a3 = 0,
and this equation has a solution if | a, a21 И 0. It remains to compute the
new constant term. Observe that this term is the number that results when
we substitute hi, h2 for *i, x2 in the left-hand member of the original quadratic.
This term is computed as follows:
"uhi + 2al2hlh2 + a22h2 + 2αλφλ + 2a23h2 + Дзз
= (дц/z! + al2h2 + ai3)hi + (αλφλ + a22h2 + a23)h2 + a13hi + a^h-i + 033
= «13^1 + огзЛ2 + озз,
since hi, h2 are chosen so as to make the two parentheses zero. Thus the
quadratic becomes
а\\У\2 + 2ax2yxy2 + a22y2 + {ai3hi + a23h2 + 033) = 0,
where the first three terms are obtained from the first three terms of the
original quadratic by replacing xlt x2 by yu y2.
When we wish to remove the first-degree terms from a quadratic by
means of a translation, the above results can be applied as follows: We begin
by writing down the new constant term in terms of hi, h2 and rearranging this
expression in the manner indicated above. The rearrangement contains two
parentheses. When we set these parentheses equal to zero, the resulting
equations are equivalent to the vector equation to be solved for hi, h2.
Moreover, the new constant term reduces to ai3hi + a23h2 + 033, and this
expression is evaluated by substituting in it the values obtained for hi, h2.
We can now write down the translated equation with the first-degree terms
missing. The vectors ai, a2, аз are obtained directly from the vector equation.
This method is used only when |ai, a2| И 0.
EXAMPLE 44.1
Apply the method described above to the solution of the problem of
Example 43.1. The new constant term is
Ahi2 + 2Ahih2 + \\h22 - 40Л, - 20Ь2 - 20
= (4Л, + 12Л2 - 20)Л, + (12Л, + Ш2 - 10)Л2 - 20Л, - 10Л2 - 20.
The equations
Ahi + \2h2 - 20 = 0, \2hi + 11A2 - 10 = 0,

Sec.44] EQUATIONS IN TWO VARIABLES 175
are equivalent to the vector equation alhi + a2h2 + аз = 0, where ai =
[4, 12], a2 = [12, 11], a3 = [-20, -10]; the solution of this equation is
h\ = —1, h2 = 2. When we substitute these values into the expression
— 20hi — \0h2 — 20, we obtain —20. Hence the translation produces the
equation
4y,2 + 24yiy2 + 11;-22 - 20 = 0.
EXAMPLE 44.2
Simplify the equation 9л:!2 + 16x22 + 18λ:ι - 96x2 + 9 = 0 and draw
the curve. The new constant term is
9Л,2 + 16Л22 + 18Л, - 96Л2 + 9
= (9Л, + 9)Л, + (16Л2 - 48)Л2 + 9Л, - 48Л2 + 9.
The equations
9Л, + 9 = 0, 16Л2 - 48 = 0,
are equivalent to ai/zi + а2Л2 + a3 = 0, where ai = [9, 0], a2 = [0, 16],
a3 = [9, —48], and the solution is hi = — 1, h2 = 3. Hence the new constant
term is 9hi — 48/z2 + 9 = —144, and the translation produces the equation
9yi2 + 16y22 - 144 = 0
or
2 2
^+^ = 1.
16 9
To draw this ellipse, we draw the vectors ui, u2 with common origin О and
then draw an equivalent pair of vectors ui, u2 with new common origin
Η = (—1,3). Next we draw the usual rectangle containing the ellipse where
the position of the rectangle is located relative to the new axes. The ellipse is
tangent to the sides of the rectangle at points (4, 0), (-4, 0), (0, 3), (0, -3)
where the coordinates of these points are the
new coordinates. The old coordinates of these
points are obtained by substituting in the
translation equations
xi = yi + hi = yi - 1,
*2 = >2 + h2 = y2 + 3.
Hence the old coordinates of the above four
points are respectively (3, 3), (-5, 3), (-1,6), - -
(-1,0). See Fig. 44.1. ( 1'0'
Note that an arbitrary pair of vectors ab Figure 44.1
(-1,6)

176 TRANSFORMATIONS OF COORDINATES (Ch. 4
a2 can be written in the component form ai = [<гц, а2\\, a2 = [θΐ2> Я22] but
that they can be the vectors ai, аг associated with a quadratic equation only
when a2i = 0i2- If а'ь а'г are the vectors forming the transpose of the
determinant |ab a2|, then α2ι = ^12 if a'i = »ь а'г = a2, and in this case
the determinant is said to be symmetric.
Questions
1. Write down a formula for the general quadratic in *i, x2. 2. Apply an arbitrary
translation to this equation. 3. Find the scalar equations which must be satisfied in
order to eliminate the first-degree terms. 4. Write down the corresponding vector
equation. 5. Show how to compute the new constant term.
Problems
1. Show that the equation 9*i2 — 16*22 + 36*i + 96*2 — 252 = 0 can be
transformed by an appropriate translation into
16 9
Draw the curve and show the old and new axes.
2. Let ai = [an, an], a2 = [an, 022], »з = [an, <Чг], * = [*ι, «]·
(a) Show that
<Jll*l2 + 2(212*1*2 + <222*22 + 2(213*1 + 2(223*2 + 033
= (»i*i + а2*г)·» + 2a3·* + a».
(b) Perform the translation χ = у + h where у = [y1} y2], h = [Αι, A2], and
show that the equation becomes
(*iyi + л2у2)'у + 2(aiAi + a2A2 + a3)-y
+ (aiAi + a2A2 + a3)-h + a3-h + a33 = 0.
(c) Show that if aiAi + агЛг + a3 = 0, the equation becomes
(*iyi + л2у2)'у + (a3-h + азз) = 0.
3. Let ai = [an, ai2, an], л2 = [an, a22, а2з], a3 = [an, а2з, азз], » = [*ъ *2;
Ilia) Show that
011*12 + 2(212*1*2 + <222*22 + 2a13*l + 2(223*2 + 033
= (<Jll*l + (212*2 + <Лз)*1 + ((212*1 + 022*2 + <J23)*2
+ (<J13*1 + 023*2 + <J33)
= (arx)*i + (аг-х)дсг + a3-x
= (ai*i + a2*2 + a3)-x.
(b) Show that ai'Uj = аг-ui, aru3 = a3-ui, аг-и3 = а3-иг.

Sec. 45] APPLICATIONS OF TRANSLATIONS 177
4. Proceed as in Problem 6 of Section 43 with respect to the equation *2 — 10* +
5 = 0.
5. Proceed as in Problem 7 of Section 43 with respect to the equation *i2 + 2x\ —
3*2 + 4 = 0. Show that for this equation | ai, aj | = 0, where ai, аг are the
vectors described in Problem 2.
6. Perform an arbitrary translation on the points (αϊ, аг), (Ь\, Ьг), and let the
new coordinates be denoted respectively by (съ cz), (dh dz). Show that
V(</, - οχγ + (d2 - с2У = V(4, - aif + (i2 - a,f.
45. APPLICATIONS OF TRANSLATIONS
Translations can be used to find the center and radius of a circle.
EXAMPLE 45.1
Simplify the equation
Л2 + x22 + 2Xl -6*2 + 9
= («l + 0)*! + (0 + *2)*2 + 2(*i - 3*2) +9=0
and find the center and radius of this circle. The new constant term is
hi2 + h22 + 2Aj - 6Λ2 + 9 = (Л! + l)hx + (h2 - 3)Л2 + hx - Ък2 + 9,
where h\ + 1 = 0, h2 — 3 = 0, or equivalently, a.\h\ + a2/z2 + аз = 0,
where ai = [1, 0], a2 = [0, 1], a3 = [1, —3]. Hence h\ = — 1, h2 = 3, and
the constant term is h\ — 3/z2 + 9 = — 1. The translation produces the
equation
У\ +У22 -1=0 or yi2 +y22 = 1.
This is a circle with center at the new origin (1, —3) and radius 1. If we
solve the translation equations for yY, y2, we obtain
У\ = xi - hi = *! + 1, y2 = *2 - h2 = *2 - 3,
and if we substitute these values in the new equation, we obtain
У!2+У22= (*1 + Ό2 + (*2 - 3)2 = 1.
This result could also be obtained by completing the square. In fact a
translation furnishes a method of completing the square.
Let us use the method of translation to complete the square and solve
the quadratic equation
ax2 + bx + с = 0.

178 TRANSFORMATIONS OF COORDINATES (Ch. 4
If we substitute χ = у + h in this equation, we obtain
a(y + h)2 + b(y + h) + с = ay2 + laky + ah2 + by + bh + с
= ay2 + (2ah + b)y + (ah2 + bh + c) = 0.
The first-degree term in у can be removed by setting
b
2ah + b = 0 or h =
2a
The new constant term then becomes
ab2 b2
ah2 + bh + с = — he
4a2 2a
_b2 b2 _ b2 b2 - Aac
4a 2a 4a 4a
and the new equation becomes
b2 — 4ac b2 — 4ac
ay2 = 0 or y2 =
4a 4a2
The solution of this equation for>> is
± Vb2 - 4ac
У =
2a
Hence
* Vb2 - 4ac
χ = h+y = ±
2a 2a
-b ± Vb2 - 4ac
2a
This is a proof of the quadratic formula. Recall that the expression b2 — 4ac
under the radical is called the discriminant and that the roots are real and
distinct, real and equal, or complex, according as the discriminant is positive,
zero, or negative.
EXAMPLE 45.2
Simplify the equation χ2 + 6χλ — 4χ-ι + 5 = 0 and draw the curve.
The new constant term is
h2 + 6hx - 4h2 + 5 = {hx + 3)/*! + (0 - 2)h2 + 3Λχ - 2h2 + 5.

Sec.45] APPLICATIONS OF TRANSLATIONS 179
If we attempt to choose hi, h2 so that the two parentheses are zero, or equiva-
lently, so that
[1, 0]*! + [0, 0]h2 + [3, -2] = а,Л! + a2A2 + a3 = 0,
we discover that this is impossible (note that |a1; a2| = 0)i However, the
quadratic equation can be simplified by translation. Thus we substitute
*i = У\ + h\, x2 = y2h2 and obtain
(yi + hi? + 6(yi + Aj) - 4(y2 + h2) + 5 \ /
= yi2 + 2(Aj + 3b - 4y2 + \ /
\ /*
(A,2 + 6Л, - 4A2 + 5) = 0. \ u* /^L^u,
The first-degree term in y\ can be removed \ **v\
by setting
Figure 45.1
hx + 3 = 0 or A! = - 3,
and the constant term can be removed by setting
A!2 + 6Л, - 4A2 + 5 = 9 - 18 - 4A2 + 5 = -4A2 - 4 = 0,
or A2 = — 1. Thus, if we choose the new origin to be the point Η =
( —3, —1), the equation becomes
yi2 - 4y2 = 0 or yi2 = 4y2.
Figure 45.1 shows a drawing of this parabola.
Questions
1. Show how to find the center and radius of the circle x\2 + дгг2 + 2*i — 6xj +
9 = 0 by means of a translation. 2. Show how to derive the quadratic formula by the
method of translations.
Problems
1. Simplify the equation 8*i2 - 12*!*2 - 8*22 - 40*! - 20*2 - 20 = 0 by a
translation.
2. Simplify the equation 4*i2 — *22 — 16*i — 6дг2 -(- 3 = 0 by a translation.
Draw the curve in its correct position relative to the old axes.
3. Simplify the equation *i2 + *22 — 4*i + 6x2 + 12 = 0 by a translation. Find
the radius of this circle and the coordinates of the center relative to the old
axes.

180 TRANSFORMATIONS OF COORDINATES (Ch. 4
4. Simplify the equation *i2 — 8xi — 6*2 + 25 = 0 by a translation. Draw the
curve in its correct position relative to the old axes.
5. Simplify the equation x22 + 16*i — 6*2 — 7 = 0 and draw the curve.
6. Solve the equation bx2 + 2cx + d = 0 by using the appropriate translation.
46. ROTATIONS
Consider two right-handed coordinate systems, one defined with respect
to an orthonormal system ui, U2 and the other with respect to Vi, V2, and
suppose that the two coordinate systems have a common origin 0. If a point X
has coordinates x\, x2 with respect to the first system and coordinates yi, y2
with respect to the second, then
OX = Ui*i + U2*2 = Vi>>i + У2У2-
Since |vi, V2I = 1 И 0, the existence and uniqueness of yi, y2 is assured.
There exist scalars c, s such that
Vi = CU! + SU2,
and since | vi | = 1, it follows that
Vi'Vi
= A
+ s2=\.
We shall show that the correct value of V2 is
v2 = — nil + oi2.
We have \2 -L Vi and | v2 \ = | Vi | = 1. That is,
νι·ν2 = (oil + su2)'( — sui + cu2) = —cs + sc = 0,
v2 · v2 = (— nil + cu2) · (— nil + cu2) = s2 + c2 = 1.
Hence vi, v2 is an orthonormal system. Moreover,
|vi, v2| =
с s
— s с
= c2 + s* = \,
and hence vi, v2 has the same orientation as Ui, U2. If w is any other vector
perpendicular to vi, then by Tll.l there exists / such that w = tv2. If Vi, w
is a right-handed orthonormal system, then
|vi, w| = |vi, <v2| = i|vb v2| = t = 1.

Sec. 46] ROTATIONS 181
Hence w = v2; i.e., v2 is unique. We now have
«1*1 + «2*2 = (CUI + SU2)yi + (-ЛЦ + CU2)y2
= (0Ί _ ■0'2)ui + (sy! + cy2)u2.
The vector equation is equivalent to
xi = <У\ - sy2, *2 = -9Ί + cy2
and these scalar equations define the rotation.
EXAMPLE 46.1
Apply the rotation for which с = %, s = % to the equation
4*!2 + 24*!*;! + 11*22 - 20 = 0.
Note that this is the equation obtained as the result of the translation of
Example 43.1. If we substitute
x\ = hi ~ Ъъ *2 = \y\ + \уг,
the equation becomes
4(Ь - Ы2 + 24(b - b2)(bi + Ы + 1Kb + Ы2 - 20
= &(bi2 ~ 24у1У2 + \6y22) + U(\2yi2 - 7у1У2 - \2yi)
+ Ш^6У12 + 24^2 + 9>22) - 20
= W?i2 + 0 - W?22 - 20
= 20V - 5y22 - 20 = 0,
and hence
2 2
y\ У2_ =
1 4
The curve is thus a hyperbola. To draw the curve, first draw the vectors
Vl = ^U! + \U2, V2 = - \Ui + |u2
constituting the new axes. Note that it is easier to draw the vectors 5vi =
4щ + 3u2, 5v2 = — 3ui + 4u2, first and then measure off unit lengths along
these vectors to form vi, v2. Next draw the asymptotes relative to the new
axes and locate the points where the curve cuts the axis y2 = 0. Then sketch
the curve. Figure 43.1 shows the curve. In Section 48 we shall show how
to choose the appropriate rotation to remove the cross-product term and
produce the simplification.

182 TRANSFORMATIONS OF COORDINATES {Ch. 4
Questions
1. Given that vi, V2 is a right-handed orthonormal system and vi = [c, s]. Obtain
an equation relating с and s. 2. Show how to obtain the components of V2 and show
that V2 is uniquely determined by vi. 3. Derive the rotation equations.
Problems
1. Show that the rotation for which с = 1/λ/2, s = Ι/λ/2 transforms the
equation *ι*2 — 2 = 0 into the equation
tf _y£ = ,
4 4
Draw the old axes and the new axes. Draw the usual rectangle, the diagonals
forming the asymptotes, and draw the hyperbola. Note that, for the curve to
be in its correct position relative to the old axes, the sides of the rectangle must
be parallel to the new axes.
2. Perform the same rotation as in Problem 1 on the equation 5*i2 + 6*1*2 +
5*22 — 8 = 0, and draw the curve in its correct position relative to the old
axes.
3. Show that the rotation x\ = cy\ — syt, *2 = sy\ + суг transforms the equation
£.2 _|_ χΛ = „2 intn th#» #>^iiiatirtn it.2 _l_ λιλ2
4. Show that the rotation for which с = a\/\^a\2 + a^, s = аъ/у/а-р + "22
transforms the equation mxi + 02*2 + 03 = 0 into the equation у/<цг + °22 У\
+ a3 = 0.
5. Show that the rotation for which с = 0, s = 1 transforms the equation
a2 b2
into the equation
b2 ^ a2
6. Apply the rotation for which с = — 1, χ = 0 to the equation of Problem 5.
7. Apply the rotation for which с = 0, s = — 1 to the equation of Problem 5.
47. FURTHER PROPERTIES OF ROTATIONS
A rotation can be visualized as follows: Draw the axes Ui, U2 with origin
0. Place a sheet of semitransparent paper over the drawing and trace the
vectors Ui, u2 on this upper sheet. Hold the point О on the upper sheet fixed,
by holding a sharp-pointed object at this point. Then turn the upper paper
through an angle about О and transfer the traces of ui, U2 (in their rotated

Sec. 47] FURTHER PROPERTIES OF ROTATIONS 183
positions) back to the lower sheet. Assign the labels vi, v2 to the transferred
vectors. The vectors vi, v2 are called the rotated vectors Ui, U2, respectively.
Any vector a can be rotated in this manner. To accomplish this, draw a on
the lower sheet and place the upper sheet in its original position. Then trace
a on the upper sheet, rotate the upper sheet through the same angle as before,
and transfer the trace of a back to the lower sheet. The vector thus obtained
is the rotated vector a, and we denote this vector by R(a), where R is written
in boldface to remind us that the result of the rotation is a vector. In particular
R(ui) = Vi = cui + su2, R(u2) = v2 = — nil + cu2.
We denote the rotation by R. A rotation R is a law that assigns a unique
vector R(a) to each vector a. Such a law is called a function, or more
specifically, a vector function. Of course a rotation is a
very special law, and we shall study its special sR(o)+(R(b)
properties.
A transformation that reverses the
orientation of the orthonormal system can be visualized
as being produced by turning the transparent
sheet over, with reverse side up, before
transferring the vectors back to the lower sheet.
Let a, b be two vectors and s, t two scalars,
and let us see how R transforms the vector
as + bt. Draw the triangle used in constructing
the sum as + bt, place the semitransparent sheet
again in its original position, and trace the
complete figure on this sheet. Then rotate again
through the same angle and transfer the figure in Figure 47.1
its rotated position to the lower sheet. All
distances and angles between lines of the figure are preserved by these
operations, and the vectors a, as, b, bt, as + bt are transformed respectively into
R(a), R(as) = sR(a), R(b), R(b/) = /R(b), R(a* + Ы) = *R(a) + /R(b).
See Fig. 47.1. We have the following definitions.
D47.1. A vector function R, which is such that R(as + bt) = sR(a.) + tR(b) for
every pair of vectors a, b and every pair of scalars s, t, is said to be linear. A linear
vector function is called a matrix.
D47.2. A rotation is a matrix which transforms the orthonormal system щ, u2 into a
right-handed orthonormal system.
When R(ui) = Vi is known, then R(a) is determined for every vector
a = [βι, a2], since R(u2) = v2 is determined by the condition that vi, v2 is an

184 TRANSFORMATIONS OF COORDINATES (Ch. 4
orthonormal system with the same orientation as Ui, U2, and since
R(a) = R(aiUi + a2u2) = aiR(ui) + a2R(u2)
= aiVi + a2v2.
EXAMPLE 47.1
Let
R(ui) = fui + ^u2 and a = 2ui — u2.
Then
R(u2) = - £ui + fu2
and hence
R(2u, - u2) = 2R(ui) - R(u2)
= 2(§и, + И) - (-FUi + fu2)
= 2ui + u2.
Let us see how to visualize the performing of a rotation such as that of
Example 46.1. Consider a point X of the given curve. The curve and the
point X are not rotated but remain fixed on the lower sheet. However,
the axes ui, u2 are rotated and become R(ui) = Vi, R(u2) = v2. The point X
acquires the new coordinates y\, y2 relative to the rotated axes.
Questions
1. Describe the way in which a rotation can be visualized. 2. What is the meaning
of R(a)? 3. What is a vector function? 4. Show (by intuitive argument) that the
interpretation of a rotation implies that R(aj + bi) = R(a)x + R(b)i. 5. Define a
matrix. 6. Define a rotation. 7. Describe how to visualize the performing of a rotation
on an equation representing a curve.
Problems
1. Let a = [2, — 1], b = [1, 2], and let R be the rotation such that R(ui) =
[H, %\, R(u2) = [-%, Ш. Compute R(a), R(b) and show that R(a)-R(b) =
a«b.
2. Let v, = [л/3/2, У2\, v2 = [-Ц, Vb/2], w, = [Ц, л/3/2], w2 = [-л/3/2,
У2\, and let R be such that R(ui) = vb R(u2) = v2. Show that R(wi) = u2,
R(w2) = —ui.
3. Let vi, v2, wi, w2, and R be defined as in Problem 2. Show that R(vi) =
Wi, R(V2) = W2.
4. Let Vl = [l/\/2, 1/V2], v2 = [-i/y/2, 1/V2], and let R(ui) = vb
R(u2) = v2. Show that R(vi) = u2, R(v2) = — щ.

Sec. 48] REMOVING THE CROSS-PRODUCT TERM 185
5. Let vi = [c, s], v2 = [ — s, c], v'i = [c, —s], v'2 = [s, c], and let R(ui) =
vi, R(u2) = v2 where c2 + s2 = 1. Show that R(v'i) = щ, R(v'2) = u2.
6. (a) Let a = [αϊ, α2], b = [b\, i2] and let vi, v2 be a right-handed orthonormal
system. Show that
(viai + ν2α2)·(νιέι + v2i2) = a«b.
(b) Show that if R is such that R(ui) = vb R(u2) = v2, then
R(a)«R(b) = a-b.
7. Let vi, v2, v'i, v'2, R be defined as in Problem 5, and let a = [αϊ, α2], b =
[*i, bt]. Show that
R(a) = [cai — sa2, sa\ + ca2] = [a· v'i, a-v'2]
R(b) = [b.v'„b-v'2].
8. Let vi, v2, v'i, v'2 be defined as in Problem 5. Show that the transpose of
I v'i, v'2| is |vb v2|.
9. Let vi, v2, R, a, b be defined as in Problem 7. Show that
|R(a),R(b)| = |vi,v2|.|a,b| = |a,b|.
Hint: Use the results of Problems 7 and 8 together with T19.1.
48. REMOVING THE CROSS-PRODUCT TERM
Consider a quadratic equation in which the first-degree terms have been
removed by a translation. The equation has the form
«ιι*ι2 + 2ai2*i*2 + a22*22 + Озз = 0,
and it can be rewritten in the form
αιι*ι2 + 2αΐ2*ι*2 + я22*22 + Озз = (<Ίι*ι + αΐ2*2)*ι
+ (Ol2*l + 022*2)*2 + «33 = 0-
If ai = [an, β12], а2 = [αι2, ягг], х = [*ι, *г], then
«11*1 + 012*2 = Λι·Χ, αι2Χ\ + 022*2 = »2**>
and hence
Oll*l2 + 2αΐ2*1*2 + 022*22 + ^33 = (»ΐ·«)*1 + (»2·«)*2 + ^33
= (а^! + а2*2)·* + озз = О-
Let
ai*i + a2*2 = A(x),

186 TRANSFORMATIONS OF COORDINATES (Ch. 4
where A is written in boldface type to remind us that the resulting expression
is a vector. We now have a law that assigns to each vector χ the vector A(x) =
ai*i + a2*2, and such a law is a vector function. We denote this function by
A. Furthermore if x, у are any two vectors and s, t any two scalars, then
A(ix + /y) = A(uiCr*i + tyi) + u2(sx2 + ty2))
= »i(sxi + tyi) + a2(sx2 + ty2)
= s{*ixi + a2*2) + '(a^ + л2у2)
= jA(x) + /A(y).
That is, A is linear and therefore a matrix. It is not, however, a rotation
except in a trivial case. We can now write the quadratic in the compact
form
αιι*ι2 + 2ai2*i*2 + a22*22 + озз = Α(χ)·χ + азз = 0.
EXAMPLE 48.1
Consider the equation of Example 46.1, namely,
4*12 + 24*1*2 + И *22 - 20
= (4*! + Ux3)xi + (12*! + \\x2)x2 - 20 = 0.
If ai = [4, 12], a2 = [12, 11], χ = [xu x2], then
4xi + \2x2 = a^x, 12*! + 11*2 = a2-x,
4*!2 + 24*,*2 + 11*22 - 20 = (a,-x)x, + (a2-x)*2 - 20
= (ai*i + λ2χ2)·χ — 20
= Α(χ)·χ - 20 = 0,
where A(x) = ai*i + a2*2-
Recall that we performed the rotation
X = *iU! + *2«2 = )>lVl + У2У2,
where
vi = [f,a v2 = [-ii]
on this equation. We now perform this rotation as follows:
A(x) = A(vij/i + v2y2) = A(vi)>! + A(v2)>>2,

Sec. 48] REMOVING THE CROSS-PRODUCT TERM 187
and since A(ui) = [4, 12], A(u2) = [12, 11], we have
A(Vl) = A(|Ul + |u2) = |A(u,) + £A(u2)
= |[4, 12] + f[12, 11] = [ψ, ψ] = 20[|, *] = 20v1;
A(v2) = A(-±Ul + |u2) = -iA(Ul) + |A(u2)
= -£[4, 12] + |[12, 11] = [ψ, -ψ] = -5[-ί, |] = -5Vl.
Hence
A0>iV! + У2У2) = >ΊΑ(νι) + >>2A(v2)
= 20j>ivi - 5y2v2
and
Α(χ)·χ - 20 = Α(νι>Ί + v2yz)'(yiyi + У2У2) ~ 20
= (20)^1 - 5y2v2)'(yiyi + У2У2) - 20
= 20V - 5y22 - 20.
The relation A(vi) = 20vi, A(v2) = —5v2, is described in D48.
D48. // A is a matrix, b a scalar, and ν a vector such that
A(v) = bv,
then b is called an eigen-value and ν an eigen-vector of A.
The reason for the success of the particular orthonormal system vi, V2
of Example 48.1 is that vi, v2 are eigen-vectors. The corresponding
eigenvalues are respectively 20, —5. We shall develop a method of finding an
eigen-value b and a unit eigen-vector ν = ηΐγ + su2 of a given matrix A.
We wish to have
A(v) = A(oi! + su2) = cui + ia2
= bv = bail + bsu2,
i.e.,
c(ai — iui) + s(a2 — bu2) = 0,
where c, s are not both zero since c2 + s2 = 1. By T17.1 this is possible if and
only if |ai — iui,a2 — bu2\ = 0. Thus
Iai — 6m, a2 — iu2| = |ab a2 — iu2| — Ь\щ, а2 — iu2|
= |aba2| - i|*i, «21 ~ b\ui,a2\ + i2|ui,u2|
= |ai,a2| - (|ai,ua| + |uba2|)6 + b2 = 0.
This quadratic equation in b is called the characteristic equation of A.

188 TRANSFORMATIONS OF COORDINATES (Ch. 4
EXAMPLE 48.2
In Example 48.1 we have ai = [4, 12], a2 = [12, 11], and hence
|ab u2| =
4 0
12 1
l*i, *г1
= 4, |u1;a2| =
=
4 12
12 11
= -
1 12
0 11
100.
= 11.
Thus the characteristic equation is
-100 - 15ό + b2 = (b - 20) (b + 5) = 0
and the roots of this equation are bi = 20, b2 = — 5. We next find c\, s\ such
that c2 + s\2 = 1 and
fi(»i - Mi) + Л («a - Ma) = cJ-16, 12] + „[12, -9]
= 4Cl[-4, 3] - 3„[-4, 3] = 0.
This vector equation is satisfied if and only if Лс\ — 3ji = 0 or c\ = (%)„.
We satisfy the additional condition c2 -\- s2 = \ as follows:
and hence s2 = *%$ or sx = ±^5- We choose the plus sign although the
minus sign could equally well be chosen. Then s\ = ^, c\ = (%)s\ = %.
Similarly,
c2(ai - Mi) + s2(a2 - i2u2) = c2[9, 12] + s2[\2, 16]
= 3c2[3, 4] + 4j2[3, 4] = 0.
Hence c2 = ( — %)s2, s2 = ±%. This time it is necessary to choose the plus
sign in order for the vectors vi = [ci, „], v2 = [c2, s2] to form a right-handed
orthonormal system. We now have
A(vi) = iivi = 20vi, A(v2) = b2v2 = -5v2
as in Example 48.1.
Α(χ)·χ - 20 = 20^2 - 5y22 - 20,
Questions
1. Show that if the first-degree terms of a quadratic in x\, хг are missing, then the
equation can be written in the form Α(χ)·χ + "η = 0, where A(x) = ai*i + лгхг.
2. Show thai the vector function A defined in Question 2 is linear and hence is a

Sec. 49] PROPERTIES OF MATRICES 189
matrix. 3. Define an eigen-value and an eigen-vector of a matrix. 4. Show that b is an
eigen-value and that there exists a corresponding eigen-vector of a matrix A if and
only if b satisfies the characteristic equation of A. 5. Illustrate how to find a unit
eigen-vector when ai = [4, 12], a2 = [12, 11], b = 20.
Problems
1. Remove the cross-product term from the equation 8*i2 — \2х\хг — 8*г2 —
20 = 0 by means of a rotation. Find the rotated axes.
2. Remove the cross-product term from the equation 5*i2 + 6*1*2 + 5*22 — 8 = 0
by means of a rotation. Find the rotated axes.
3. Show that if A(w) = bw, where w^ 0, and if ν = w/1 w |, then A(v) = bv.
That is, show that if there exists a nonzero eigen-vector corresponding to the
eigen-value b, then there exists a unit eigen-vector corresponding to b. Show
that —v is an eigen-vector.
4. Let ai = [011,021]) »2 = [012,022], a'l = [011,012], a'2 = [021,022]; i.e., let
|a'i, a's| be the transpose of |ai, аг|. Compute the components of the vector
A(x) and show that these components are a'i-x, aV«. Thai is, show that
А (ж) = [a'i-x,aVx].
5. Let ai, аг, a'i, a'2 be defined as in Problem 4. Show that
A(x)-y = (aVx)* + (a'2-x)jr2 = A'(y)-x,
where A'(y) = a'iyi + л'гуг.
6. Let А(дг) = ai*i + агдгг, where ai = [оц, 012], аг = [αΐ2, ац]. Compute | ai, иг |,
| ui, аг |, | ai, аг |, and show that the characteristic equation of A is
(оц022 — αΐ22) — («и + агг)* + b2 = 0.
Show that the discriminant of this equation is
on2 — 2ац022 + а222 + 4αΐ22 = (оц — агг)2 + 4αι22 έ 0.
49. PROPERTIES OF MATRICES
If b is a real root of the characteristic equation of a matrix A, then it
follows from T17.1 that there exist c, s not both zero such that
c(ai — bill) + s(a2 — bu2) = 0,
or equivalently, such that
A(v) = by,

190 TRANSFORMATIONS OF COORDINATES (Ch. 4
where ν = [с, s], and it should be clear from the method of Example 48.2
how to satisfy the additional condition c2 + s2 = 1. One may wonder,
however, whether the roots are necessarily real and whether we were lucky
to find eigen-vectors vi, v2 forming an orthonormal system. To answer these
questions, we need to study some additional properties of matrices.
Let A be an arbitrary matrix. The following notations display respectively
the vectors ai, a2 determining A and the components of these vectors.
a r ι Γ011 °121
A = [a,, a2] =
1-021 022-1
The transpose A' of A is defined in precisely the same way as the transpose of
a determinant is defined. Thus
А' Г ' Ί P11 °211
A = [a u a 2\ = \
L0l2 022-1
In terms of the transpose we have the following:
A(x) = [a11; a2i]xi + [al2, a22]x2
= [вц*1 + 012*2, α21*1 + 022*2]
= [а'гх, a'2-x];
i.e., we have T49.1.
T49.1. A(x) = [a'i-x, a'2-x], where [а'^а'г] is the transpose of Α..
If x, у are any two vectors, then
A(x)-y = [л\-х,л'2-х]-[Уиу2]
= (aV»))i + (а'г'Х^г = (»Ί>Ί +a'2>>2)·*
= A'(y)·*.
Thus we have T49.2.
T49.2. A(x)-y = A'(y)-x/or every two vectors x, y.
Recall that a matrix defined in terms of a quadratic equation is such
that ai2 = a2x and hence such that A' = A. Such a matrix is said to be
symmetric. We have T49.3.
T49.3. A(x)«y = A(y)«x for every x, у if A is symmetric.

Sec. 49] PROPERTIES OF MATRICES 191
We show next that the roots of the characteristic equation of a symmetric
matrix are real. To obtain this equation, we compute the following:
|ai,u2| =
e„ 0
0l2 1
= an, |ui,a2| =
1 Д12
0 a22
= a22,
|a1; a2| =
an a12
α12 a22
α11α22 — a12
Hence the characteristic equation is
(aua22 - ai22) - («и + 022)* + b2 = 0,
and its discriminant is
(«и + a22)2 — 4(aua22 — αλ22) = αλ2 + 2aua22 + a22 — 4αΜα22 + Aa12
= an2 - 2аца22 + а22 + 4α122
= (on - огг)2 + 4αι22.
Since this discriminant can never be negative, the roots are real. Thus we
have T49.4.
T49.4. The roots of the characteristic equation of a symmetric matrix are always
real.
Note that the roots are equal if and only if the discriminant is zero and
hence if and only if au = a22 and ai2 = 0. In this case
*i = onUi, a2 = аци2,
A(x) = anUi*i + anu2*2 = aux,
and hence every vector χ is an eigen-vector. Also in this case the quadratic
equation is
aii(*i2 + *22) + 033 = 0-
If 0ц and дзз have opposite signs, then this equation represents a circle, but
if an and дзз are both positive or both negative, then there are no real values
of χι, x2 which satisfy the equation.
If A is a symmetric matrix, then there exist b\, Vi such that | vi | = 1 and
A(vi) = biv\. There also exists a vector y2 snch that vi, v2 is a right-handed

192 TRANSFORMATIONS OF COORDINATES {Ch. 4
orthonormal system. Moreover, since |vi, V2I = 1 И 0, there exist x, у such
that
A(v2) = xvi + yv2-
However,
νι·Α(ν2) = ν2·Α(νι) = iivrv2 = 0
= vi · (xvi + yv2) = x,
and hence
A(v2) = yv2-
We let jv = i2. Then
A(vi) = iivi, A(v2) = i2v2.
That is, vi, v2 are eigen-vectors forming a right-handed orthonormal system.
Questions
1. Define the transpose of a matrix. 2. Show that A(x) = [x*a'i, x*a'2] if [a'i, a'2]
is the transpose of A. 3. Show that A(x)«y = Α'(γ)·χ and that Α(χ)·γ = Α(γ)·χ if
A is symmetric. 4. Show that the characteristic equation of a symmetric matrix has
real roots. 5. Discuss the case in which the roots are equal. 6. Show that a symmetric
matrix has eigen-vectors vi, V2 forming a right-handed orthonormal system.
Problems
1. Let 61, i2 be eigen-values and vi, V2 the corresponding eigen-vectors of a
matrix A. Show thai
νι·Α(ν2) — ν2·Α(νι) = (b2 — *i)vi'V2 = 0,
and hence that vi ± v2 if b\ у* Ьг.
2. Let A = [ai, a2], A' = [a'i, a'2] and let bi, b2 be any two vectors. Show lhat
A(bi) = [bi-a'i, b,-a'2l, A(b2) = [b2-a'b b2-a'2]
and that (see T19.1)
|A(bi), A(b2)| = |ab a2|-|bb b2|.
3. Let А, В be two matrices and let С be such that C(x) = A(B(x)). Show that С
is a vector function and that it is linear, and hence that it is a matrix. Let С be
denoted by AB. Show that
AB = [ab a2][bb b2] = , ■
La2bi a2D2J
Note the analogy with the product of two determinants.

Sec. 50]
RECAPITULATION
193
4. Let R = [vi, V2] be a rotation and wi, W2 an orthonormal system. Show lhat
(see Problem 6 of Section 47)
R(wi)«R(wi) = wi-wi = 1,
R(wi)-R(w2) = wi-w2 = 0,
R(w2)-R(w2) = 1,
and hence that R(wi), R(w2) is an orthonormal system. Show that (see
Problem 2 of this section)
|R(wi), R(w2)| = |vi, v2| · |wi, w2| = |wb w2|
and that R(wi), R(w2) has the same orientation as wi, w2.
5. Let vi, v2 and wi, w2 be right-handed orthonormal systems, and let R =
Kb v2]> S = [wi, w2] be rotations. Show that
RS = [R(w,), R(w2)]
is a rotation. Write down the formula for this rotation analogous to that of
Problem 3.
6. Find the matrix A associated with the equation x2 + у2 — 25 = 0. Find the
eigen-values and eigen-vectors.
50. RECAPITULATION
If we perform the translation χ = у + h (where χ = [χι, Χ2], у =
bi'^Lh = [Λι, Λ2]) on the quadratic
aii*i2 + 2a12*i*2 + 022*22 + 2αι3*ι + 2α23*2 + "зз
= (an*i + βΐ2*2)*1 + («12*Ί + 022*2)*2 + 2(α13·<1 + 023*2) + «33
= (a!*! + a2*2)*(ui*i + u2*2) + 2(ai3U! + α23«2)·(«ι·<ι + «2*2) + озз
= Α(χ)·χ + 2а3'Х + α33 = О,
where А(х) = а^ + а2*2, а! = [а1Ь а12], а2 = [αλ2, α22], а3 = [а13, я23].
The equation becomes
А(у + h)-(y + h) + 2a3-(y + h) + a33
= (A(y) + A(h)).(y + h) + 2a3-(y + h) + язз
= A(y)-y + (A(h)-y + A(y)-h + 2a3-y) + (A(h)-h + 2a3-h + a33)
= A(y)-y + 2(A(h) + a3)-y + (A(h) + 2a3)-h + язз = 0,

194 TRANSFORMATIONS OF COORDINATES (Ch. 4
since A(y)-h = A(h)«y. If |ai, аг| = 0, we cannot in general eliminate the
first-degree terms in yi, y2 by a translation, and in this case we first perform
a rotation. An example of this will be considered later. If |ai, a2\ И 0, we
can solve
A(h) + a3 = ai/z! + а2Л2 + a3 = 0 or A(h) = —a3
for hi, h2 (i.e., for h). The constant term is then
(A(h) + 2a3)'h + 033 = (-a3 + 2a3)«h + 033 = a3«h + 033
and the equation becomes
A(y)«y + a3-h + азз = 0.
Next perform the rotation Uiyi + И2У2 = ViZi + V2Z2, where Vi, v2 are
eigen-vectors forming an orthonormal system and bi, b2 are the corresponding
eigen-values. Then
A(vi) = biVi, A(v2) = b2v2,
and the equation becomes
А(у)*У + a3«h + a33 = Afc^i + z2v2) · (ziVi + z2v2) + a3«h + 033
= (zibiVi + z2b2v2)'(ziVi + z2v2) + a3«h + азз
= bizi2 + b2z2 + a3«h + дзз = 0.
EXAMPLE 50.1
Simplify the equation
17*!2 - \2Xlx2 + 8*22 + 18*! - 4*2 = 0
and draw the curve. In Problem 1 of Section 43 we found
a, = [17, -6], a2 = [-6, 8], a3 = [9, -2], h = [-§, -£]
and the translated equation. We compute the following
|a1; u2|=17, |Ul,a2| = 8, |a,,a2|=100.
The characteristic equation is
100 - 256 + b2 = (b - 5)(b - 20)
and the roots are bi = 5, b2 = 20. The simplified equation is
W + b2z22 + a3-h + азз = W + 20z22 -5 = 0,
or

Sec. 50]
RECAPITULATION
195
To draw the curve, we find the new axes. Thus we find Vi = [c\, si] such
that
fi(ai - Mi) + ^i(»2 - *iu2)
= Cl([17, -6] - 5[1, 0]) + .,([-6, 8] - 5[0, 1])
= Cl[12, -6] + л[-б, 3] = 2ci[6, -3] - л[б, -3] = 0.
Hence 2ci — Ji = 0, C\2 + Si2
W
1, Ci = \/y/b, Si = 2/y/b (where
plus sign is chosen). Thus
Vl
" LV5' Vs
]■
V2 = [v]'i;\
Draw the vectors Ui, u2 with origin 0,
and then draw equivalent vectors Ui,
u2 with origin #=(-%, -)%).
Next draw the vectors v5 Vi = [1, 2]
ν 5 v2 = [—2, 1] with origin Η and Figure 50.1
measure off unit lengths along these
vectors to obtain vi, v2. Draw the ellipse relative to this last axis system.
Figure 50.1 shows the drawing. The unit is 2 cm.
EXAMPLE 50.2
Simplify the equation
9л2 + 24*!*2 + 16*22 + Юл + 5*2 + 6
= (9*! + \2χ2)χλ + (12*! + 16*2)*2 + 2 (5*i + - *2 J + 6 = 0
rve. We have
ι, = [9,12], a2 = [12,16], a3 = [δ, ^J ·
and draw the curve. We have
Ι*ι, U2I = 9, |ui,a2|
16,
|aba2| = 0.
Since|a1; a2| = 0, we perform the rotation before, the translation. The
characteristic equation is
0 - 25b + b2 = b(b - 25) = 0,

196 TRANSFORMATIONS OF COORDINATES (Ch. 4
and the roots are bi = 0, b2 = 25. The corresponding c\, s\ are such that
^(a! - ^ui) + ^(a2 - i2u2) = ti[9, 12] + c2[12, 16]
= Ъф, 4] + 4JJ3, 4] = 0.
Hence Si = — (%)сь c2 + sf = (2Мб)с12 = 1- Let us choose C\ = + У-
Then., = -y5,Vl= [y5, -%],y2 = [%, У5], A(v,) = 0, A(v2) = 25v2.
9Xl2 + 24Xlx2 + \6x22
= Α(χ)·χ
= A(vi>>i + \2у2)' (у\У\ + v2>>2) = 2bv2y2-{\iyi + v2y2)
= 25y22.
The rotation equations are
*1 = |^1 + %У2, Х2 = - hi + 5У2,
and hence
Юдц + 5x2 = 10(bi + Ы + 5(-Ь + Ьг)
= 5n + \0y2.
The quadratic becomes
25y22 + 5У1 + \0y2 + 6 = 0.
We now perform the translation y\ = z\ + A1; y2 = z2 + h2, and obtain
25(z2 + h2)2 + 5(Zl + A0 + 10(z2 + h2) + 6
= 25z22 + (50A2 + 10)z2 + 5z! + (25Л22 + 10A2 + 5hx + 6).
We choose h2 so that
50A2 + 10
and h\ so that
25Л22 + 10A2 + bh + 6
The equation becomes
25z22 + 5Zl = 0 or z22= (-£)zi.
To draw the curve, draw ui, u2 with origin О and then draw the axes Vi, v2
also with origin 0. Plot the new origin (—1, — У,) with respect to the axes
Vi, v2 and draw equivalent vectors vi, v2 at the new origin. Draw the curve
with respect to this final axis system. Figure 50.2 shows the drawing. The
unit is 5 cm.
= 0 or h2 = -\
= +5hx +5 = 0 or A, = -1.

Sec. 50] RECAPITULATION 197
u2
v2
^- ·-",
V' N.
v,
Figure 50.2
Questions
1. Show that the quadratic equation
αιι*ι2 + 2αΐ2*ι*2 + <J22*22 + 2aitxi + 2ацХ2 + агг = О
can be written in the form Α(χ)·χ + 2a3«x + агг = 0. 2. Show that if χ = у + h,
this equation becomes
A(y)-y + 2(A(h) + a3)-y + (A(h) + 2a3)-h + a33 = 0.
3. Show that if |ai, аг| ^0, lhen there exists an h such that A(h) + a3 = 0 and
that for this h the quadratic becomes
A(y)«y + a3«h + a33 = 0.
4. Show that if b\, bz are eigen-values and vi, V2 are corresponding eigen-vectors
forming a right-handed orthonormal system, then the rotation \i\y\ + игуг =
vi^i + V2Z2 transforms the quadratic into the equation
bizi2 + i2*22 + a3-h + агг = 0.
5. What is the order of procedure when | ai, аг | =0?
Problems
1. Simplify the equation
β*,2 - 12*!*2 - 8*22 - 40*1 - 20*2 - 20 = 0
by a translation and rotation. If you choose h = 10, vi = [3/vTO, -1/л/Т0],
the equation becomes V/2 — z22/2 = 1. To draw the unit vector vi, draw

198 TRANSFORMATIONS OF COORDINATES (Ch. 4
VTOvi = [3, —1] and measure off a unit length along this latter vector to
form vi. If you choose *i = —10, vi = [1/λ/ΐΟ, 3/λ/ΐΟ], the equation
becomes — z\2/2 + z^/2 = 1. Draw all three sets of axes and draw the curve
in its correct position relative to the original axis system.
2. Simplify the equation
5*i2 + 6*i*2 + 5*22 - 8*i - 24*2 + 24 = 0.
Find the coordinates of the new origin relative to the original axis system and
find the rotated axes. A drawing is not required.
3. Show that with an appropriate translation and rotation the equation
9*i2 + 24*i*2 + 16*22 - 20*i + 15*2 + 25 = 0
becomes one of the following:
V = ±Z2, Z22 = ±Zb
Draw all three sets of axes and draw the curve.
4. Show that by appropriate translation and rotation the equation
*i2 + 2*i*2 + *22 - 4*i - 5 - 0
becomes one of the following:
zi2 = ±V2zt, *i2 = ±y/2zv
5. Let A be the vector function such that
А (ж) = ai*i + a2*2 + аз*з
for every vector χ — [*i, *2, *з]. Show that A is linear and hence that it is a
matrix.
6. Define the transpose A' = [a'i, а'г, а'з] of A. Show that
A(x) = [x-a'i, «-а'г, *«а'3]
and that
А(ж)-у = А'(у)-ж
for every pair of vectors χ, γ.
7. Let A(u,) = [4, 12, 0], A(u2) = [12, 11, 0], A(u») = [0, 0, 5], ν = [с, s, 0].
Find c, s, b such that (see Example 48.1)
A(v) = bv.
Show that из is an eigen-vector and find the corresponding eigen-value. Find
b\, Ьг, bz, vi, V2, V3 such that vi, V2, V3 is a right-handed orthonormal system and
A(vi) = iivi, A(v2) = *2V2, A(v3) = *3ν3.

Sec. 51] QUADRATIC EQUATIONS Ш THREE DIMENSIONS 199
8. Let A be such that A' = A and let
A(m) = ab A(u2) = a2, A(u3) = *3u3,
where i3 is real. Use a result of Problem 6 to show that
ai-u3 = A(u3)-ui = О, аг-и3 = О.
Show that this result implies that ai, Лг are in the plane of ui, U2, that A defines
a two-dimensional symmetric matrix transforming each vector in the plane of
ui, иг into another vector in this plane, and that the remaining eigen-values of
A are real.
51. QUADRATIC EQUATIONS IN THREE DIMENSIONS
The general quadratic in three dimensions has the form
on*i2 + a22*22 + дзз*з2 + 2012*1*2 + 2013*1*3 + 2д2з*2*з
+ 2au*i + 2α24*2 + 2аз4*з + «44
= («11*1 + «12*2 + 013*з)*1 + (θΐ2*1 + «22*2 + Я23*з)*2
+ (θΐ3*1 + 023*2 + 033*з)*3
+ 2(aU*l + 024*2 + 034*з) + 044 = 0.
If
»ι = [βιι, βΐ2, β13], a2 = [012,022,023], a3 = [β13, а23, взз]>
»4 = [θΐ4, 024, 034], * = [*1, *2, *з],
then
Oll*l + Ol2*2 + 013*3 = Λχ·Χ, Οΐ2*1 + 022*2 + «23*3 = а2'Х,
013*1 + 023*2 + «33*3 = »3'Χ, «14*1 + «24*2 + «34*3 = а4'Х,
and hence the quadratic can be written in the form
(ai-x)*i + (a2-x)*2 + (а3-х)*з + 2a4-x + a44
= (ai*, + a2*2 + a3*3)«x + 2a4-x + a44 = 0.
Let A be the vector function that assigns to each χ the vector
A(x) = ai*i + a2*2 + a3*3.
The proof that A is linear is almost identical with that for the corresponding
two-dimensional vector function. That is, A is a matrix. We can now write the
quadratic in the form
Α(χ)·χ + 2a4-x + a44 = 0.

200 TRANSFORMATIONS OF COORDINATES (Ch. 4
To remove the first-degree terms from this quadratic, we perform the
translation
x = [*i, *2, *з] = ЬиУ2,Уз] + [hi, h2, h3] = у + h.
As in the two-dimensional case, A(y)-h = A(h)-y. We shall assume this
result in this section and prove it in Section 52. The translation thus
produces the equation
Α(χ)·χ + 2a4-x + Я44
= A(y + h)-(y + h) + 2a4-(y + h) + a44
= A(y)-y + (A(h)-y + A(y)-h + 2a4-y) + (A(h)-h + 2a4-h + a44)
= A(y)-y + 2(A(h) +a4).y + (A(h)-h + 2a4-h + a44) = 0.
We shall consider only the case |ai, аг, аз| Ά 0. We can then determine
h so that
A(h) + a4 = uihi + a2/z2 + а3Л3 + a4 = 0,
or
A(h) = -a4.
For this value of h the new constant term is
(A(h) + 2a4)'h + a44 = a4«h + a44,
and the equation becomes
A(y)«y + a4-h + a44 = 0.
We hope to be able to remove the cross-product terms by means of a
rotation
У = У1Щ + y2U2 + У3Щ = ZiVl + Z2V2 + Z3V3,
where vi, V2, V3 is an orthonormal system. That is, we want this orthonormal
system to be such that Vi, V2, V3 are eigen-vectors of A. If b is a real
eigenvalue and ν = [/, m, n] is a unit eigen-vector of A, then
A(v) = ai/ + Я2т + *3" = bo = Wui + bmvi2 + Ьпиз,
or equivalently,
/(a! - iui) + m(a2 - bu2) + n(a3 - bu3) - 0,
where /, m, η are not all zero, since I2 + m2 + r? — 1. These conditions can
be satisfied if and only if
I a! - oui, a2 - bu2, a3 - iu3| = 0.

Sec. 51] QUADRATIC EQUATIONS IN THREE DIMENSIONS 201
Thus
|a! - iu,,a2 - bu2, a3 - iu3| = 0
= I a,, a2, a31 - (| u,, a2, a31 + | a,, u2, a31 + | a,, a2, u31 )b
+ (|al5 u2, u3| + |uba2, u3l + !u,, u2, a3|)i2 - |u,, u2, u3 | b3.
This is the characteristic equation of A, and the roots b\, b2, i3 of this cubic
are the eigen-values. In this section we shall assume that these eigen-values
are real and that the corresponding eigen-vectors vi, v2, v3 can be chosen so
as to form a right-handed orthonormal system. Then
A(vi) = iivi, A(v2) = b2v2, A(v3) = i3v3
and the rotation transforms the quadratic as follows:
A(y)-y + a4-h + a44
= A(z,v, + z2v2 + ζ3ν3)·(·ζινι + z2v2 + z3v3) + a4-h + a44
= (zibiVi + z2b2v2 + z3i3v3)«(z!Vi + z2v2 + z3v3) + a4«h + a44
= bizi2 + b2z2 + b3z32 + a4-h + a44 = 0.
EXAMPLE 51.1
Simplify the equation
5*,2 + 11*22 + 2*32 - 16*,*2 + 20*,*3 + 4*2*3 + 20л:, + 4*2 + 4*3 - 16
= (5*, - 8*2 + Ккз)*! + (-8*, + 11*2 + 2*з)*2
+ (10*, + 2*2 + 2*3)*з + 2(10*, + 2*2 + 2*3) -16 = 0.
We have
a, = [5, -8, 10], a2 = [-8,11,2], «β = [10, 2, 21, a4 = [10, 2, 2].
We determine h so that A(h) + a4 = 0, and find h = [0,0, —1]. The
translation χ = у + h produces the new constant term a4-h + a44 = —18
and the equation
5?i2 + Wy22 + 2y32 - 16^,^2 + 20^з + 4y2y3 - 18 = 0.
To find the characteristic equation, we make the following computations:
|ab a2,a3| = -1458, |u,, u2, u3| = 1,
|u,,a2, a3| = 18, |a,, u2, a3|= -90, |a,,a2, u3| = -9,
|ai,u2,u3| = 5, |u,,a2, u3| = 11, | u,, u2, a3| = 2.

202 TRANSFORMATIONS OF COORDINATES (Ch. 4
The characteristic equation is
-1458 + 8U + Ш2 - b3 = 0.
In Example 53.1 we shall show how to solve this equation. For the present
it is sufficient to verify by direct substitution that the roots are b\ — 9,
i>2 = — 9, i3 = 18. Thus the rotation produces the equation
9Zl2 - 9z22 + 18z32 - 18 = 0,
or
2 2 2
•Zl *2 *3 _
2 2 1
Let us find an eigen-vector corresponding to the eigen-value bi — 9;
i.e., let us find /, m, η such that I2 + m2 + n2 = 1 and
/(a! - 9ui) + m(a2 - 9u2) + я(а3 - 9u3)
= /[-4, -8, 10] + m[-8, 2, 2] + я[10, 2, -7] = 0
or equivalent y, such that
-4/ - 8ет = -Юл, -8/+2m = -2n, 10/+ 2m = 7n.
The first two of these equations imply that / = л/2, m = n, and since there
exists a solution of all three equations, these values must satisfy the third
equation. It is easy to check that they do. We must also have
2 η 2
I2 + m2 + n2 = - + n2 + n2 = — = 1,
4 4
and hence η — ±%. We choose η — % and obtain / = }/&, m = %, η = %,
and the eigen-vector [J^, %, %]. Similarly, we find v2 = [~ H, ~ И, %]
corresponding to b2 — — 9. It is easy to check that
V3 = v,Xv2 = [f, -|, \\
is an eigen-vector corresponding to i3 = 18.
Questions
1. Show that the general quadratic equation in three dimensions can be written
in the form А (ж)· ж + 2a4«x + α44 = 0. 2. Show that the translation χ — у + h
transforms this equation into
A(y)-y + 2(A(h) + a4)-y + A(h)-h + 2a4-h + au = 0.

Sec. 51] QUADRATIC EQUATIONS Ш THREE DIMENSIONS 203
3. Show how to eliminate the first-degree terms by a proper choice of h and show
what the resulting equation becomes. 4. Show how to obtain eigen-values of a matrix
A. 5. Show that if the eigen-values are real and the eigen-vectors form a right-
handed orthonormal system, then the general quadratic can be written in the form
*i*i2 + *2^22 + b)Z32 + a4-h + ait = 0.
Problems
1. By a translation remove the first-degree terms from the equation
*i2 + *22 + X)2 ~ 4*! + 4*2 + 2*з = 0.
Find the center and radius of this sphere.
2. Show that with an appropriate rotation the equation 4*i*2 + *з2 — 1 = 0
becomes 2yi2 — 2угг + Уз2 ~ 1 = 0 or a similar equation obtained by
interchanging the coordinates yi, уг, уг- Note that in giving the answer, we have also
given the eigen-values. Find corresponding eigen-vectors forming a right-
handed orthonormal system.
3. Given that the eigen-values of the matrix
A =
6 0 0
0 0 3
L0 3 0.
are 6, 3, —3. Find eigen-vectors forming an orthonormal system.
4. Given the orthonormal system vi = []4, гЛ, гЛ\, V2 = [гЛ, Η, —2Λ\, V3
\-2А,гЛ,-хЛ\ and the matrix
A =
-2 2 2
2 4 1
2 1 4
(a) Given that an eigen-value is 6, show that vi is a corresponding eigen-vector.
(b) Show that νι·Α(ν2) = 0 = νι·Α(ν3). See Problem 8 of Section 50.
(c) Show that А(уг) = Зуз and А(уз) = 3v2.
(d) Let w = cv2 + sv3. Show that A(w) = iw if b = 3, с = \/у/2, s =
\/V2, and also if b = -3, с = -1/V2, s = 1/V2.
Given the curve хг = *i3 — 3*i2 + 5. Plot the points for which x\ = —2, —1,
0, 1, 2, 3, and connect these points by a smooth curve. Estimate the point at
which the curve crosses the *i-axis. Note that the *i-coordinate of the point of
crossing is a root of the equation х\г — 3*i2 + 5 = 0.

204 TRANSFORMATIONS OF COORDINATES (Ch. 4
52. THREE-BY-THREE MATRICES
We shall use the following alternative notations for a matrix A.
A = [αϊ, o2, a3] =
Oil Ol2 Ol3
021 022 023
L«31 a32 033
The expression on the right is made up of three rows and three columns and
is called a three-by-three (or 3 X 3) matrix. Section 48 is concerned with
2 X 2 matrices. There are 2X3 matrices, 4X1 matrices, etc., but they
will not be considered in this text. The transpose of A is the matrix
A' = [a'b a'2, a'3] =
~0ц 021 «ЗГ
0l2 022 «32
L0l3 a23 033J
We can write A(x) in terms of the transpose as follows:
A(x) = [an, a2i, 03i]*i + [oi2, 022, o32]*2
+ [αΐ3, 023, азз]*з
= [θΐ1*1 + 012*2 + 013*3, 021*1 + 022*2
+ 023*3, 031*1 + 032*2 + 033*3]
= [Л·*, a'2-x,a'3-x].
Hence for any two vectors x, у we have
A(x)-y = [а'х'Х, а'2-х, *'з-*]-[уиУ2,уз]
= (а'гхЫ + (а'г'хЬ'г + (а'з-эфз
= (a'lJVi + a'2>>2 + а'3>>з)-х = A'(y)-x.
The conditions 012 = 021, 013 = 031, агз = 032 imposed on a matrix A
associated with a quadratic equation imply that A' = A; i.e., that A is
symmetric. A symmetric matrix is such that
A(x)-y = A(y)-x
for any two vectors x, y.
The characteristic equation of a 3 X 3 matrix is a cubic with real
coefficients, and it can be proved that every such cubic has at least one real root.
The proof is beyond the scope of this text, but the result can be made plaus-

Sec. 52] THREE-BY-THREE MATRICES 205
ible as follows: A cubic polynomial has the form a^x3 + οι*2 + ^2* + 03,
where a0 И 0. We consider the case in which во is positive and show that
the polynomial has a negative value when χ is a numerically large negative
number (say, χ = —b) and has a positive value when χ is a large positive
number (say, χ = с). We then indicate why it is reasonable to suppose that
there is a value of χ between —b and с for which the polynomial is equal to
zero. This value is the real root of the cubic
equation a$x + axx2 + a^x + 03 = 0. The
reasoning can be pictured with the aid of
Fig. 52.1, which shows the graph of a typical
cubic polynomial. The polynomial can be
written in the form
*2 =
a0x'
3 + αιχ2 + a2x + a3
"'(
oo -\ r- — +
χ x2 ' x3/
Root
Figure 52.1
When χ is a very large positive or
numerically large negative number, every term in
the parenthesis with the exception of До is
very small. In fact χ can be chosen so large
that the parenthesis has the same sign as
a0; i.e., so large that the parenthesis is
positive. For such an χ the polynomial has
the same sign as x; i.e., it is positive when χ
is positive and negative when χ is negative.
It is reasonable to believe that the
polynomial cannot pass from negative to positive
values without taking on the value zero (Fig. 52.1 shows the graph of a
typical cubic polynomial). The point at which it is zero is the root.
If A is a symmetric matrix, then its characteristic equation has at least
one real root Й3 and there is a corresponding unit eigen-vector V3. Let us
assume tentatively that there exist vectors Vi, V2 such that vi, V2, V3 is a
right-handed orthonormal system (see Problems 4 and 5 of Section 30).
Then there exist scalars x, y, ζ such that
A(vi) = xvi + yv2 + zv3.
Moreover,
ζ = ν3·Α(νι) = νι·Α(ν3) = b3Vi'V3 = 0.
Let χ = Сц,у = C2i- Thus
A(vi) = cnVi + C21V2 = cb

206 TRANSFORMATIONS OF COORDINATES (Ch. 4
and similarly, there exist ci2, C22 sucn that
A(v2) = C12V1 + C22V2 = c2.
Then Α(νι)·ν2 = C21 = Α(ν2)·νι = ci2 and A(xiVi + x2v2) = *iCi + x2c2;
i.e., A defines a two-dimensional symmetric matrix that transforms a vector
in the plane of Vi, V2 into another vector in this plane. We have seen that
such a matrix has real eigen-values b\, b2 and corresponding eigen-vectors
Wi, W2 forming a right-handed orthonormal system in the plane. That is,
wi> w2) V3 is a right-handed orthonormal system such that
A.(wi) = biwu A(w2) = b2w2, A(v3) = b3v3,
and bi, b2, Ьз are real eigen-values. It remains to show how to construct the
vectors vi, v2. If V3, Ui are linearly independent, then v3 X ui И 0 and
vi = V3 X U1/IV3 X Ui I is such that |vi| = 1 and Vi«V3 = 0. Let v2 =
V3 X Vi- Then vi, V2, V3 is an orthonormal system and this system is right-
handed, since
ν2·ν2 = 1 = ν2· ν3 X vi = I v2, v3, vi I = I vi, v2, v31.
If V3, Ui are linearly dependent, then V3, U2 are linearly independent, and
we can proceed as in the previous case.
Questions
1. Define the transpose A' of a 3 X 3 matrix A. 2. Show that if A' = [a'i, a'2, a'3],
then A(x) = [λ'ι·χ, л'2'х, л'з'х] and that A(x)«y = A'(y)«x for every two vectors
x, y. 3. Define a symmetric matrix and show that A (x) · у = A (y) · χ if A is symmetric.
4. Discuss the existence of a real root of a cubic equation. 5. Show that if a
symmetric matrix has one real eigen-value, then it has three real eigen-values, and that
the corresponding eigen-vectors can be chosen so as to form a right-handed ortho-
normal system.
Problems
1. Let vi, V2, V3 be a right-handed orthonormal system and let R be the rotation
such that R(u0 = vi, R(u2) = v2, R(u3) = v3. Let x = [*i, дг2, дгз], у =
[уиУьУз]· Show that
R(x)'R(y) = x-y.
See Problem 6 of Section 47.
2. Let R be defined as in Problem 1 and let wb w2, wj bea right-handed ortho-
normal system. Show that R(wi), R(w2), R(w3) is a right-handed orthonormal
system. See Problem 4 of Section 49.

Sec. 53] CUBICS, HIGHER-DEGREE EQUATIONS 207
3. Let vi, V2, v3, wi, w2, w3 be defined as in Problem 2 and let R = [vi, V2, v3],
S = [wi, w2, w3]. Show that
RS
Vi'Wi V l'W2 V l'W3
v'i'Wl v'2*W2 vVw3
Lv'j'Wl v'3-W2 v'3-W3
and that this product is a rotation. See Problem 5 of Section 49.
4. Show that [v'i, v'2, v'3]' = [vi, V2, v3] and that
"1 0 0"
R'R = 0 1 0 = [ui, u2, u3].
.0 0 1.
5. Show that R'R(x) = χ for every vector x.
6. Let R = [иг, ui, — u3], S = [112, u3, ui]. Show that
RS(ui) = R(S(ui)) = R(u2) = ui, RS(u2) = -u3, RS(u3) = u2.
Compute SR(ui), SR(u2), SR(u3), and show thai RS ^ SR.
7. (a) Show that for every vector χ there exist scalars y\, уг, уг such that
x = viyi + У2У2 + Узуз,
where vi, V2, v3, R are defined as in Problem 1.
(b) Show that
(c) Show that
K'W = Ьиуг, yd and RR'(«) = »·
RR'
1
0
0
0
1
0
0-
0
1.
= [ui, u2, u3].
53. CUBICS AND HIGHER-DEGREE EQUATIONS
In Sections 51 and 52 we have been concerned with matrices whose
characteristic equations are cubics. We shall show how to solve a cubic
when its coefficients are integers and it has at least one rational root. Thus
we consider an equation
до*3 + a\x2 + a2x + a3 = 0
having integers as the coefficients a0, аь а2> аз and having at least one rational
root. A rational number is a ratio χ = p/q of two integers p, q. We assume
that the fraction p/q is in its lowest terms; i.e., that p and q have no factors
in common other than +1 and — 1. We shall show that if p/q is a root, then

208 TRANSFORMATIONS OF COORDINATES (Ch. 4
the numerator p must be a factor of the constant term 03, and the
denominator q must be a factor of the leading coefficient a0. Hence there are usually
only a few rational numbers that can possibly be roots, and we need only to
substitute these numbers into the equation to see which ones satisfy. To show
that this is the case, we substitute χ = p/q and note that the resulting
equation is equivalent to
= a0p3 + 0lp2q + a2pq2 + a3q3 = 0,
and hence to
Р(<*ор2 + <*ipq + a2if) = -a^q3.
Thus p times the integer a0p2 + axpq + α2^ is equal to — a^q3; i.e., p is a
factor of — Дз03. It follows that p must be a factor of 03, since p and q have only
the factors +1, —1 in common, and the product — a^q3 can be decomposed
into its prime factors in only one way unless it is zero. We omit the proof
of the uniqueness of such a factorization. Similarly
q{a\p2 + a2pq + az<f) = -"op3,
and hence q must be a factor of a0. Analogous results hold for equations of
higher degree. Thus, if the fraction p/q (reduced to lowest terms) is a root
of an equation of degree n, then p must be a factor of the constant term and
q a factor of the leading coefficient.
Suppose that we have found a root r of, say, a quartic:
a0x* + αγχ3 + a2x2 + a3x + a4 = 0.
Let us see how to find the remaining roots. If we divide the left-hand member
of the quartic equation by χ — r, we obtain a quotient and a remainder.
Thus
αλχ* + αυ3 + a2x2 + a3x + aA b4
= b0x3 + bix2 + b2x + b3-\
where bix3 + b\x2 + b2x + 63 denotes the quotient and Й4 the remainder.
If we multiply both sides of this equation by χ — r, we obtain
a0x4 + aix3 + a2x2 + a3x + a4 = (box3 + b^x2 + b2x + b3)(x — r) + b4.
If we substitute χ = r in this last equation, the right-hand member (and
hence the left-hand member) becomes 0 + b4 = b4. Hence, if г is a root of
the given quartic, the remainder b4 must be zero and the quartic can be
written in the factored form

Sec.53] CUBICS, Η IG HE R-DE G RE Ε EQUATIONS 209
a0x* + a^x3 + a2x2 + a3x + a4 = (b0x3 + b^x2 + b2x + b3)(x - r) = 0.
The remaining roots are therefore roots of the cubic
blX* + blX2 + b2x + b3 = 0.
A similar result holds for an equation of an arbitrary degree.
EXAMPLE 53.1
Solve the characteristic equation
-b3 + Ш2 + 81* - 1458 = 0
of Example 51.1. The possible numerators are the factors of the constant
term -1458, namely, 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458,
and the negatives of these numbers. The possible denominators are the factors
of the leading coefficient, namely, +1 and —1. Since a fraction can be
assigned a positive denominator and its sign determined by the sign of the
numerator, we need consider only the denominator +1. We discover that 9
is a root and
-b3 + 18b2 + 8\b - 1458
= -b2 + 9b + 162.
b - 9
The candidates for the roots of this quadratic are 1, 2, 3, 6, 9, 18, 27, 81, 162,
and their negatives. We find that
-b2 + % + 162
= -b + 18.
b + 9
Hence the cubic can be written in the factored form
-b3 + Ш2 + 81* - 1458 = - (b - 9)(b + 9)(b - 18) = 0
and the roots are 9, — 9, 18.
EXAMPLE 53.2
Solve Ъх* + Ίχ3 - \Ъх2 - 35* - 10 = 0. If the equation has a rational
root, then the numerator must be a factor of 10, namely, 1, 2, 5, 10, — 1, —2,
— 5, or — 10, and the denominator must be a factor of 3. The positive factors
of 3 are 1 and 3. The candidates for rational roots are 1, 2, 5, 10, }/&, %, %,
1%, and their negatives. We test these candidates and discover that —2 is a
root. We perform the division
Ъх* + 7x3 - \Ъх2 - 35* - 10
= 3*3 + χ2 - 15* - 5.
* + 2

210 TRANSFORMATIONS OF COORDINATES [Ch. 4
The remaining roots must satisfy the equation obtained by setting this cubic
equal to zero. The candidates for the roots of the cubic are 1, 5, }/§, %, and
their negatives. We find that — \^ is a root and that
3x3 + x2 - 15* - 5
; = Ъх2 - 15
x + l
= з(х + Vs)(* - VI).
Hence
Ъх* + 7*3 - \Ъх2 - 35* - 10 = (χ + 2)(3λ^ + χ2 - 15* - 5)
= 3(* + 2)(χ + |)(* + Vl)(x - VI)
is a factorization of the quartic and the roots are — 2, — }/&, v5, — V 5.
A famous problem that remained unsolved for about 2000 years is that
of duplicating the cube. That is, we are given a segment representing an
edge of a cube and we are required to construct another segment
representing an edge of a cube with twice the volume. If we choose the length of
the given segment as a unit, then the given cube has volume 1. If the segment
to be constructed has length * relative to the chosen unit, then * must be such
that the corresponding cube has volume 2, and hence * must satisfy the
equation
x3 - 2 = 0.
If the segment of length * can be constructed, then * is said to be a constructible
root of this equation. A constructible root does not have to be rational, but
it can be proved that if a cubic has a constructible root, then at least one of its
roots must be rational. The proof involves only elementary mathematics
but is too long to justify inclusion in this text. The only candidates for
rational roots of the above cubic are ±1, ±2. Since none of these numbers is a
root, the cubic has no rational root and hence no constructible root. The
duplication of the cube is therefore an impossible construction with ruler and
compass. In Section 73 we shall use a similar method to show the impossibility
of trisecting the general angle.
Questions
1. Show that if a cubic with integer coefficients has a rational root reduced to
lowest terms, then the numerator is a factor of the constant term and the denominator
is a factor of the leading coefficient. Discuss the generalization of this result. 2. Show
that a quartic can be factored when one of its roots is known. Discuss the generalization.

Sec. 54] THE INTERSECTION OF A PLANE WITH A CONE 211
Problems
1. (a) Plot the points of the curve *2 = 2*i3 + 5*i2 — 4*i — 3 for which *i =
— 4, —2, —1, 0, 2, and sketch the curve. Use 1 cm as a unit on the *i-axis
and 1 dm (i.e., 10 cm) as a unit on the *2-axis.
(b) Solve the equation 2*3 + 5*2 — 4* — 3 = 0. If necessary correct your
drawing of part (a) in the light of your answer to (b).
2. (a) Plot the points of the curve *2 = *i4 + 2*i3 — 5*i2 — 4aci + 6 for which
xi = —4, —2, —1, 0, 2, and sketch the curve. Use the same units as in
Problem 1.
(b) Solve the equation *4 + 2*3 — 5*2 — 4* + 6 = 0 and correct your
drawing of part (a) if necessary.
3. Solve the equation — x3 + 6*2 + 9x — 54 = 0. See Problems 3 and 4 of
Section 51.
4. Simplify the equation *i2 + xi* — 3*з2 — 2*i*2 + 6*1*3 + 6*2*з — 6 = 0 by a
translation and rotation. Find eigen-vectors forming an orthonormal system.
5. Show that the equation 8*3 — 6* + 1 =0 has no constructible root. This
equation is concerned with the trisection of an angle. See Section 73.
6. Show that -4/2 and "°УЪ are irrational numbers.
54. THE INTERSECTION OF A PLANE WITH A CONE
We shall use a rotation and a translation to show that the intersection of
the cone „
V + *22 - xi = 0
with the plane
— sx\ + с*з — b = 0
(where c2 + s2 = 1) is a conic section. If we perform the rotation
xi = cyi - sy3, *2 = y2, *3 = -0Ί - 9"з,
the equations become respectively
(θΊ - ■О'з)2 - (m + 0"з)2 + Уг = (f2 - s2)y2 - Acsyxyz
-(c2-s2W+y22
= 0
and
-s(cyi - sy3) + c(syi + cy3) - b = y3 - b = 0.
First consider the case с = s = 1/л/2. The equations of the cone and plane
become
-2yiys + У2 =0, Уз - b = 0.

212 TRANSFORMATIONS OF COORDINATES [Ch. 4
We now perform the translation
У\ = Zl, У2 = Z2, Уг = *3 + Ь
and obtain . . , 2 «
-2Zl(z3 + 6) + z22, z3 = 0,
and these equations are equivalent to
-2iz1+z22 = 0, z3 = 0.
This latter pair of equations represents a parabola in the plane z$ = 0.
With a little algebra one can show that the intersection of the plane and
cone is an ellipse when c2 — s2 > 0 and hyperbola when c2 — s2 < 0.
However, we shall content ourselves with considering special cases in which the
equations are comparatively simple. Thus, when с = 0, s = 1, the equations
of the plane and cone (in terms of the coordinates y\, y2, уз) become
respectively 2_i_2i_2n l η
-yi + У2 + yz =0, y3 - b = 0.
We again perform the translation yi = zlt y2 = z2,y3 = z% + b, and obtain
-zl2 + (zz + b)2 + z22 = 0, z3 = 0,
and these equations are equivalent to
b2 = z,2-z22, z3 = 0.
This pair of equations represents a hyperbola in the plane z$ = 0. When
с = 1, s = 0, we obtain the equations
yi2 ~ уъ + У22 = 0, уз - b = 0,
and the translation y\ = z\, y2 = z2, yz = z$ -\- b produces the equations
zi2 ~ (z3 + b)2 + z22 = 0, z3 = 0.
The equivalent equations
z,2 + z22 = b2, z3 = 0,
represent a circle in the plane Z3 = 0. When с = -\/b/2, s = У2, we have
У* аЛ l 2 , 2
— V3)1)3 - -y3 + y2
Ы-уз^)2 . 2 , 2 n . 0
= 2уъ + y2 = 0, y3 - b = 0.
We perform the translation, y\ = z\ + iv3, ^2 = z2> >*з = z3 + b, and
obtain
— - 2(z3 + b)2 + z22 = 0, Z3 = 0.

Sec. 54] THE INTERSECTION OF A PLANE WITH A CONE 213
Figure 54.1
Figure 54.2
The equivalent equations
+ z22 = 2b2, z3 = 0,
represent an ellipse in the plane z$ = 0.
Figure 54.1 shows a parabolic cross-section and an elliptic cross-section
of a cone. The parabolic cross-section is obtained when the intersecting plane
is parallel to a ruling. Figure 54.2 shows a hyperbolic cross-section.
Questions
1. Discuss the intersection of the cone xi2 + *22 — *з2 = О with the plane
— sx\ + схг — b = 0, where c2 + s2 = 1. Consider the cases (а) с = 1/λ/2 =
s; (b) с = 0, s = 1; (с) с = 1, s = 0; (d) с = Vl/2, s = У2. Perform the
appropriate rotation and translation for each case.
Problems
1. Discuss the intersection of the cone 4(*i2 + *22) — *s2 = 0 with the plane
— sx\ + cx3 — b = 0, where c2 + s2 = 1. Consider the cases (а) с = 1/λ/5,
s = 2/\/5; (b) с = 0, s = 1; (с) с = l/\/2 = s. Perform the appropriate
rotation and translation for each case.

TRIGONOMETRY
5
55. COSINES AND SINES
Trigonometry is concerned with relations between distances and angles.
An angle will be defined in terms of a pair of rays, and a ray is defined as
follows: If a is any vector and С is any point, then the set of points X for
which CX has the form a/, where / is a non-negative number, is called a ray
and С is called the origin of the ray. The ray is determined by the origin С and
by the direction and sense of a, but is independent of the magnitude of a.
Thus, if χ- = | a | / then
CX = a/ = (a/|a|)x, where χ ^ 0.
That is, the unit vector a/1 a | and the origin С determine the same ray as a
and C. It turns out to be necessary to consider two different kinds of angles.

Sec. 55]
COSINES AND SINES
215
D55.1. An oriented angle is an ordered pair of rays, and an unoriented angle is an
unordered pair of rays.
A pair of nonzero vectors a, b determines a pair of rays, which in turn
determines two oriented angles denoted by (a, b), (b, a) and one unoriented
angle denoted by {a, b}, where {a, b} = {b, a}. We have
<*·Μ-(ϊίτϊϊϊ> '-^-fe'iiil-
Moreover (a, b) = (b, a) if and only if
a b
Ы=]ь\'
In Section 1 we have seen that the direction and sense of a vector in a
plane (say, a unit vector v) can be specified by specifying the angle that ν
forms with the unit vector щ, where ui extends horizontally to the right. A
measure will be assigned to this angle and will be interpreted as the amount
of rotation required to carry ui into v. We can see intuitively that an opposite
rotation is required to carry ν back into ui. That is, we are concerned with
an oriented angle (ui, v), since this angle is different from (v, ui). If the
measure is positive, then the rotation is in a positive or counterclockwise
sense, and if the measure is negative, the rotation is in a negative or clockwise
sense. A given position ν can be obtained by many different rotations. Thus
suppose that the position ν results from a rotation through 30 °. The same
position can be obtained by rotating through 390°; i.e., by rotating ui all
the way around through 360° and then rotating it an additional 30°.
Similarly, this position can be obtained by rotating through 750°. The position
can also be obtained by a clockwise rotation; for example, a rotation through
-330° or through -690°.
Thus every oriented angle is assigned a measure in degrees. The measure
is not unique, but if a given angle has both the measure A ° and the measure
B°, then A must differ from В by a multiple of 360. Conversely, if an angle
has measure A ° and В differs from A by a multiple of 360, then B° is a
measure of that angle. The angle (щ, ui) has measure 0° and also the measures
360°, 720°, -360°, etc. One of the measures of (m, u2) is 90°. We assume
that to each real number A there corresponds a unique unit vector ν such
that (ui, v) has measure A °. We shall consider how the degree measure of
(ui, v) is related to the components of ν in the directions of ui, иг- These are
the simplest relations between distance and angle.

216
TRIGONOMETRY
[Ch. 5
If A is a given real number and ν = [с, s] is the unique unit vector such
that (ui, v) has measure A", then the components c, s of ν are called
respectively the cosine and sine of A °, and we write
с = cos A", s = sin A°.
We thus have a law that assigns to each number A the unique number
cos A". Such a law is called a function (or more specifically, a scalar function).
This particular function is called the cosine. Similarly, the sine is the law that
assigns to each number A the number sin A °. The sine and cosine are called
trigonometric functions. The numbers cos A °, sin A ° are related by the equation
c2 + s2 = (cos A0)2 + (sin A0)2 = vv = 1.
It is customary to use the simpler notation
(cos A °)2 = cos2 A °, (sin A °)2 = sin2 A °,
and in terms of this notation we have T55.1.
T55.1. cos2^° + sin2^0 = 1.
If (ui, v) has measure A" and ν is a unit vector, then
ν = [cos A °, sin A °] = ui cos A ° + иг sin A °.
The following intuitive discussion will help us visualize cos A" and sin A".
With the aid of a protractor we can draw ν so that (ui, v) has measure
A". We can then draw the vectors ui cos A°, иг sin A", whose lengths are
respectively |cos^4°|, |sin Л°|. These lengths can be estimated by
measurement, and the algebraic signs of cos A °, sin A °, can readily be
determined by noting that these numbers are the components of v. In this way
we can obtain approximate values of cos A ° and sin A ° for a given number
A. Figure 55.1 shows the construction for the angles of measures 50° and
140°. Note that cos 50° sin 50°, sin 140° are positive, whereas cos 140° is
negative.
In Problem 1 you will be asked to estimate the cosines and sines of 0°,
10°, 20°, 30°, 40°, 120°, 210°, 330°, 360°. The constructions can be
combined into a single figure as follows: Since the tip of ν lies on a circle with
center at the origin and radius 1, we begin by drawing this circle. Choose
1 dm (10 cm) as a unit. An ordinary sheet of typewriter paper is wide
enough to contain such a circle. Then draw the radii, making the required
angles with щ. Drop perpendiculars from the extremities of these radii to
the line of the axis Ui, thus forming a series of right triangles. The lengths of
the sides of these triangles give | cos A ° |, | sin A ° | for the various numbers A.
Figure 55.2 shows the construction but is about a quarter the size of the figure

Sec. 55]
COSINES AND SINES
217
u2 sin 140
u2 sin 50°
210
υ ι cos 140 и, cos 50
Figure 55.1
Figure 55.2
you are asked to draw. It is possible to measure to the nearest hundredth
of a decimeter (nearest millimeter) and hence to obtain two-place accuracy
by such a drawing. However, unless you have had experience in drawing
and in using a protractor, you will probably not achieve this degree of
accuracy. Do the best you can. At a later point we shall show how to check
your answers with tables.
Questions
1. What is a ray? 2· What is an angle? 3. Distinguish between an oriented angle
and an unoriented angle. 4. Consider an oriented angle of the form (щ, ν), where ν is
a unit vector in the plane of щ, иг- What relations are there between the various
degree measures assigned to such an angle? 5. Name various degree measures assigned
to the angle (щ, ui). 6. Name one degree measure assigned to (щ, иг). 7. Does a
degree measure of (щ, ν) determine ν and its components? 8. Define cos A", sin A".
9. Define the functions sine and cosine. 10. Show that cos2 A" + sin2 A" = 1. 11. How
can cos A" and sin A° be estimated by geometrical construction? 12· Describe how to
combine various constructions of this type into a single figure. 13. How are the
algebraic signs of cos A" and sin A" determined?
Problems
1
Estimate the sine and cosine of 0°, 10°, 20°, 30°, 40°, 120°, 210°, 330°, 360°
(two-place accuracy) by an appropriate construction. Arrange your results in a
table. Be sure that the proper algebraic signs are attached.
Draw an equilateral triangle with one vertical side and with the opposite
vertex to the left of this side. Let the sides of this triangle be each of length 1 and
choose 4 cm as a unit. Draw the horizontal altitude of this triangle. Use this

218
TRIGONOMETRY
[Ch. 5
figure to find the exact value of sin 30°. Solve the equation cos2 30° +
sin2 30° = 1 for cos 30° and attach the appropriate algebraic sign to the
solution. Check your answers with Problem 1.
3. Draw an equilateral triangle with its bottom side horizontal and draw the
vertical attitude. Use this figure to find exact values of cos 60°, sin 60°.
4. Draw the figure used in estimating cos 45°, sin 45°. The hypotenuse of this
triangle is of length 1 and the remaining two sides are of equal lengths. Let χ
denote the common length. Use the pythagorean theorem to show that 2x2 = 1.
Solve for χ and find exact values of cos 45°, sin 45°.
5. Compute cos 90°, sin 90°. Note that u2 = [0, 1].
56. ROTATION THROUGH AN ANGLE
A right-handed orthonormal system vi = [c, s], v2 = [ — s, c] defines a
rotation R. This is the rotation through the angle (ui, vi). Since R is a
matrix, we have the alternative notations
R= [vi,v2]= [' J·
The transpose of R is the matrix
R' = [v'i, v'2] =
where v'i = [c, — s], v'2 = [s, c]. Since
vVv' = 1 = v'2V2, νΊ·ν'2 = 0,
|νΊ,ν'2| = |νι,ν2|= 1,
it follows that v'i, v'2 is a right-handed orthonormal system and hence that
R' is a rotation.
Let a = [αϊ, a2], b = [bi, b2], be any two vectors. Then
R(a)-R(b) = (β,ν, + α2ν2)·(έινι + i2v2)
= a\b\ + a2i2 = a«b.
That is, a rotation preserves scalar product. In particular, if Wj, w2 is any ortho-
normal system, then
R(w,)-R(w,) = wi-wi = 1, R(w,)-R(w2) = wrw2 = 0,
R(w2).R(w2) = 1,
.;:

Sec. 56] ROTATION THROUGH AN ANGLE 219
and hence R(wi), R(w2) is also an orthonormal system. To determine the
orientation of this system, note that T49.1 implies that
R(wi) = [v'i-Wi,vVwi], R(w2) = [v'i-w2, v'2-w2],
and hence by T19.1 we have
|R(w,),R(w2)| =
Vl'Wl V i'W2
v'2-Wi v'2-w2
|v i, v2| · |wb Wi\
= |vb v2| · \wu w2| = \wi, w2|.
Therefore R(wi), R(w2) has the same orientation as Wj, w2. That is, R
transforms an orthonormal system Wj, w2 into the orthonormal system R(wi), R(w2)
and preserves the orientation.
If Wi, w2 is right-handed, it defines a rotation S that transforms an
arbitrary vector χ = [xi, x2] into the vector
S(x) = *lW! + *2W2.
Moreover, R(wi), R(w2) is also right-handed and defines a rotation that we
denote by RS. The rotation RS transforms χ into
RS(x) = ^(wO + *2R(w2) = R(xlwl + x2w2), = R(S(x)),
and hence is the resultant of the rotation S followed by the rotation R. We
call this resultant the product of R and S. It is given by the formula of T56.1.
T56.1
Γν'ι-Wi v'i-w2"l
RS = [vi, v2][wb w2] =
Lv2'Wi V2'W2J
2'Wj V2'W2J
Note the analogy between the product of two determinants and the product
of two rotations, but also note that the determinants | Vi, v2|, | v'i, v'2| are
equal, whereas the rotations [vi, v2], [v'i, v'2] are not in general equal.
If (ui, Vi) has measure A°, then R is called the rotation through A", and if
vi = [f, s], then с = cos A°, s = sin A°. Similarly, if (щ, Wi) has measure
B°, then S is the rotation through B°, and we have
Vi = [cos A °, sin A °], v2 = [ — sin A °, cos A °],
w, = [cos B°, sin B°], w2 = [- sin B°, cos B°],
v'i = [cos A °, - sin A °], v'2 = [sin A °, cos A °].
T56.2 follows from T56.1.

220
TRIGONOMETRY
[Ch. 5
T56.2
"cos B° - sin B°
.sin B° cos B°
Γ cos Λ ° -sin^°"irco
Lsin4° cos4°JLsii
[°
Lsi
cos A" cos B" — sin A° sin B° cos A° sin 5° — sin A" cos Z?°
sin Λ0 cos B° + cos Λ0 sin B° cos Л0 cos fi° - sin A" sin fi°
Since the right-hand member of these equalities is unchanged if A is
interchanged with B, it follows that
RS = SR;
i.e., that rotations in two dimensions are commutative. Compare Problem 6 of
Section 52.
We assume that measures are assigned to angles in such a way that the
rotation S through B° followed by the rotation R through A" results in the
rotation through B° + A" = A" + B°. Then the angle (ub R(w0) has
measure A ° + 5° and
R(wj) = [cos (A0 + B°),sin (A0 + B°)]
= [cos A ° cos B° — sin A ° sin B°, sin A ° cos B° + cos A ° sin B°].
Hence we have T56.3.
T56.3
cos (A ° + B°) = cos A ° cos B° - sin A ° sin B°,
sin (A° + B°) = sin A ° cos B° + cos A ° sin B°.
Since [v'i, v'2]' = [vb v2], it follows from T56.1 that
Lv2'Vi v2'V2J L0 1.
[ui, u2],
where [щ, u2] is denoted by I and is such that
R'R(x) = I(x) = *!U! + ^2u2 = χ
for every vector x. That is, R transforms χ into R(x) and R' transforms R(x)
back into x. The resultant I = R'R is called the identity rotation or the rotation
through 0 °.
If R is the rotation through A" and R' the rotation through B°, then
R'R = I is the rotation through A" + B°, and hence A + В must be either

Sec. 56] ROTATION THROUGH AN ANGLE 221
zero or a multiple of 360. That is, the measure B° of the angle (ui, v'i) must
differ from — A" by at most a multiple of 360°, and one of the measures of
this angle must be — A". Hence
v'i = [cos — A°, sin — A0] = [cos A", — sin A°]
and we have T56.4.
T56.4
cos — A° = cos A", sin — A° = — sin A°.
It follows from T56.3 that
cos (B° - A") = cos -A°cosB° - sin -4°sin B°,
sin (B° - A") = sin -A°cos B° + cos - A" sin B°,
and these results together with T56.4 imply T56.5.
T56.5
cos (B° - A °) = cos A ° cos B° + sin A ° sin B°,
sin (B° - A °) = - sin A ° cos B° + cos A ° sin B°.
Recall that PI to P5 of Section 9 define a commutative group with
respect to the operation +. The vector 0 is called the identity of this group, and
it has the property that a + 0 = a. The vector — a is called the inverse of a,
and it has the property that a + (—a) = 0. This group is called an additive
group, since the operation is that of addition. A group in which the operation
is that of multiplication is called a multiplicative group. The two-dimensional
rotations form a multiplicative group and also a commutative group. The
product RS of two rotations is a rotation and RS = SR. It is not difficult to
show that (RS)T = R(ST). The identity is the rotation I, and it has the
property that RI = R. The inverse of R is the transpose R', and it has the
property that R'R = RR' = I.
Questions
1. Show that the transpose R' of a rotation R is a rotation. 2· Show that a rotation
preserves scalar product. 3. Show that a rotation transforms an orthonormal system
into another orthonormal system with the same orientation. 4. Show that the product
RS of two rotations is the resultant of the rotation S followed by the rotation R and
that this resultant is a rotation. 5. Obtain a formula for the product RS analogous to
the formula for the product of two determinants. 6. What is meant by the rotation

222
TRIGONOMETRY
[Ch. 5
through A"? 7. Let R be the rotation through A" and S the rotation through B".
Obtain a formula for RS in terms of the cosines and sines of A° and B". 8. Show that
RS = SR. 9. Obtain formulas for cos (A° + B"), sin (A° + B°). 10. Show that
R'R = [m, u2]. Discuss. 11. Obtain formulas for cos — A", sin — A". 12. Obtain
formulas for cos (B° — A"), sin (B° — A"). 13. Discuss the group properties of
rotations.
Problems
1. Let J = [иг, — ui] and let the product JJ be denoted by J2. Then J is the
rotation through 90° and J2 is the rotation through 90° + 90° = 180°.
Show that
J2(u,) = J(u2) = -u,, J2(u2) = -u2) J2(«) = -x
for every vector x. Compute cos 180°, sin 180°. Note that —ui = [—1,0|.
2. Let JJ2 = J3. Then J3 is the rotation through 270°. Show that J3(ui) = -u2,
J3(u2) = щ. Compute cos 270°, sin 270°.
3. Let JJ3 = J2J2 = J4. Then J4 is the rotation through 360°. Show that J4 = I.
Compute cos 360°, sin 360°.
4. Let R = [vi,_v2], S = [wb w2J, where v, = [V5/2, И], v2 = [~У2, Vl/2],
wi = [И, л/3/2], w2 = [-V3/2, И].
(a) Compute v'b v'2. (b) Show that R2 = S. (c) Show that R3 = RS = J.
(d) Let R be the rotation through A". Show that S is the rotation through
2A° and J is the rotation through ЗА", (e) Show that ЗА is equal to 90 or
90 + 360 or 90 + 720, etc., and hence that A = 30 or 150 or 270, or else
differs from one of these numbers by a multiple of 360.
5. Let
l/VI -l/VI
R =
Л/VI l/VI.
Show that R2 = J and that if R is the rotation through B", then В = 45 or
225 or differs from one of these numbers by a multiple of 360.
Substitute В = 90 in T56.5 and show thai cos (90° — A") = sin A",
sin (90° - A°) = cos A".
Substitute В = 180 in T56.5 and show that cos (180° - A") = -cos A",
sin (180° - A") = sin A".

Sec. 57]
REDUCTION FORMULAS
223
57. REDUCTION FORMULAS
In Problem 1 you will be asked to prove the following, called reduction
formulas.
T57.1. (a) cos (90° - A") = sin A", sin (90° - A") = cos A",
(b) cos (180° - A") = - cos A", sin (180° - A0) = sin A0,
(c) cos (180° + A0) = - cos A°, sin (180° + A") = - sin A°,
(d) cos (360° - A") = cos A", sin (360° - A0) = - sin A0,
(e) cos (360° + A") = cos A", sin (360° + A0) = sin A".
The values of cos A", sin A" for A between 0 and 90 are obtained from
Table 1 (Appendix) in the columns in which the word "value" appears at
the top. For the present we ignore those columns in which the symbol "log"
appears at the top. The table is accurate to three places. If A is between 0
and 45, inclusive, then A is found in the column on the left of the table. And
cos A° is found in the same line as A in the column headed "cos"; similarly
for sin A". U A is between 45 and 90, then A is found in the column on the
right, and cos A" is found in the same line as A in the column in which the
symbol "cos" appears at the bottom; similarly for sin^4°. The column
headed "cos" has the symbol "sin" at the bottom, and the column headed
"sin" has the symbol "cos" at the bottom. If A appears in the column on the
right, then 90 — A appears in the same line as A in the column on the left.
Thus cos A" is the same entry in the table as sin (90° — A") and sin A°
is the same entry as cos (90° — A°). This is in agreement with reduction
formulas (a).
If A is between 90 and 180 (say, A = 165), we note that 180 - 165 = 15,
or 180 — 15 = 165. Hence by formulas (b) we have cos 165° = — cos 15°,
sin 165° = sin 15°. If A is between 180 and 270 (say, A = 240), we note
that 240 - 180 = 60, or 180 + 60 = 240. Hence by (c) we have cos 240°
= - cos 60°, sin 240° = - sin 60°. If A is between 270 and 360 we use (d).
We have assumed that measures can be assigned to angles of the form
(ui, v) so that the various measures assigned to a given angle differ by
multiples of 360°; to each number A there corresponds a unique unit vector ν
such that (ui, v) has measure A"; the angle (ui, Ui) has measure 0°; the
angle (ui, U2) has measure 90°; and the product of a rotation through A"
and a rotation through 5° is a rotation through A" + B°. We make the
further assumption that if A is between 0 and 90, then cos A ° and sin A °
are both positive. It may seem obvious that such assignments are possible,
but the proof of the possibility is beyond the scope of this text. Since sin A"
is positive for A between 0 and 90, and 180 — A is between 90 and 180 when

224
TRIGONOMETRY
[Ch. 5
A is between 0 and 90, it follows from reduction formulas (b) that sin A" is
positive for A between 0 and 180. Since 180 + A is betwen 180 and 360 when
A is between 0 and 180, it follows from (c) that sin A° is negative for A
between 180 and 360. Similarly, by (b), cos^4° is negative for A between 90
and 180, and by (c) it is also negative for A between 180 and 270. By (d) it
is positive for A between 270 and 360.
You will be asked to check your answers to Problem 1 of Section 55 by
means of Table 1 and to correct when there is disagreement. A correct
answer is the nearest two-place number to the corresponding three-place
number obtained from the table. Note that the nearest two-place number to
0.356 is 0.36, whereas the nearest two-place number to 0.354 is 0.35. To
obtain the nearest two-place number to 0.355, you can toss a coin to decide
whether to call it 0.35 or 0.36. This process is called rounding off.
Questions
1. Describe how to find sines and cosines of various angles by means of the
Appendix tables. 2. Illustrate how to extend the tables to find sines and cosines of
angles with measures greater than 90°. 3. Fill out the following table by inserting a
plus sign when cos A" or sin A" is positive and a minus sign when it is negative.
A
cos A"
sin A"
0 to 90
90 to 180
180 to 270
270 to 360
Show how these algebraic signs are determined by the reduction formulas and the
assumption that cos A", sin A", are positive when A is between 0 and 90. 4. Describe
the process of rounding off.
Problems
1. Prove T57.1. See Problems 6, 7 of Section 56.
2. Check your answers to Problem 1 of Section 55 and correct when there is
disagreement.
3. Use the corrected results of Problem 2 together with the reduction formulas
to make out a complete table for cos A", sin A" for A = 0, 10, 20, etc., up
to 360.
4. Consider Problem 4(e) of Section 56. Use the table of Question 3 (not
Problem 3) to show that either A = 30 or differs from 30 by a multiple of 360.

Sec. 58]
ANGLES
225
5. Apply a similar method to Problem 5 of Section 56.
6. Substitute В = A in T56.3 and use T55.1 to show that cos 2A° = cos2
A" - sin2 A" = 1 - 2 sin2 A" = 2 cos2 A" - 1, sin 2A° = 2 sin A" cos /1°.
7. Use the result of Problem 6 to show that sin A" = ± y/{\ — cos 2Λ°)/2,
cos A" = ± \/(l +cos2^°)/2.
8. Given cos 45° = 1/V2 = sin 45°, cos 30° = V3/2, sin 30° = У2.
Substitute В = 45, A = 30 in T56.5 and find cos 15°, sin 15°.
9. Find cos 75°, sin 75°.
10. Find cos 114°, sin 231° with the aid of Table 1 and reduction formulas.
11. In Table 1 find A between 0 and 90 such that cos A" is nearest to 0.6.
58. ANGLES
Thus far we have assigned measures only to oriented angles of the form
(ui, w) where w is a unit vector in the plane of Ui, иг- In this section we
shall consider angles (a, b) defined by an arbitrary pair of nonzero vectors
a, b in two-dimensional or three-dimensional space. We remove the
restriction that the two rays have a common origin, by assigning to the angle
between an arbitrary pair of rays the same measure as to the angle between an
equivalent pair of rays (i.e., the rays determined by an equivalent pair of
vectors) with the common origin 0. Thus there is no loss in generality in
assuming that a, b are unit vectors with common origin 0. We consider first
the case in which (a, b) is an oriented angle in the plane of Ui, иг- The
measure assigned to (a, b) is defined to be the same as that assigned to (ui, w)
if and only if there is a two-dimensional rotation R such that
R(m) = a and R(w) = b.
Then R must be the rotation [vi, V2], where Vi, V2 is the right-handed ortho-
normal system such that vi = a = R(ui), and w must be the unit vector
[c, s] such that c, s are the scalars satisfying the vector equation
R(w) = R(aii + su2) = cv\ + sv2 = b.
Note that if S is an arbitrary two-dimensional rotation, then (S(a), S(b)) is
assigned the same measure as (a, b), since
SR(Ul) = S(a) and SR(w) = S(b).
Let a, b, с be unit vectors in the plane of ui, U2, and let us show that a
measure of (a, b) plus a measure of (b, c) is a measure of (a, c). We express
this result symbolically by the equation
(a, b) + (b, c) = (a, c).

226
TRIGONOMETRY
[Ch. 5
Since the addition of angle measures is required to be related to products of
rotations, we begin by showing that if a measure of (a, b) is A", then the
rotation through A ° transforms a into b. Thus, if A ° is a measure of (a, b)
and R, w are such that R(ui) = a, R(w) = b, then A" is a measure of
(ui, w); and hence if S is the rotation through A °, then S(ui) = w and
S(a) = SR(Ul) = RS(Ul) = R(w) = b.
Conversely, if S(a) = b, then
S(Ul) = SR'(a) = R'S(a) = R'(b) = w,
and hence S is the rotation through A °. Similarly, if a measure of (b, c) is
5° and Τ is the rotation through B°, then T(b) = с Finally, since
TS(a) = Т(Ь) = с
and TS is the rotation through A° + B°, it follows that A" + B° is a
measure of (a, c). This establishes the result. This result holds only for oriented
angles in a plane.
If a is any unit vector, then 0° is a measure of (a, a), since if R(ui) = a,
then (R(ui), R(ui)) = (a, a). If A° is a measure of (a, b), then — A" and
360° — A" are measures of (b, a), since
(a, b) + (b, a) = (a, a).
If J = [иг, — uj, then J is the rotation through 90° and hence 90° is a
measure of (a, J(a)). The product JJ is denoted by J2 and is the rotation
through 180°. Since
J2(u,) = J(u2) = -u,, J2(u2) = J(-u,) = -u2,
it follows that if a = [ab a2], then
J2(a) = ai(-u) + a2(-u2) = -a.
Hence 180° is a measure of the angle
(a,J2(a)) = (a, -a).
This angle is called a straight angle.
Let us attempt to generalize these concepts of angle measure to angles in
three dimensions. If a, b are unit vectors, then there exists a three-dimensional
rotation R and a unit vector w = cui + m2 such that R(ui) = a, R(w) =
b, and w satisfies the additional requirement that s is non-negative. Namely,
let
R = [vi, v2, v3], vi = a, h = Ь - evi, i = |h|,

Sec. 58]
ANGLES
227
where с is such that Vi'h = 0. If s И 0, let
h h
V2 = —Τ = -' V3 = Vi X V2,
h 8
and if s = 0, let Vi, v2, V3 be any orthonormal system such that Vi = a. In
either case,
h = sv2, Ь = cvi + h = cvi + jv2, w = oil + su2,
R(w) = cR(u!) + jR(u2) = cvi + sv2 = b.
Also note that b-b = 1 = с2 + s2 = ww. Thus R is the desired rotation
and w is the desired unit vector. The components c, s of w can be expressed
directly in terms of a, b as follows:
a»b = vi«(cvi + jv2) = c,
a X b = vi X (cvi + jv2) = svl X v2 = jv3,
|aX b| = |jv8| = |j|= s.
We can now state that there exists a rotation R such that R(ui) = a and
R(w) = b, where w = (a»b)ui +|a X b|u2. Since a»b = b»a and
|aXb| = |bXa|,it follows that there also exists a rotation S such that
S(ui) = b and S(w) = a. Because of this last result, an angle in three-
dimensional space is usually regarded as an unoriented angle {a, b); i.e.,
{a, b) is regarded as the same angle as {b, a). Since s is non-negative, there
exists a unique number С between 0 and 180 such that C° is a measure of
(ui, w). We assign the unique measure C° to the unoriented angle {a, b).
Then
с = a-b = cosC0, s = |a X b = sin C°.
When we are concerned with angles in a plane (not necessarily the plane
of Ui, u2), these angles can, however, always be assigned orientations relative
to any chosen orthonormal system in that plane. Then the addition formula
(a, b) + (b, c) = (a, c) applies to such angles. For example, if R = [vi,
v2, V3] transforms Ui, w respectively into a, b, where w = (a»b)ui +
I a X b|u2, then the orthonormal system vi, v2, where Vi = a, assigns an
orientation to (a, b) in such a manner that one of the measures of this angle
is between 0° and 180°. These results will be used in proving the angle sum
theorem.

228
TRIGONOMETRY
[Ch. 5
Questions
1. How do you remove the restriction of a common origin for two rays forming an
angle? 2. When (a, b) is an oriented angle in the plane of щ, иг and a, b are unit
vectors, how do you find a rotation R and a vector w such that R(ui) = a and
R(w) = b? 3. Show that (a, b) and (S(a), S(b)) have the same measure. 4. State the
meaning of the symbolic equation (a, b) + (b, с) = (а, с) and prove this result.
5. Discuss the measures of the angles (a, a), (a, J(a)), (a, —a). 6. When a, b are unit
vectors, how do you find a three-dimensional rotation R such that R(ui) = a,
R(w) = b, where w = (a-b)ui + a X b|u2? 7. Discuss the measure of unoriented
angles. 8. Show how to assign an orientation to an angle so that one of its measures is
between 0° and 180°.
Problems
1. Let a = [И, V3/2], b = [0, 1]. Find R, w such that R(m) = a, R(w) = b.
Find a measure of (a, b).
2. Let a = [H, %\, b = [1,0]. Find R, w such that R(m) = a, R(w) = b.
Find a measure of (a, b).
3. Let a = [Уз, Ц, Ц], b = [Ц, H, -%}■ Find R such that R(m) = a, R(w) =
b, where w = (a-b)ui + a X b|u2- Find the measure of (a, b).
4. Let a, b be nonzero vectors. Compute
a b la b I
ИПьГ ΙμΤχΠ4Ι
in terms of cos C°, sin C°, where C° is the measure of (a, b). Show that
a-b = |a|-|b|cosC°, |a X b| = |a| · |Ь|sin C°.
5. Let А, В, С be respectively the points (0,-2,1), (1,0,3), (3,1,1); let
A", В", С be measures of the angles (AB, AC), (ВС, ΒΑ), (CA, CB),
respectively; and let v3 be the unit vector [-Ц, %, —H]. Show that
AB X AC = [-6, 6, -3] = v3\AB\ ■ \AC\sin A"
= BCX BA = v3\BC\ -\BA\sinB°
= CAXCB = v3\CA\ ■ |C£|sinC°.
Find the measures of the angles of the triangle. Find the area of the triangle.
6. Let А, В be points of the plane of щ, иг and let A", B" be measures of
(ui, О A), (ui, OB), respectively. Show that B° — A" is a measure of (OA, OB)
and that
OAOB = \OA\-\OB\cos{B° - A").

Sec. 59] THE LAWS OF SINES AND COSINES 229
7. Let X be a point, η a unit vector in the plane of щ, иг, and let A", T" be
measures of (ui, n), (ui, OX), respectively. Let С be such that ОС = en,
where с is non-negative, and let |ОЛГ| = r. Show that the distance from X to
the line through С with normal η is
± n-CX ±η.(δχ- ОС) = ±(rcos(T° - A°) - c).
|n|
Show that the distance from the origin to the line is с
59. THE LAWS OF SINES AND COSINES AND THE ANGLE
SUM THEOREM
If a, b are nonzero vectors, then a/1 a |, b/1 b | are unit vectors, and we
have seen that
a b a»b
= — = с = cos C°,
a |b| |a|-|b|
a b a X b
X = — = vis = V3 sin С ,
|a| |b| |a|-|b|
where С is between 0 and 180, where C° is a measure of the oriented angle
(a, b), and where the orientation is relative to a properly oriented ortho-
normal system in the plane of the angle. It follows that
a-b =|a|-|b|cosC°, a X b = v3|a| · | b|sin C°.
We shall use these results to find relations between the sides and angles of a
triangle.
Let А, В, С be vertices of a triangle; let
e=|BC|, b=\CA\, c = \AB\
denote the lengths of the sides; and let A", B°, C° denote measures of the
oriented angles
(AB, AC), (ВС, ΒΑ), (CA, CB),
respectively. We shall show that we can choose an orientation in the plane
of the triangle such that each of the numbers А, В, С can be chosen to lie

230
TRIGONOMETRY
[Ch. 5
between 0 and 180. We show first that A B X AC = ВС Χ ΒΑ = CA X СВ.
Thus
BCX ΒΑ = (AC - AB) Χ ΒΑ = AC Χ ΒΑ = -AC X AB = AB X AC,
CA X CB = CA X (ЛЯ - ЛС) = CA X AB = AB X AC.
We define the orientation of (AB, AC) relative to an orthonormal system
oriented so that
AB X AC = v3\AB\ -\AC\ sin A" = v3cb sin A",
where sin A ° is positive. We then have
AB X AC = v3cb sin A" = ВС X BA = v3ac sin B°
= CAXCB = v3ba sin C°.
Hence sin^4°, sin B°, sin C° are all positive, and each of the numbers
А, В, С can be chosen to lie between 0 and 180. Moreover,
cb sin A ° = ac sin B° = ba sin C°,
and if we divide each of the members of these equalities by abc, we obtain the
following result, called the law of sines.
T59.1
B° sin C°
Since
we have
ACAB = \AC\ -\AB\ cos A" = be cos A",
a2 = ВС- ВС = (АС - AB) ■ (AC - AB)
= AC-AC - 2ACAB + ABAB
= b2 - 2bc cos A" + c2.
Similar expressions can be obtained for b2, c2. Hence we have the following
result, called the law of cosines.

Sec. 59] THE LAWS OF SINES AND COSINES 231
T59.2. a2 = b2 + c2 - 2bc cos A °,
b2 = c2 + a2 - 2ca cos B°,
c2 = a2 + b2 - lab cos C°.
We consider next the sum of the angles of a triangle. We have
(AB, AC) + (ВС, ΒΑ) + (CA, CB)
= (AB, AC) + (CA, CB) + (ВС, ВА)
= (AB,^C) + (J2(C^),J2(CS)) + (ВС, ~BA)
= (AB, AC) + (AC, ВС) + (ВС, ВА)
= (AB, ВС) + (ВС, ВА) = (AB, ΒΑ) = (AB, -AB).
This symbolic addition means that A" + B° + C° is a measure of (AB,
— AB) and hence equal to 180° plus perhaps a multiple of 360°. However,
since each of the numbers А, В, С is between 0 and 180, it follows that
180 - 360 < 0 < A + В + С < 540 = 180 + 360,
and hence the only possibility is that A + В + С = 180. Thus we have
T59.3.
T59.3. The sum of the degree measures of the angles of any triangle is 180 °.
Questions
1. Derive the formulas a · b = | a | · | b | cos C°, a X b = v31 a | · | b | sin C°, where
С is a measure of the oriented angle (a, b), С is between 0 and 180, and V3 is a
properly chosen unit vector. 2. Show that if А, В, С are vertices of a triangle, a, b, с
are lengths of the sides, and А", В", С are measures of the angles, then AB X AC =
—* —* —> —*
ВС Χ ΒΑ = CA X CB and
vzcb sin A" = vzac sin B" = vzba sin C°.
3. Prove the law of sines. 4. Prove the law of cosines. 5. Prove the angle sum theorem.

232
TRIGONOMETRY
[Ch. 5
Problems
1. Let a = 1, 6 = 2, r = 3 be sides of a triangle. Substitute these values in the
law of cosines and solve for cos A", cos B", cos C°. Find А, В, С
2. Let a = \, b = 2, С = 45° be two sides and the included angle of a triangle.
Find с by the law of cosines. Find A, B by the law of cosines.
3. Let A" = 60°, B° = 30°, с = 2 be two angles and the included side of a
triangle. Find С by the angle sum theorem. Show that the law of sines implies
that
sin A" sin B"
Find a, b.
4. Let a = λ/2, с = 2, A" = 30° be two sides and an angle not included. Show
that the law of sines implies
^o sin A°
sin Cr = с
a
Show that С can be either 45 or 135. Draw the two triangles corresponding to
these two values of C. This is called the ambiguous case.
5. Let a = \,c = 2, A" = 30°. Show that there is only value of С Find this value
and find В and b.
6. Let a = 1, с = 4, A" = 30°. Show that no triangle can be formed.
7. Suppose that B° = 90°. Show thai С = 90° - A" and that the law of
sines becomes
sin A" _ 1 _ cos A"
a b с
60. TRIANGLE SOLUTION
We now study various triangle solution problems. First, given the three
sides a, b, с of a triangle, it is required to find the angles. We use the
abbreviation SSS to denote this problem. Any one of the angles can be obtained by
the law of cosines. Thus if we solve the third equation of T59.2 for cos C°, we
obtain
a2 + b2 - c2
cos C° =
lab
We can obtain С from the Appendix tables, and in a similar manner we can
find B. Then A can be found by T59.3. Next, given two sides and the
included angle (say, b, c, A) it is required to find the remaining side and
remaining angles (abbreviation SAS). We can find a by the first equation of
T59.2 and then proceed as in SSS to find В, С Finally, given two angles and

Sec. 60]
TRIANGLE SOLUTION
233
the included side (say, В, C, a), it is required to find the remaining angle
and remaining sides (abbreviation ASA). By T59.3 we obtain A. We now
know A, B, C, a, and if we solve the equations of T59.1 for b, c, we obtain
sin B° sin C°
b = a 1 с = a
sin A" sin A"
where sin A °, sin B", sin C° are obtained from the tables.
When one of the angles, say, (CA, CB), has the measure 90°, the
computation in the triangle solution is somewhat simplified. We have
cos C° = cos 90° = 0, sin C° = 1,
A + В + 90 = 180, В = 90 - A,
sin B° = sin (90° - A") = cos A".
The third equation of T59.2 becomes
Л — „2 ι i,2.
с = a -+- b ,
i.e., the Pythagorean theorem. When С = 90, the law of sines becomes
sin A ° cos A ° 1
a be
These equations can be transformed in various ways as follows:
a b sin A ° a cos A ° b
sin A" = -· cos A" = -ι = -. — = -.
с с cos A" b sin A" a
sin A ° cos A °
a = с sin A ° = b 1 b = с cos A ° = a ■
cos A" sin Ac
1 1
с = a = b
sin A" cos Л0
To simplify the computation further, we introduce four new trigonometric
functions called the tangent, cotangent, secant, and cosecant. The values of these
functions for a given number A are denoted respectively by tan A °, ctn A °,
sec A °, esc A °, and are defined as follows:
sin A" cos A" 1
tan A" = · ctn A" = > sec^4c
esc A" ■■
sin A'
cos A ° sin Л ° cos Лс
1

234
TRIGONOMETRY
[Ch. 5
These formulas define the functions. If A = 90 or A = 270, then cos A ° = 0
and tan A °, sec A ° are not defined. Similarly, ctn A °, esc A ° are not defined
when A = 0 or A = 180. The alternative notation ctn A" = cot A" is
sometimes used. In terms of these new functions the above formulas become
a b a b
sin A ° = - · cos A ° = - · tan A ° = - < ctn 4 ° = - ·
с с b a
a = с sin A" = b tan Л0, b = с cos Л ° = a ctn Л °,
с = a esc Л ° = b sec Л °.
When two of the sides of the right triangle are given, these formulas together
with the tables enable us to find A, and when A and one of the sides are given,
we can find the remaining sides.
We have the following reduction formulas of T60.1.
T60.1. (a) tan (90° - A0) = ctn A0, ctn (90° - A0) = tan A0,
(b) sec (90° - A") = esc A0, esc (90° - A") = sec A0,
(c) tan (180° - A0) = - tan A0, ctn (180° - A0) = - ctn A0,
(d) sec (180° - A0) = - sec A0, esc (180° - A0) = esc A0,
(e) tan (180° + A") = tan A", ctn (180° + A") = ctn A",
(f) sec (180° + A") = - sec A", esc (180° + A") = - esc A",
(g) tan (360° - A") = - tan A", ctn (360° - A") = - ctn A",
(h) sec (360° - A0) = sec A0, esc (360° - A0) = - esc A",
(i) tan (360° + A") = tan A", ctn (360° + A") = ctn A",
(j) sec (360° + A0) = sec A0, esc (360° + A0) = esc A0.
We shall prove two of these formulas. Thus
sin(90°-^°) cos4°
tan (90° - A") = = = ctn A",
cos (90° - A") sin A"
1 1
esc (180° + A") = = = -esc A".
sin (180° + A") -sin4°
The remaining formulas are similarly proved.
Questions
1. Discuss triangle solution for the cases SSS, SAS, and ASA. 2. What does the
law of sines become when С = 90? 3. Define tan A", ctn A", sec A", esc A". 4. Find
equations relating sin A", cos A", tan A", ctn A", sec A", esc A" to the sides of a
right triangle.

Sec. 61]
RADIAN MEASURE
235
Problems
Solve the following triangles; find sides to three-place accuracy and angles to
the nearest degree:
1. a = 3, b = 4, с = 5.
2. a = 1, b = 2,C = 60.
3. a = 1, b = \,C = 60.
4. A = 50, В = 60, с = 100.
5. α = 5, b = 12, С = 90.
6. α = 10, b = 8, Л = 90. Note that the law of sines becomes
± _ sin B° _ cosB"
a b с
7. A = 30, В = 90, с = 100.
Additional problems in triangle solution are given in Sections 62 and 66.
61. RADIAN MEASURE
To compute the length of circular arc subtended by a given angle, we
introduce a new measure of the angle called radian measure. It is defined in
such a way that if an arc of a circle with unit radius is subtended by an angle
with radian measure θ (Greek letter theta) then the length of the arc is also Θ.
If, instead, the circle has radius r, then the angle of θ radians subtends an arc
r times as great; i.e., the length of the arc is
L = r0.
It follows that
θ = L/r,
and hence the radian measure can be defined alternatively as the length of
arc subtended by the angle divided by the radius of the circle. A circle of
radius r has the circumference 2irr. An arc of this circle subtended by an
angle with degree measure 180 has the length equal to half the circumference;
i.e., L = irr. The radian measure of this angle is therefore
L яг
θ = - = — = 7Γ.
r r
In converting degree measure into radian measure we assume that the two
measures are proportional to one another. The proof of the fact is, however,
beyond the scope of the text. To find the length of an arc subtended by an

236
TRIGONOMETRY
(Ch. 5
angle whose measure is given in degrees, we first convert the degree measure
into radian measure and then use the radian measure to compute the arc
length. The conversion is accomplished as follows: Since the two measures
are proportional to one another, there exists a constant of proportionality к
such that θ = kA where A is the degree measure and θ the radian measure of
a given angle. Since the angle whose degree measure is 180 has the radian
measure π, we have π = A;180, or к = 7г/180. Thus we have T61.1.
T61.1. The radian measure θ of an angle whose degree measure is A is given by the
formula
_ Απ
~ 180
Figure 61.1 shows the right triangle used to estimate the values of cos A",
sin A °, where A is small. Note that for a small angle, sin A ° is approximately
equal to the length of the arc sub-
| tended by the angle; i.e., to the radian
sin A° measure of the angle. For smaller
I angles the percentage of error is
smaller. The radian measure of the
Figure 61.1 angle of 1° is тг/180 =0.017. The
approximation 0.017 to sin 1° turns
out to be accurate to three places. We now apply the formula cos2 A" +
sin2^4° = 1 to obtain the approximation
— (0.017)2 = 1 (three-place accuracy)
to the cos 1 °. We can check these approximations as follows: In Problem 8
of Section 57 you obtained exact values for cos 15°, sin 15°. We can use our
approximations for cos 1 °, sin 1 ° in conjunction with T56.3 to obtain
approximate values of cos 15°, sin 15 °, and these latter approximate values
can be checked with the exact ones. Thus we can compute the cosine and
sine of 1° + 1°, 1° + 2°, 2° + 3°, 5° + 5°, 5° + 10°, starting with the
approximate values. We can continue this procedure and obtain a table of
cosines and sines of the angles from 1 ° to 45°. The table can be extended to
angles given in degrees and minutes as follows: An angle of 1 min has the
radian measure π/180 X 60 = 0.000284, and this is an approximation to
sin 1' = 0.00029. We do not propose to perform these computations, but if
we did, it is very possible that our approximations to cos 15°, sin 15°, would
not be accurate to three places owing to accumulation of errors. The purpose
of the present discussion is to show one method by which a more accurate

Sec. 61]
RADIAN MEASURE
237
table can be obtained. More refined and less laborious methods have been
discovered, but these methods will not be discussed in this text.
If the measure of an angle is θ radians, then the values of the sine and
cosine are written sin б and cos 6, respectively; i.e., without the degree
symbol. For example, the sine of 2° is written sin 2°, whereas the sine of 2
radians is written sin 2. As another example we have
7Г
cos 90° = cos— = 0.
2
In general
cos θ = cos A °, sin θ = sin A °
if θ = Απ/ISO. In the problems you will be asked to use these relations to
construct a table giving the values of cos Θ, sin θ for various numbers Θ.
Questions
1. If a circle has a unit radius, what is the length of arc subtended by an angle of
θ radians? 2. If a circle has radius r, what is the length of arc subtended by an angle
of θ radians? 3. Define radian measure. 4. Show how to obtain the radian measure of
an angle whose degree measure is 180. 5. Assuming that θ = kA, where A is the degree
measure and θ the radian measure of a given angle, show how to compute k. 6. Show
that θ = Απ/\&0. 7. Show how to approximate sin 1 °, cos 1 °. 8. Show how to check
these results, using exact values for cos 15°, sin 15°. 9. Describe how to obtain a
table of sines and cosines where the angle measures are given in degrees and minutes.
Problems
1. Complete the following table (nearest hundredth), giving degree measures of
angles, radian measures, sines, and cosines.
A
θ
cos
sin
0
30
0.52
0.87
0.50
45
60
90
135
180
3.14
225
270
315
360
2. Assume that the area £ of a sector of a circle is proportional to the radian
measure θ of the arc subtended by the sector; i.e., £ = кв. Assume that the
area of a circle of radius r is 7rr2. Determine к and write down the resulting
formula for the area of the sector.
3. A circle has radius r and a sector of this circle subtends an arc of length L.
Show that the area of the sector is Lr/2.

238
TRIGONOMETRY
(Ch. 5
4. A right circular cone with slant height h and base of radius r is made of paper.
If the cone is cut along an element and the paper spread out so that it is flat,
then the resulting figure is a circle with a sector removed. Find the length of
arc of the sector that remains and the area of this sector; i.e., find the lateral
area of the cone. If you have difficulty in
visualizing this problem, cut out a circle from a
sheet of paper, cut out a sector from this circle,
and then bend the paper to form a cone by
bringing the edges of the sector together.
5. Figure 61.2 shows a sector contained in a right
triangle and in turn containing a triangle. The
circle has radius 1, and the sector subtends an
angle of θ radians. Show that the base and
altitude of the interior triangle are respectively
1, sin Θ, and the base and altitude of the exterior triangle are respectively 1 and
tan Θ. Find the areas of the sector and two triangles and show that
Figure 61.2
sin θ < θ < tan θ =
sin θ
and hence that
cos б
θ cos θ < sin θ < θ.
Show that λ/ΐ — θ2 < cos θ, and hence that
0Vl - Θ2 < sin Θ < θ.
Note that these inequalities show that the error in approximating sin б by б is
less than the difference θ — θ\/\ — θ2.
6. Let Χ, η, С, г be defined as in Problem 7 of Section 58 and let the measures of
(ui, n), (ui, OX) be respectively θ radians and a (Greek letter alpha) radians.
Show that the distance from X to the line through С with normal η is ±(r cos
(Θ - a) - c).
62. GRAPHS AND INTERPOLATION
Let us plot the curve whose equation is
*2 = Sin Χγ °.
The following two-place table gives values of sin χλ ° for various values of
x\-
XI
X2 =
sin
*1°
0
0
30
0.50
45
0.71
60
0.87
90
1.00
120
0.87
135
0.71
150
0.50
180
0

Sec. 62] GRAPHS AND INTERPOLATION 239
This table can readily be extended to values of x\ up to 360 or even beyond
and also extended to negative values of χχ. We plot the points (*i, хг)
corresponding to pairs of entries in this table and connect these points by a
smooth curve. This is the curve whose equation is x-i = sin xx °. It is called
the graph of the sine and is shown in Fig. 62.1. The unit of length along the
*i-axis is 7г/180 times that chosen for the *2-axis. Hence, if we use the unit of
the *2-axis to measure distances along the *i-axis, the resulting measures
■ Sin X|
Figure 62.1
are the radian equivalents of the corresponding degree measures. That is, if
we reinterpret Fig. 62.1 as having the same unit of length along the xi-axis as
that along the *2-axis, then this reinterpreted figure is the curve whose
equation is
x2 = sin χι,
where x\ is measured in radians. The graphs of the cosine, tangent, and
secant are similarly constructed and are shown in Figs. 62.2, 62.3, and 62.4.
Consider the problem of using Table 1 to find an approximate value of,
say, sin 41.4°. We wish to find the ^-coordinate of the point of the curve
*2 = sin*i° for which χλ = 41.4. The table gives the values of sin 41 °,
sin 42°, and enables us to locate the points (41, sin 41°), (42, sin 42°). The
portion of the curve lying between these points can be approximated rather
closely by a straight-line segment connecting the points, and hence sin 41.4°
can be approximated by the ^-coordinate of the point of this segment for
Figure 62.2

240 TRIGONOMETRY (Ch. 5
X2 = ton χ, X2 = sec X|
ϋ
-1
-2
-3
-4
0 '
к З*
7 χ 2
Π
2χ
Figure 62.3 Figure 62.4
which χι = 41.4. An equation of the line through two points (αϊ, a2), (b\, b2)
is given by the two-point formula of Section 18, namely,
(bi - αι)(*2 - a2) = (b2 — α2)(*ι - Oi)·
Since we are concerned with the case bi И αϊ, we can solve for x2 and obtain
*i — "ι
*2 = «2 + (*2 - Я2)
bi - αϊ
This is called an interpolation formula, the value of x2 obtained from the formula
is called the interpolated value, and the method is called interpolation (or more
specifically, straight-line interpolation). The interpolated value of sin 41.4° is
41.4 - 41
x2 = sin 41° + (sin 42° - sin 41°)
42-41
= sin 41° + 0.4(sin 42° - sin 41°) = 0.656 + 0.4 X 0.013 = 0.661.
When αϊ, bi are successive entries in the table, the interpolated value
usually has the same degree of accuracy as the table. When we are using a
three-place table, the interpolated value should be rounded off to three
places. If we were to write a four-place number for the interpolated value,
we would be claiming accuracy to the nearest ten thousandth, our claim
would probably be incorrect, and hence our answer would probably be
wrong. The interpolated value always lies between the two values between

Sec. 62] GRAPHS AND INTERPOLATION 241
which we are interpolating. Thus
sin 41° < sin 41.4° < sin 42°.
For the cosine the corresponding inequalities are reversed. Thus the
interpolated value of cos 41.4° is
*2 = cos 41 ° + 0.4(cos 42° - cos 41 °) = 0.755 - 0.4 X 0.012 = 0.750
and
cos 41° > cos 41.4° > cos 42°.
As a further example let us find tan 27°35'. The interpolated value is
27M - 27
x2 = tan 27° + (tan 28° - tan 27°) ——
28-27
= tan 27° + Μ (tan 28° - tan 27°) = 0.510 + ^0.022 = 0.523.
When xi lies outside the interval from a\ to b\, the value of x% given by
the interpolation formula is called the extrapolated value, and the method is
called extrapolation. We shall not, however, be concerned with extrapolation.
Suppose that we are given the value of, say, sec A ° and are required to
find A. We can solve the two-point formula for x\. Thus
*2 — "2
x\ = a\ + (bi - αϊ)
£>2 — Д2
For example, if we are required to find A between 0 and 90, such that sec A °
= 2.04, we note that
sec 60° < 2.04 < sec 61°.
Hence the interpolated value of A is
2.04 - sec 60
χ = 60 + (61 - 60)
sec 61 — sec 60
0.04
= 60 + = 60.7.
0.06
Questions
1. Describe how to sketch the curve хг = sin x\". 2. Show how the units can be
chosen so that this curve can be reinterpreted as the curve whose equation is хг = sin
χι, where x\ is measured in radians. 3. Discuss the geometric interpretation of straight-
line interpolation and show how to obtain the interpolation formula. 4. Discuss the
degree of accuracy of an interpolated value. 5. Show how to find A such that sec
A" = 2.04.

242
TRIGONOMETRY
(Ch. 5
Problems
1. Find cos 71.3°, sin 118.2°.
2. Find A between 0 and 90 such that tan A" = 3.68.
3. Given a = 20.0, b = 30.0, с = 42.1. Find А, В, С (accuracy to the nearest
thousandth).
4. Given a = 20.0, b = 30.0, C° = 53.2°. Find с
5. Given a = 20.0, B° = 37.2°, С = 54.7°. Find b.
6. Given a = 30.0, A" = 29.6°, B° = 90.0°. Find b, с
7. Given a = 20.0, с = 30.0, £° = 90.0. Find A, C, b.
63. CONSTRUCTING A SLIDE RULE
We shall show how to construct and operate a very simple slide rule that
can be used to perform multiplication and division. This slide rule consists
1 2 4 8 16 32 64 128 256 512 1024
τ—ι 1 1 1 1 1 1 1 1 1 1—|—'—
1 2 4 8 16 32 64 128 256 512 1024
Figure 63.1
of a fixed scale (lower scale in Figs. 63.1 to 63.5) and a sliding scale (upper
scale). The two scales are identical. To multiply m times n, place the 1 (left
end) of the sliding scale directly above the m of the fixed scale. The product
mn is then found on the fixed scale directly below the η of the sliding scale.
To construct such a slide rule, choose 1 dm as a unit and (on each scale)
locate the point 2 at a distance of 0.3 dm (3 cm) to the right of the 1. This
1 2 4 8 10 100
τ 1 1 1—ι ι Ι ι ι I 1 1 1—I I I I I I 1 1 1—I I I I I I—
1 2 4 8 10 20 40 80 100 1000
Figure 63.2
choice for the position of the 2 determines the positions of all the numbers.
The reason for the choice will appear later. To perform various
multiplications by 2, place the 1 of the sliding scale directly above the 2 of the fixed
scale. Then, directly below the 2 of the sliding scale, we find 2X2 = 4.

Sec.63] CONSTRUCTING A SLIDE RULE 243
Hence 4 is 0.6 dm to the right of 1. Directly below the 4 of the sliding scale we
find the 8 of the fixed scale, and hence 8 is 0.9 dm to the right of 1. In a
similar manner we locate 16, 32, 64, 128, 256, 512, 1024. These points are spaced
0.3 dm apart, and a simple count shows that 1024 is 3 dm to the right of 1.
To perform these constructions, you will need to make two scales, each 3 dm
in length. The diagonal of an ordinary sheet of typewriter paper is a little
over 3 dm in length. Figure 63.1 shows the construction but is about half
the size of the figure you are asked to make.
The point 1000 should be located slightly (actually about 1 mm) to the
left of 1024. We assign the label 1000 instead of 1024 to the point 3 dm to the
1 2
1
4 5
2
8 10
4 5
8 10
50
80 100
100
200
1000
Figure 63.3
right of 1, and in so doing, we are committing only a small error. Assume for
the moment that 10 has been located, and let us perform various
multiplications by 10. We place the 1 of the sliding scale above the 10 of the fixed
scale. Then 100 and 1000 on the fixed scale are respectively below 10 and 100
on the sliding scale. Hence 1, 10, 100, 1000 must be equally spaced, and
they must be 1 dm apart, since 1 and 1000 are 3 dm apart. It should now be
clear why the points 1, 2 were chosen 0.3 dm apart. When the 1 of the sliding
scale is above the 10 of the fixed scale, the points 20, 40, 80 of the fixed scale
are respectively below 2, 4, 8 of the sliding scale and, for example, 80 is 1.9
dm to the right of 1. The multiplications by 10 are indicated in Fig. 63.2,
but some of the points located in Fig. 63.1 are omitted in Fig. 63.2. Figure
63.3 indicates multiplications by 5. The 1 and 2 of the sliding scale are
respectively above the 5 and 10 of the fixed scale. Thus 5 is 0.3 dm to the
left of 10 and 0.7 dm to the right of 1. The 50 of the fixed scale is below the 10
of the sliding scale and is 1.7 dm to the right of 1.
1
2
4 5
1 2
7 8 10
4 5
78 10
49 81 100
100 ί
1000
Figure 63.4
We assign the label 49 instead of 50 to the point 1.7 dm to the right of 1
and the label 81 instead of 80 to the point 1.9 dm to the right of 1. In so

244
TRIGONOMETRY
(Ch. 5
doing, we commit slight errors. Figure 63.4 indicates multiplication by 7.
The 1, 7 of the sliding scale are respectively above 7, 49 of the fixed scale.
Thus 1, 7, 49 are equally spaced, and this locates the 7. Figure 63.5 indicates
multiplication by 3. The 1, 3, 9, 27 of the sliding scale are respectively above
the 3, 9, 27, 81 of the fixed scale. Thus 1, 3, 9, 27, 81 are equally spaced, and
this locates the 3 and 9. The 6 of the fixed scale is located below the 2 of the
sliding scale. We have now located 1,2, 3, 4, 5, 6, 7, 8, 9, 10.
The scale from 1 to 10 can be further refined by locating such points as
1.1, 1.2, 1.3, etc. The labels on the finer divisions must be ascertained by
counting, since there is not room on the slide rule to include all the labels.
3 4 5 6 7 8910 27 100 200
Ι ι Hi '
3 4 5 6 78910 27 81 100 1000
Figure 63.5
Problem
1. Locate the points labeled 1,2, 3, 4, 5, 6, 7, 8, 9, 10 on a slide rule scale by the
method described in this section.
64. THE SLIDE RULE AND LOGARITHMS
Let us choose the left-hand end point of a slide rule scale (i.e., the point
labeled 1) as an origin. If a point labeled m is to the right of this origin, then
its distance from the origin is its coordinate, and this coordinate is called the
logarithm of m. We have chosen our unit so that the point labeled 10 has the
coordinate (i.e., logarithm) 1, and when the unit is so chosen, the number 10 is
called the base of the logarithms. The coordinate of the point labeled m is then
called the logarithm to the base 10 of m, and we denote this coordinate by logio
m. Since the coordinate of 10 is 1, we have
log10 10 = 1.
Recall that to obtain the product m times n, we place the origin of the sliding
scale above the point labeled m on the fixed scale and then locate the product
mn on the fixed scale directly below the point η of the sliding scale. By this
mechanical device we have added the coordinate of η (i.e., logio n) to the
coordinate of m (i.e., logio m), and this sum is the coordinate of mn (i.e., logio
mn). Thus
logio ™ + logio η = logio mn.

Sec. 64] THE SLIDE RULE AND LOGARITHMS 245
To compute the product mn, we look up logio m and logio л in a table of
logarithms (Table 2) and add these logarithms. The number whose
logarithm is equal to this sum is the product mn. Let us see how to compute a
quotient χ = m/n. Since xn = m, we have logio χ + logio " = logio mi and
hence
logio - = logio x = logio m ~ logio "·
Thus m/n is the number whose logarithm is the difference logio m ~ logio "·
To compute this quotient with a slide rule, we imagine how the rule should
be set up to compute the product xn = m. The origin of the sliding scale is
placed above the χ of the fixed scale, and the product m on the fixed scale
is below the η of the sliding scale. Hence to compute the quotient, we place
the η of the sliding scale above the m of the fixed scale and locate the quotient
x = m/n on the fixed scale below the origin of the sliding scale.
EXAMPLE 64.1
Find 3.7 X 1.9 and 3.7/1.9 by logarithms. The computation is as
follows:
logio 3.7 = 0.568 logio 3.7 = 0.568
logio 1.9 = 0.279 logio 1.9 = 0.279
logio 3.7 X 1.9 = 0.847 = sum logio 3.7/1.9 = 0.289 = difference
3.7 X 1.9 = 7.0 3.7/1.9 = 1.9
(accuracy to nearest tenth).
Actually, logio 7.0 = 0.845, logio 1.9 = 0.279 (three-place accuracy), where
0.845, 0.279 are the numbers in the table nearest 0.847, 0.289 respectively.
Three-place accuracy can be achieved by interpolation.
To locate the points labeled 1, 2, 3, · · · 10 on a slide rule scale, we
computed the coordinates (i.e., the logarithms) of these labels. In Problem 1
you will be asked to round off these numbers to two places and arrange your
results in a two-place table of logarithms. There is a question as to whether
logio 3 should be rounded off to 0.47 or 0.48. Table 2 gives logio 3 = 0.477,
and hence you should choose 0.48. You should settle logio 6 similarly.
Table 2 gives logio 7 = 0.845, and there is a question as to whether your
answer 0.85 is correct to the nearest hundredth or whether the answer shquld
be 0.84. A four-place table gives logio 7 = 0.8451, and hence the answer
should be 0.85.
There are certain logarithms that can now be computed exactly. Since

246 TRIGONOMETRY (Ch. 5
the coordinate of the point labeled 1 is zero,
logic 1 = 0.
We have already seen that logio 10 = 1. We also have the following:
log10 100 = log10 102 = log10 10 X 10 = log10 10 + log10 10 = 1 + 1 = 2,
logio 1000 = logio 103 = log10 10 + log10 10 + log10 10 = 3,
logio 104 = 4 log10 10 = 4.
logio Ю-2 = logi0 1/Ю2 = log10 1 - logio 102 = 0 - 2 = -2,
logio Ю-3 = logio 1/Ю3 = 0 - 3 = -3,
and in general
logio 10" = я,
where η is any integer positive, negative, or zero.
Logarithms of numbers such as 190, 0.0037 are found as follows:
logio 190 = log,0 1.9 X Ю2 = logio 1.9 + log10 Ю2 = 0.279 + 2,
logio 0.0037 = logio 3.7 Χ 10-3 = logio 3.7 + logio Ю-3 = 0.568 - 3.
The integer 2 is called the characteristic, and 0.279 the mantissa of logio 190.
Similarly, —3 is the characteristic and 0.568 the mantissa of logio 0.0037. A
table of logarithms such as Table 2 gives logarithms of numbers from 1 to 10.
To find the logarithm of an arbitrary positive number M, we write that
number in the form
Μ = m\0n
where 1 ^ m < 10 and η is an integer. Then
logio Μ = logio m + log10 Ю" = logio m + Щ
where logi0 m is the mantissa and η the characteristic of logio Λ/· For example,
logio 190 = log,o 1.9 X Ю2 = logio 1.9 + 2,
log,o 0.0037 = log,o 3.7 Χ 10-3 = logio 3.7 - 3.
The mantissas of logio 190, logio 0.0037 are respectively logio 1-9, logio 3.7,
and the characteristics are respectively 2, — 3.
To compute the quotient 190/0.0037, we find logio 190 - logio 0.0037,
and since the difference 0.279 — 0.568 of the mantissas is negative, we write
logio 190 in the form 1.279 + 1 instead of 0.279 + 2. The computation is
as follows:
logio 190 = 1.279 + 1
log,o 0.0037 = 0.568 - 3
difference = 0.711 + 4 = log,0 190/0.0037
190/0.0037 = 51,400 (accuracy to the nearest thousandth).

Sec. 64] THE SLIDE RULE AND LOGARITHMS 247
A slide rule is also constructed only for numbers from 1 to 10 and decimal
points are computed mentally. In Section 63 we located various points on
the scale from 1 to 1000 as a device for completing the scale from 1 to 10.
If the scale from 10 to 100 were completed, it would be a replica of the scale
from 1 to 10 except that the decimal point on each label would be moved
one place to the right. That is, the labels would be 10, 20, 30, etc., instead of
1, 2, 3, etc. In the scale from 100 to 1000 the decimal point on each label is
moved two places to the right. We could also extend the scale to the left of 1
by reproducing the scale from 1 to 10 and moving the decimal point on each
label one place to the left. To compute the quotient 190/0.0037 on a slide
rule, we start by attempting to compute the quotient 1.9/3.7; i.e., we place
the 3.7 of the sliding scale above the 1.9 of the fixed scale. We discover that
the origin of the sliding scale is to the left of the origin of the fixed scale and
hence is not above any number on the fixed scale. In such a case we imagine
that the sliding scale is assigned the labels 0.1, 0.2, · · · 0.9, 1 instead of
1,2, · · · 9, 10; i.e., the decimal point on each label is moved one place to
the left. Then the right-hand end of this scale becomes the origin and the
rule is set up to perform the division 1.9/0.37. Since the point 5.1 is under the
new origin, the result of this division is 5.1. The quotient 190/0.0037 is thus
1.9 X 102/0.37 X 10-2 = 5.1 Χ ΙΟ2 Χ 102 = 51,000. Similarly, to compute
the product 5.1 X 3.7, we again regard the right-hand end of the sliding
scale as the origin. That is, we first compute the product 5.1 X 0.37 by
placing the right-hand end of the sliding scale above the 5.1 of the fixed
scale. We find 1.9 under 0.37, and hence 5.1 X 0.37 = 1.9 and 5.1 X 3.7 =
19. Thus either end of either scale can be regarded as an origin, and it is
customary to assign the label 1 instead of 10 to the right-hand end of each
scale. For further information about slide rules the reader is referred to the
instructions that are furnished with a slide rule.
Questions
1. Describe how a coordinate system is set up on a slide rule scale so that the
coordinate of a label is equal to the logarithm of the label. 2. How is the unit chosen
when the base is 10? 3. Explain why logio 10 = 1, logio 1=0, logio mn = logio m +
logio n. 4. Show that logio m/n = logio m — logio n. 5. Explain the equation logio 10"
= n, where η is any integer positive, negative, or zero. 6. Define characteristic and
mantissa.

248
TRIGONOMETRY
(Ch. 5
Problems
1. Use the results of Problem 1 of Section 63 to construct a two-place table of
logarithms. Check your table with Table 2 and correct if necessary.
2. Using your two-place table, compute logio 12 = logio 3X4, logio 14, logm 15,
logio 16, logio 18.
3. Using your two-place table, find approximate values of logio 11 and logio IV.
Note that 11 X 11 and 7 X 17 are both approximately 12 X 10. Find
approximate values of logio 13 and logio 19. Note that 13X13 and 9 X 19 are
both approximately 10 X 17.
4. Use the results of Problems 2 and 3 to make out a two-place table of the
logarithms of the numbers 1.1, 1.2, 1.3, · · · 1.9. Check with Table 2 and correct
if necessary. Add the corresponding points to your slide rule scale of Problem 1
of Section 63 but do not label these points. It is given that logio 1.8 = 0.2553
(four-place accuracy).
5. Compute 272 X 0.456 (three-place accuracy) by means of logarithms. Before
consulting the table, make out a scheme of logarithmic computation such as
the following
logio 272 = +2
logio 0.456 = -1
sum = +1 = logio 272 X 0.456
272 X 0.456 =
You will then have a place to put each number as you find it in the table.
Similarly compute 272/0.456 and 0.456/272. Make out a scheme before
consulting the table.
6. Compute 2722 = 272 X 272 by logarithms. Also compute 2723 and 272 ~3 =
1/2723.
7. Compute χ = λ/θ.456 by logarithms. Note that logio χ Χ χ = logio 0.456.
8. Compute -^0.456.
65. EXPONENTS
In Section 64 we introduced the concept of a logarithm. We assume that
there is a function called the logarithm which assigns to each positive number
m a unique number logio *n and that to each real number χ there corresponds
a unique positive number m such that logio m = x. We assume further that
logio 1=0, logio Ю = 1) and logio mn = logio m + logio η for every pair of
positive numbers m, n. The proof of the existence of such a function is beyond

Sec. 65]
EXPONENTS
249
the scope of this text. In this section we use logarithms to perform
computations involving exponents and to develop properties of exponents.
If a is any positive number, we can compute a2 and a3 as follows:
logio a2 = logio aa = log10 a + log10 a = 2 log10 a
logio a3 = logio «a2 = log10 a + log10 «2 = log10 а + 2 logio a
= 3 logio a,
and hence a2 is the number whose logarithm is 2 logio a and a3 is the number
whose logarithm is 3 logio a· In general, for any positive integer n,
logio an = η logio a.
Moreover,
logio o_n = logio Va" = logio 1 - logio a" = 0 - η log10 a
= —n logio a
and
logio o° = logio 1=0 = 0 logio ".
If we wish to compute χ = α = \Лг, we observe that x3 = a, logio *3 =
3 logio * = logio a, and hence
logio oH = logio * = 1 logio a.
In general if χ is any real number (say, χ = — л/2), then logio a1 =
χ logio ai and hence a1 is the number whose logarithm is χ logio a· Let us take
this as a definition of ax and let us show that this definition leads to the usual
properties of exponents. Thus
log10 aV = logio ax + logio ay = χ logio " + у log10 a
= (x + y) logio a = logio ax+v,
and hence
αχαν = α*+ι/
Since logio a° = 0 logio α = 0 = logio 1 and logio a1 = 1 logio a> it follows
that
a0 = 1 and a1 = a.
Since
logio β*/*" = logio ax - log10 a" = χ log10 a - > log10 a
= (x - y) logio a = logio a*-",
it follows that
ax
— = ax~y

250
TRIGONOMETRY
(Ch. 5
and in particular that
1 a° η
— = — = α0-" = α-".
α" α"
Since
logio (α*)" = У logio я* = *y logio α = logio «*",
we have
{ax)y = axy.
If л is a positive integer and χ = α1/η, then
x» = (al,n)n = anln = a1 = a,
and hence
α'/η = χ= ry-^
If m, л are positive integers, then
β™/« = (β»/»)- = (^)™
= (am)1/n = v^·
Therefore our definition of ax does lead to the usual properties of exponents.
EXAMPLE 65.1
A musical instrument is usually tuned so that A has the frequency of 440
vibrations per second. On a well-tempered scale the frequency of one
halftone above A is 440 X 2Иг, and the frequency three halftones above A
(i.e., the С one octave above middle C) is
/ = 440 X 2M2 = 440 X 2й.
Hence
log,0/ = logio 440 + log,0 2й = logio 440 + \ log,,, 2.
The computation of/ is
logio 2 = 0.301
\ log10 2 = 0.075
logio 440 = 0.643 + 2
sum = 0.718 + 2 = log10/
/ = 522 (interpolated value).
Let us solve the equation
10* = m

Sec. 65]
EXPONENTS
251
for x, where m is a given positive number. The solution is
log10 m = logm 10* = χ log10 10 = x.
That is, logio m = χ is the solution of the equation, and hence logio m is the
exponent of the power to which 10 must be raised to produce m. More
generally we have D65.1.
D65.1. // a, m are positive numbers and α И 1, then the logarithm to the base a of m
(written logo m) is the exponent of the power to which a must be raised to produce m;
i.e., the number χ such that ax = m.
The equation ax = m is solved as follows:
logio m = log10 ax = χ log10 a,
and hence
Note that
ι logio m
loga m = χ = y-Ξ
logio a
\n„ „ iogioa ι i„„ ι logio 1 η
log0 a = , = 1, logo 1 = , = 0.
logio α logio α
If a slide rule scale is constructed in such a manner that the distance from
the origin to the point labeled a is chosen as a unit, then the coordinate of a
point labeled m is logo w. From now on we shall omit the subscript 10 on
a logarithm when it is understood that the base is 10. That is, we shall write
logio ™ = log m.
Questions
1. State all the conditions imposed on the logarithm (i.e., the function) when the
base is 10. 2. Explain the equation logio a" = η logio a, where a is positive. Consider
the cases in which η is a positive integer, a negative integer, η = 0, and η = И·
3. Define ax in terms of logarithms. 4. Show that this definition implies that aV =
a*+vt a" = l,_a' = a, az/a» = a*-», a~x = \/az, (a1)" = a3», a11" = y/'a, amln =
(Va)m = y/am if m, η are positive integers. 5. Show how to solve the equation
101 = m for χ when η is a positive number. 6. Define loga m. 7. Show how to compute
log,, m.
Problems
1. In calculus, logarithms to a base called e are used, where с is a number that we
shall describe later. Compute loge 2 and loge 2.72, given that logio e = 0.434.
2. Problems 2, 3, and 4 are concerned with showing relations between logarithms
to the base e and certain areas. Figure 65.1 shows the curve хг = \/x\. The area

252
TRIGONOMETRY
(Ch. 5
bounded by this curve, the *i-axis, the line x\ = 1, and the line x\ = m is
called the area from 1 to m and is equal to loge m. We shall indicate why this is
the case. This problem is concerned with estimating the area from 1 to 2;
i.e., log» 2. Figure 65.1 shows ten shaded rectangles included in this area.
Thus the sum of the areas of these rectangles is less than the area from 1 to 2.
Each rectangle has the width Ho and the height such that the top right-hand
corner is on the curve. Thus the left-hand rectangle has the height *2 = 1/1.1 =
0.91, and the area 0.91 X (Ho) = 0.091. Each rectangle is extended to include
an unshaded portion so that the extended rectangles include the area from 1 to
2, and the sum of the areas of the extended rectangles is greater than the area
from 1 to 2. The left-hand extended rectangle has the height 1 and the area
0.100. Find the sum of the areas of the shaded rectangles and the sum of the
areas of the extended rectangles. Split the difference between these two sums
and you should obtain the result 0.695. The area from 1 to 2 is 0.693 (accuracy
to the nearest thousandth). You can use the accompanying table.
X\
lAi
1.0
1.00
1.1
0.91
1.2
0.83
1.3
0.77
1.4
0.71
1.5
0.67
1.6
0.63
1.7
0.59
1.8
0.56
1.9
0.53
2.0
0.50
3. (a) Plot the points of the curve *2 = \/x\ for which x\ = \i, И, 1, 2, 4, 5, 10.
Choose 1 cm as a unit for both axes. Connect the points by a smooth curve.
The portion of the curve joining the points (5, И) and (10, Ho) can be
approximated by a straight line drawn with a ruler. Draw the lines x\ = 1,
*i = 2 and shade the area from 1 to 2.
(b) Plot the points of the curve ДС2 = V*i, for which x\ = Ho, H> H, 1,2, %, 5.
Choose 2 cm as a unit for the *i-axis and Η cm as a unit for the дгг-axis.
Connect these points by a smooth curve and shade the area from 1 to 2.
The two curves should be the same, but if the area from 1 to 2 of Fig. (b)
were transferred to Fig. (a), it would occupy the position of the area from

Sec. 65]
EXPONENTS
253
2 to 4 where figures a, b, c, are respectively the figures you are requested
to draw in parts a, b, с of this question. Moreover, the area from 1 to
2 of Fig. (a) should equal the area from 1 to 2 of Fig. (b) and in turn
equal the area from 2 to 4 of Fig. (a), since Fig. (b) is obtained from Fig.
(a) by multiplying all horizontal distances by 2 and dividing all
vertical distances by 2, and the result of the two transformations should not
change areas. Hence
Area from 1 to 2 + area from 1 to 2 = area from 1 to 2 + area from 2 to 4
= area from 1 to 4;
loge2 + bge2 = loge4.
(c) Plot the points of the curve хг = l/*i for which x\ = Ив, Ио, Уъ, Η, Η,
1, 2. Choose 5 cm as a unit for the *i-axis and \i cm as a unit for the дгг-axis.
Connect the points by a smooth curve and shade the area from 1 to 2.
The curve should be the same as that of Fig. (a), but if the area from 1 to 2
of Fig. (c) were transferred to Fig. (a), it would occupy the position of the
area from 5 to 10. The area from 5 to 10 of Fig. (a) should be equal to
the area from 1 to 2 of Fig. (a), since Fig. (c) is obtained from Fig. (a) by
multiplying horizontal distances by 5 and dividing vertical distances by
5. Hence
Area from 1 to 5 + area from 1 to 2 = area from 1 to 5 + area from 5 to 10
= area from 1 to 10;
log. 5 + log. 2 = log. 10.
The above discussion suggests why these areas behave as logarithms do.
Since log. e = 1, the base e must be such that the area from 1 to e is 1. The area
of the left-hand shaded rectangle is less than the area from 1 to 1.1, and this in
turn is less than the area of the extended rectangle. That is,
TV(0.91) < log. 1.1 <TVl,
and hence
0.91 < 10 log. 1.1 = log. (II)10 < 1.
Thus we should expect с to be somewhat larger than (l.l)10. Greater accuracy
can be achieved by forming 100 rectangles, each of width J-foo and all
included in the area from 1 to 2, and 100 extended rectangles including this
area. The left-hand shaded rectangle would have height 1/1.01 = 0990 and
area 0.990/100, and the left-hand extended rectangle would have height 1 and
area Moo. The area from 1 to 1.01 would be between these two areas. That is,
0-990 , , Λ< 1
100-<logel.01<-.
0.990 < 100 log. 1.01 = log. (1.01)100 < 1.

254
TRIGONOMETRY
(Ch. 5
Since 1/1.001 = 0.999, it follows by similar reasoning that
0.999 , ΛΛ„ 1
1Ш< bge LOOK—·
0.999 < 1000 log. 1.001 = log. (1.001)1000 < 1.
This suggests that it must be possible to approximate e by (1 + l/n)n, by
taking η very large. Compute (1.01)100, (1.001)1000, and (1.0001)10·000, given
that logio 1.01 = 0.0043214, log10 1.001 = 0.0004341, logio 1.0001 =
0.0000434.
5. The area from 1 to m can be computed very accurately for any number m,
and hence we can obtain accurate tables for logarithms to the base e. We can
use these tables to find logarithms to the base 10. Show that
■ bg„ m
logio m =
log, io
6. Given log. 2 = 0.693, log. 3 = 1.10, log, 10 = 2.30, compute logio 2, logio 3
by the formula of Problem 5. Perform the divisions 0.693/2.30, 1.10/2.30 by
means of logarithms.
66. TRIANGLE SOLUTION BY LOGARITHMS
Suppose that we are given sin 43° = 0.682 and cos 43° = 0.731 and
wish to find tan 43° = sin 43°/cos 43° = 0.682/0.731. Then
log (tan 43°) = log (sin 43°) - log (cos 43°) = log 0.682 - log 0.731.
It is customary to omit the parentheses and write log tan 43°, log sin 43°,
log cos 43° instead of log (tan 43°), log (sin 43°), log (cos 43°). The
computation of tan 43° is
log sin 43° = log 0.682 = 1.834 - 2
log cos 43° = log 0.731 = 0.864 - 1
difference = 0.970 - 1 = log tan 43°
tan 43° = 0.933.
Note that log sin 43° = 1.834 — 2 is formed by adding 1 to the mantissa
and subtracting 1 from the characteristic so that the above subtraction will
result in a positive number. The logarithms of the sine, cosine, tangent, etc.,
of the various angles are given directly in Table 1 (Appendix) and can be
used in triangle solution.

Sec. 66] TRIANGLE SOLUTION BY LOGARITHMS 255
EXAMPLE 66.1
Given a = 42.7, 5° = 53.2°, C° = 32.7°, find b. By angle sum we have
A" = 94.1°, where sin 94.1° = sin (180° - 94.1°) = sin 85.9°, and by the
law of sines we have b = a sin 5°/sin A". Hence
log b = log a + log sin 5° — log sin A".
Before consulting the table, it is advisable to write down a scheme of
logarithmic computation such as the following:
log a = log 42.7 = +1
log sin В = log sin 53.2° =
sum =
log sin A ° = log sin 85.9° =
difference = = log b
b =
You will then have a place to put each number as you find it in the table
It requires no additional writing to make out such a scheme in advance, and
unless this is done, the computation tends to be haphazard and you are
much more likely to make mistakes. The computation is now accomplished
by filling in the scheme. Thus
log a = log 42.7 = 0.630 + 1
log sin 5° = log sin 53.2° = 0.903 - 1
sum = 1.533
log sin A ° = log sin 85.9° = 0.999 - 1
difference = 0.534 + 1 = log b
b = 34.2.
We find с by a similar procedure. It is not convenient to use logarithms in
problems involving the law of cosines. There are alternative formulas to the
law of cosines which are adaptable to logarithmic computation, but we shall
not take the time to develop them. Logarithms are adaptable to right
triangle solution with the exception of those problems involving the
Pythagorean theorem.

256
TRIGONOMETRY
(Ch. 5
Problems
1. Given log cos 21 ° = 0.970 - 1. Show that
log sec 21 ° = -log cos 21 ° = 0.030.
Find log esc 21 °.
2. Given a triangle in which A° = 27.2°, B° = 39.4°, с = 496. Find C, a, b by
means of logarithms.
3. Given b = 17.6, с = 21.2, Л° = 90°. Find £, C, a.
4. Given i = 17.6, с = 21.2, C° = 90°. Find ^, B, a.
5. Given Л0 = 23.2°, B° = 90°, a = 13.7. Find C, i, с
6. Given A" = 23.2°, B° = 90°, i = 13.7. Find в, с.
7. Given A" = 23.2°, B° = 90°, с = 13.7. Find a, b.
67. IDENTITIES
The reduction formulas T57.1 hold for all values of the number A. Such
equations are called identities. Similarly, the formula for cos (A0 + B°)
(T56.3) holds for all values of A and В and is also called an identity. Let us
attempt to prove the formula
/l + cos 2 Лс
cos A" =
If we square both sides, we obtain
1 + cos 2A° 1 + cos (A0 + A")
cos2 A" = =
2 2
We now apply T56.3 and obtain
1 + cos A" cos A° — sin A° sin A"
cos2 A" =
2
1 + cos2 A" - sin2 A" 1 - sin2 A" + cos2 A"
~ 2 ~ 2 '
and from this we obtain the obviously correct equation
cos2 A" + cos2 A"
cos2 A" =

Sec. 67]
IDENTITIES
257
We now question whether this procedure establishes the correctness of the
formula to be proved. Let us test the formula for special values of A. If
A = 90, then 0 0
cos A = cos 90 = 0,
and
1 + cos 2A° 1 + cos 180° 1 - 1
—2 V—ϊ V"^0·
and hence the formula is correct for A = 90. However, if A = 180, then
cos A" = cos 180° = -1,
whereas
1 + cos 2A° 1 + cos 360° 1 + 1
and hence the formula is not correct for A = 180. This example shows that
the above procedure cannot be counted on to establish an identity. Actually,
what the procedure shows is that if the formula to be proved is correct, then
the final formula is also correct.
We need to proceed in the reverse direction. Thus we start with the
obviously correct equation
cos2 A" + cos2 A"
cos2 A" =
2
and from this it follows that
1 - sin2 A" + cos2 A" 1 + cos2 A" - sin2 A"
cos2 A0 =
2 2
1 + cos A" cos A" — sin A° sin A°
2
and hence by T56.3 we have
1 + cos (A0 + A") 1 + cos 2ЛС
cos2 A0 =
2 2
We now take square roots of both sides and obtain
/1 + соз2Лс
cos A" = ±
This formula is correct, and the method of proof used does establish its
correctness. However, the reader may wonder how we could ever think of
this proof. The answer is simple. The thing to do is to start with the identity
to be established and work toward an obviously true equation. Then take a

258
TRIGONOMETRY
(Ch. 5
fresh sheet of paper and copy the steps in the reverse order, and as you do so,
you should check that each step is reversible and should add appropriate
explanations. An even simpler procedure is to write the identity to be proved
at the bottom of a page and work toward the top.
As a further example let us prove that the identity
tan A" + ctn A" = sec A" esc A"
holds except when the formula is meaningless. First read the proof backward;
i.e., from bottom to top. Then reread the proof the way it is intended; i.e.,
from top to bottom. In the rereading, check that each step is actually
reversible. We start with the identity
sin2 A" + cos2 A" = 1,
and from this we obtain
/sin A" cos Л°\
sin2 A" + cos2 A" = созЛ°зтЛ0( h ) = 1.
Vcos A" sin A°/
We now divide by cos A" sin A" and obtain
sin A" cos A" 1 11
cos A" sin A" cos A" sin A" cos A" sin A"
and this is equivalent to
tan A" + ctn A" = sec A" esc A".
This identity can also be established by starting with the left-hand side
and working toward the right. Thus
sin A° cos A°
tan A" + ctn A" = +
cos A" sin A"
The fractions on the right are added by reducing to the common denominator
cos A" sin A°. Thus
sin A" cos A"
tan A" + ctn A" = 1
cos A" sin A"
sin2 A0 cos2 A0
cos A° sin A° cos A° sin A°
sin2 A" + cos2 A" _ 1
cos A° sin A° cos A" sin A"
1 1
cos A" sin A"
= sec A" esc A".

Sec. 67]
IDENTITIES
259
Thus identities are not always proved by working backward. We gave
the method of working backward first in order to show how a very tempting
but incorrect method of proof can be converted into a correct method. The
following is a list of the more important trigonometric identities. Those that
have not already been established will be proved in the problems.
T67.1. sin (A0 + B°) = sin A° cos 5° + cos A° sin 5°.
T67.2. cos (A0 + B°) = cos A" cos 5° - sin A° sinB".
T67.3. sin (A0 - B°) = sin A° cos 5° - cos A" sin 5°.
T67.4. cos (A0 - B°) = cos A" cos 5° + sin A" sin 5°.
T67.5. sec A" = 1/cos A", cos A" = 1/sec A".
T67.6. esc A ° = 1 /sin A °, sin A ° = 1 /esc A °.
T67.7. tan A ° = sin A °/cos A °.
T67.8. ctn A ° = cos A °/sin A °.
T67.9. ctn A" = 1/tan A", tan A" = 1/ctn A".
T67.10. соз2Л° + sin2 A" = 1.
T67.ll. 1 + tan2 A" = sec2 A".
T67.12. 1 + сШ2Л° = сзс2Л°.
T67.13. sin 2A° = 2 sin A" cos A".
T67.14. cos 2A° = cos2 A" - sin2 A" = 1 - 2 sin2 Л0 = 2 cos2 Л0 - 1.
tan Л0 + tan 5°
T67.15. tan (A0 + B°) =
T67.16. tan (A0 - B°) =
T67.17. tan2^
1 - tan A" tan 5°
tan A" - tan 5°
1 + tan A" tan 5°
2 tan A"
1 - tan2 A0
T67.18. sin|^° = ±V(1 - созЛ°)/2.
T67.19. соз|Л° = ±V(1 +созЛ°)/2.
T67.20. tan \A° = ±V(1 - cos Л°)/(1 + cos A0).
T67.21. 1ап|Л° = (1 - cos A°)/sin Л°.
T67.22. tan \A° = sin Л°/(1 + cos A").
T67.23. sin C° + sin £>° = 2 sin |(C° + £>°) cos |(C° - £>°).
T67.24. sin C° - sin £>° = 2 cos |(C° + £>°) sin |(C° - £>°).
T67.25. cos C° + cos D° = 2 cos |(C° + D°) cos |(C° - £>°).

260
TRIGONOMETRY
(Ch. 5
T67.26. cos C° - cos D° = -2 sin |(C° + D°) sin |(C° - £>°).
sin C° - sin D° tan §(C° - D°)
T67.27.
sin C° + sin D° tan |(C° + D°)
Problems
1. Prove T67.9.
2. Write down T67.10; divide each member of this equation by cos2 A° (divide
each term in the left member by cos2 A°), and use this result to prove T67.ll.
3. Prove T67.12.
4. Substitute В = A in T67.1 and use the resulting equation to prove T67.13.
5. Prove T67.14.
6. Prove that
sin A° cos B° + cos A° sin B°
tan (A° + B°)
cos A° cos B° — sin A° sin B°
Divide both numerator and denominator by cos A° cos B° and use the
resulting equation to prove T67.15.
7. Prove T67.16.
8. Prove T67.17.
9. Prove that cos B° = ±V(l +cos2£°)/2 and use this result to prove
T67.19.
10. Prove T67.18.
11. Prove T67.20.
12. Prove that
1 - cos2£ 2 sin2 В
sin 2B 2 sin В cos В
— tan В
and use this result to prove T67.21.
13. Prove T67.22.
14. Prove that sin (A° + B°) + sin (A° - B°) = 2 sin A° cos B°.
15. Solve the equations A + В — С, A — В — D for А, В in terms of C, D;
substitute these values in the equation of Problem 14 and use the resulting
equation to prove T67.23.
16. Prove T67.24.
17. Prove T67.25.
18. Prove T67.26.
19. Prove T67.27.

Sec. 68]
INVERSE FUNCTIONS
261
Supplementary Problems
Establish the following identities:
1 — cos θ sin θ
sin б 1 + cos θ
2. (sin θ + cos б)2 = 1 + sin 20.
1 tan0
3.
cos 0 + ctn 0 1 + sin 0
sin 20 cos 20
4. — = sec 0.
sin 0 cos θ
5. sin πιθ cos я0 = И sin (m + л)0 + И sin (m — n) 9.
6. (tan 0 + ctn 0)2 = sec2 б + esc2 0.
2 ctn0
7. sin 20
1 +ctn20
1
8. ctn A° esc A° = — „
sec A — cos A
9. sin A° tan Λ° = sec A° — cos A°.
cos 2?" sin i?°
10. -0 + -z = cos B° + sin Я".
1 - tan B° 1 - ctn B°
11. tan^°= ± . sin^° .
VI - sin2 A°
12. sin Λ° = ± . tan^° .
VI + tan2 A°
13. cos^° = ± ----- ^_.
VI + tan2 A°
68. INVERSE FUNCTIONS
Suppose that in the equation x2 = sin xu we are given the value of x2
and wish to solve for x\. Since sin X\ lies between — 1 and +1, we must have x2
between — 1 and +1. If x2 is such a number and x\ is a corresponding number
that satisfies the equation, then π — xiy x\ + 2π, Χ\ + 4π, etc., are also
solutions. There is, however, always one and only one solution χγ that
satisfies the additional conditions
7Γ 7Γ
- - й χι й —
2 2
and this solution is denoted by
X\ = sin-1 x2.

262
TRIGONOMETRY
(Ch. 5
The notation is suggested by the analogous problem of solving the equation
*2 = axi for Χι, where xu x2, and a are numbers and α И 0. The solution of
this equation is
*i = i-)*2 = a-1*2.
Note, however, that sin-1 x2 does not mean 1/sin x2 = esc x2. If we wished to
indicate the reciprocal of sin χ by a negative exponent, we would have to
write (sinx2)—' = 1/sin x2. The notation sin-1 x2 = arcsin x2 is also used.
We now have a law that assigns to each number x2 of the interval — 1 ^
x2 ^ 1 a unique number Χι = sin-1 x2. This law is a function called the
inverse nne (or arcsine). Similarly, when —1 ^ x2 = 1» there is a unique
number x\ satisfying the equation x2 = cos x\ together with the additional
conditions
0 й χι й π,
and this number is denoted by
X\ = COS-1 X2.
The inverse cosine (or arccosine) is the function (law) that assigns to each
number x2 of the interval — 1 ^ x2 = 1 the number Xi = cos-1 x2. For
example, let us find sin-1 0.5, sin-1 —0.5, cos-1 0.5, cos-1 —0.5. Since
sin π/6 = 0.5, sin — π/6 = —0.5, and — π/2 ^ π/6 ^ π/2, —π/2 ^
— π/6 ^ π/2, we have sin-1 0.5 = π/6, sin-1 — 0.5 = —π/6. Similarly,
cos π/З = 0.5, cos 2π/3 = -0.5, and 0 й π/3 й π, 0 й 2π/3 ^ π, and
hence cos-1 0.5 = π/3, cos-1 — 0.5 = 2π/3. Note that the conditions
— π/2 ^ χι ^ π/2 are appropriate for the inverse sine but inappropriate for
the inverse cosine, since there are two values of x\ (namely, π/3, —π/3)
satisfying the conditions cos Χι = 0.5, —π/2 ^ χγ ^ π/2, and there is no
number satisfying the conditions cos Xi = —0.5, —π/2 ^ χλ ^ π/2.
Similarly, the conditions 0 ^ χι ^ π are appropriate for the inverse cosine but
inappropriate for the inverse sine.
An alternative way of describing the inverses of trigonometric functions is
the following: If —1 ^ x2 ^ 1, then there are many numbers X\ such that
sin Xi = x2. That number X\ for which \x\\ is a minimum is X\ = sin-1 x2.
There are also many numbers x\ such that cos x\ = x2. Of these numbers,
the one for which \x\\ is a minimum and x\ is non-negative is X\ = cos-1 x2.
The inverses of the remaining trigonometric functions are similarly defined.
Thus, if дсг is any real number, then X\ = tan-1 x2 is that number X\ such that
tan X\ = x2 and | X\ | is a minimum. If x2 is a real number not in the interval
— 1 ύ *2 = 1> then *i = sec-1 x2 is that non-negative number Χγ such that

Sec. 68]
INVERSE FUNCTIONS
263
sec X\ = *2 and | X\ j is a minimum. In any case, X\ is chosen so that | JCi | is a
minimum, and when there are two numbers X\ for which | jci | is a minimum,
the positive one is chosen. However, x\ = sec-1 x-i is sometimes defined as
that number xx such that sec χλ = χ-ι and 0 ^ X\ < π/2, or π ^ X\ < Ътг/2.
These two definitions do not agree, and it is necessary to specify which
definition is being used.
EXAMPLE 68.1
Find cos 2 tan-1 л/3· Let tan-1 л/з = θ. Then tan 0 = л/з, ~ V2 =
θ = π/2. Thus θ = тг/3 = tan-1 л/з. 2 tan-1 л/з = 2тг/3, and hence
_ι Г 2ηΓ 1
cos 2 tan V 3 = cos — =
3 2
EXAMPLE 68.2
Show that cos-1 χ = sec-1 1/x. Let cos-1 χ = θ. Then cos θ = χ, 0 ^
б ^ π, sec б = 1/cos θ = l/x, and hence cos-1 χ = б = sec-1 1/x.
EXAMPLE 68.3
Show.that tan cos-1 χ = y/l — x2/x. Let θ = cos-1 x. Then cos θ = x,
О ^ б й τ, sin б = +л/1 - cos^fl = Vl - x2, and hence
sin θ Vl - *2
tan cos χ = tan б = =
cos θ χ
EXAMPLE 68.4
Show that tan-1 3 - tan-1 4 = tan-1 (-1/13). Let tan-1 3 = a
(alpha), tan-1 4 = β (beta). Then tan a = 3, -π/2 й а й т/2, tan /3 = 4,
- тг/2 й β й π/2, and
tan a — tan /3 3—4 1
tan (a - β) = = = ;
1 + tana tan β 1 + 12 13
hence tan-1 3 - tan-1 4 = a - β = tan-1 (-Из)· N«te that 0 < a,
β < π/2, and - тг/2 < a - β < 0 < тг/2.
EXAMPLE 68.5
Prove that sin-1 χ + sin-1 у = sin-1 {xy/l — y2 + ул/l — x2). Let
sin-1 χ = a, sin-1 у = β. Then sin a = x, cos a = \/l — я2, sin β = у,
cos β = \/l — У> and sin (a + /3) = sin a cos β + cos a sin β = xy/l — у
+ ji\/l — *2> and from this the result follows.

264
TRIGONOMETRY
(Ch. 5
Questions
1. What restrictions are imposed on *2 in order for sin-1 *2 to be defined? 2. Define
sin-1 *2. 3. Define cos-1 *2. 4. Define the functions inverse sine and inverse cosine.
5. Define tan-1 хг and sec-1 хг and state the values of хг for which these functions
are defined.
Problems
1. Find cos sin-1 ( — %), tan 2 sin-1 ( —И)·
2. Find sec tan-1 1, esc tan-1 ( — 1).
3. Show that tan-1 Ц + tan-1 У, - π/4.
4. Show that cos"1 И + cos"1 Из = cos-1 (-"^s)·
5. Show that cos-1 { — %) — cos-1 % — cos-1 0.
6. Show that sin"1 l/λ/ΪΟ + cos"1 2/VI = тг/4.
Prove the following identities:
7. tan-1 χ = ctn-1 1/дс.
8. tan sin-1 χ =■ x/V\ — xi.
9. cos sin ' χ = y/\ — x2.
10. tan-1 χ + tan-1>> = tan-1 (* + y)/(\ — xy).
11. cos ' χ + cos ' у = cos ' (xy — λ/(1 — дг2)(1 — у2)).

ALTERNATIVE
COORDINATE
SYSTEMS
69. POLAR COORDINATES
Polar coordinates are defined as follows: Let ν be any unit vector and r
any scalar. Then there exists a unique point X (plane analytic geometry)
such that
OX = rv.
Moreover, the unit vector ν is uniquely determined by the radian measure θ
of the angle (ui, v). Hence Θ, r determine the point X. The numbers б, г are
called polar coordinates of X, and X is called the point (0, r). However, X does
not determine its polar coordinates uniquely. For example, if r = 0, then X is
the origin О no matter what choice is made of 0 (i.e., of v). Moreover, if X is
the point (r, 0), then
OX = rv = (-r)(-v)

266 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
where (see T57.1c) θ + π is a measure of
(ub — v). That is, θ + π, —r are polar
coordinates of X. This point also has the
coordinates θ + 2π, r. Note that the coordinate
r can be negative; i.e., ν and OX can have
opposite senses. Figure 69.1 illustrates the
Figure 69.1 plotting of such a point, namely, the point
(π/ό, — 3). It is customary to measure the
angle (ub v) in radians because there are many problems involving both
polar coordinates and the calculus, and in these problems radian measure is
much more convenient than degree measure.
If we have an equation relating θ and r, then the set of points whose polar
coordinates satisfy the equation is a curve.
EXAMPLE 69.1
Plot the curve
2 cos Θ.
We begin by making a table giving the values of r corresponding to various
values of Θ. This is done with the aid of the tables obtained in Problem 1 of
(¥--U)
(¥-o)
(¥--1.4)
(^-1.4)
Figure 69.2
Section 61. Figure 69.2 shows the curve.
θ
r
0
2
π/6
1.7
тг/4
1.4
7Г/3
1.0
тг/2
0
Зтг/4
-1.4
7Г
-2
5тг/4
-1.4
Зтг/2
0
7тг/4
1.4
2тг
2

Sec. 69]
POLAR COORDINATES
267
Note that some of the points plotted are assigned two pairs of coordinates.
In fact, as θ varies from zero to 2π, the curve is traced twice.
The curve of Fig. 69.2 looks like a circle. To show that it is, we find its
equation in cartesian coordinates. Relations between the two coordinate
systems are obtained as follows:
OX = [χι, x2] = rv = r[cos 0, sin Θ] = [r cos б, г sin Θ],
and hence
X\ = r cos 0, *2 = r sin Θ.
We also have
*i2 + *22 = r2 (cos2 θ + sin2 0) = r2,
Г = ±Vxi2 + *22,
and 0 is such that
— = cos 0, — = sin Θ.
r r
The equation of Example 69.1 is
r = 2 cos θ or r2 = 2r cos Θ.
Note that there is a point of the curve r = 2 cos θ for which r = 0, namely,
(π/2, 0), and hence that the multiplication by r does not introduce any
extraneous point.
Since
r2 = x2 + *22, r cos θ = χι
the equation becomes
*i2 + *22 = 2*.
By the method of Section 45 this can be converted into
(Ж1 - l)2 + x22 = 1,
and this equation represents a circle with radius 1 and center at (1, 0)
(cartesian coordinates). The polar coordinates of the center are (0, 1).
In Section 13 we obtained the formula
n-CX
|h| = ±—-
where CX = OX — ОС for the distance from the point X to the line
containing the point С and having the normal n. We shall specialize this formula

268 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
to the case in which η is a unit vector and С is such that ОС = en, where с is
non-negative. We also let OX = rv. Then
|h| = ±n-CX = Мл-OX- n-OC)
— ±(rn-v — cn-n) = ±(rvn — c).
If X, С are respectively the points (0, г), (а, с), then (ub v), (ub n), (n, v)
have respectively the radian measures θ, α, θ — a, and
vn = cos (0 — a).
Hence
|h| = ±(rcos (0 - a) - c).
In particular the distance from the line to the origin (Θ, 0) is +c. That is,
we are concerned with the line whose normal is the unit vector η and whose
distance from the origin is c. This line is the set of points (0, r) for which
| h | =0. The equation is
r cos (Θ — a) — с = 0.
EXAMPLE 69.2
Find the equation of the line whose normal is — Ui and whose distance
from the origin is 3. Find the distance from (71-/З, 2) to this line. Since the
measure of (ui, — Ui) is π radians and since
cos (Θ — π) = cos (π — θ) = — cos θ,
an equation of the line is — r cos θ — 3 = 0 or
r cos θ + 3 = 0.
The distance from (71-/З, 2) to this line is
= +i2cos^+ 3j = 1 + 3 = 4.
EXAMPLE 69.3
Consider the line whose normal is —щ and whose distance from the origin
is c. Its equation is
r cos θ + с = 0.
A point (0, r) moves so that its distance from the origin, divided by its
distance from this line, is always e. Find an equation of the locus. The
equation is

Sec. 69]
POLAR COORDINATES
269
r cos θ + с
We can solve this equation for r (or —r) and obtain
ce
ce
1 — e cos θ
or — r
1 + e cos θ 1 — ί cos (0 + π)
according as the plus sign or minus sign is chosen. Since (0, r) and
(θ + π, — r) represent the same point, these equations are equivalent. The
curve is the conic whose focus is the origin, whose directrix is the given line,
and whose eccentricity is e. The polar coordinate equation of a conic is the
most convenient one for solving certain problems related to planetary motion.
EXAMPLE 69.4
Plot the curve
cos 2Θ.
If we were to assign to θ the values that are multiples of π/4, we would
discover that cos 2Θ changes so rapidly that we should not know how to
connect the points by a curve. We therefore choose multiples of π/8 and
construct the accompanying table.
π/8
0.71
π/4
0
3π/8
-0.71
π/2
-1
5π/8
-0.71
3π/4
0
7 π/8
0.71
π 9π/8 5π/4 11 π/8 3π/2 13π/8 7π/4 15π/8 2 π
1 0.71 0 -0.71 -1 -0.71 0 0.71 1
The curve is shown in Fig. 69.3. It is instructive to see how complicated the
(ψ.-.π)ι
(¥-71) 1
π ll/
πι 41
(¥..71)
(¥.--7i Л
(¥
-ι)
UT..71)
' (f.-7l)
hit
/
\ (I^1}
^(¥.-■71)
(0,1)
(f.-1)
Figure 69.3

270 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
equation of this curve is in cartesian coordinates. If we multiply both sides of
the equation by r2, we obtain
r3 = r2 cos 20 = ^(cos2 θ - sin2 Θ) = r2 cos2 θ - r2 sin2 θ
2 2
= *1 - *2 ·
Since (r3)2 = (r2)3, we can square both sides and obtain
(*12 + *22)3 = (V - *22)2·
Consider a line through the origin; i.e., a line whose distance с from the
origin is zero. The equation is
r cos (Θ — a) = 0,
and this equation is satisfied if θ — a = — π/2 or θ — a = π/2; i.e., if
7Γ 7Γ / 7Г\
θ = a or θ = a ^ =(o! Ι + 7Γ.
2 2 V 2/
Since (a — π/2, r), (a — π/2 + π, —r) represent the same point, we need
consider only θ = a — π/2. Let ν be such that the measure of (ui, v) is
a — π/2. Then ν _L η and the line is the set of points X for which there
exists an r such that OX = rv. An alternative form of the equation of the
line is
τ
θ = a — -z = constant,
and conversely an equation of the form θ = constant represents a line through
the origin. An equation of the form
r = a = constant ^ 0
becomes
2 2,2 2
r = x\ + x2 = ar
in cartesian coordinates, and this represents a circle with center at the origin
and radius a.
Questions
1. Define the polar coordinates of a point. 2. Do these coordinates determine the
point? 3. Does the point determine its polar coordinates uniquely? 4. Find various
relations between the polar coordinates and the cartesian coordinates of a point.
5. Let C, X be respectively the points (a, c), (0, r),where с is non-negative. Let η, ν be
unit vectors such that ОС — en, OX — rv. Show that the distance | h | from X to

Sec. 69]
POLAR COORDINATES
271
the line having the normal η and passing through С is
|h| = ±n-CX = ±(rv-n - c) = ±(rcos(0 - a) - c).
6. Show that the distance from the origin to the line is с and that an equation of
the line is
r cos (0 — a) — с = 0.
7. A point (0, r) moves so that its distance from the origin, divided by its distance from
the line r cos 0 + с — 0, is always e. Show that an equation of the locus is
1-е cos 0
8. Show that a line through the origin has an equation of the form 0 — constant. 9.
Show that a circle with center at the origin has a equation of the form r — constant.
Problems
1. Plot the following curves:
(a) r = 2 sin 0; (b) r sin 0 + 3 = 0;
(c) r = -, where a - 2, e - \;
1-е cos0
(d) r = 0 (spiral).
For each curve make out a table in which 0 is assigned the values 0, 7г/6, 7г/4,
π/3, π/2, 37г/4, π, 5π/4, 3π/2, 7π/4, 27τ.
2. Plot the curve
r - sin 20.
Assign to 0 the values 0, π/β, π/4, 3π/8, etc., up to 2π.
3. Convert equations (a), (b) of Problem 1 into cartesian coordinates.
4. Find an equation of the line that contains the point С — (5, 2π/3) and has
the normal ОС.
5. Use T56.5 to convert the equation
r cos (0 — a) — с — 0
into cartesian coordinates.
6. The point (a, a) lies on the circle with center at the origin and radius a. Assume
that the tangent to the circle at (a, a) is perpendicular to the radius to this
point. Find a polar coordinate equation of the tangent. Show that if (a, a) has
the cartesian coordinates (αϊ, аг), then an equation of the tangent in cartesian
coordinates is
αιχι + <22*2 — a2 = 0.
Use the result of Problem 5.

272 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
70. CYLINDRICAL COORDINATES AND SPHERICAL
COORDINATES
We shall study two new coordinate systems in solid analytic geometry,
namely, cylindrical coordinates and spherical coordinates. Cylindrical
coordinates are defined as follows: Let X be any point and let Υ be the point of
the plane of Ui, U2 for which YX is a minimum; i.e., let Υ be the projection
of X on the plane of Ui, U2. Then YX'tii = 0 = YX-u.2, and hence there
exists a scalar x$ such that YX = хи$. Since Υ is in the plane of Ui, U2,
there exist a unit vector ν in this plane and a scalar r such that OY = rv.
Hence
OX = OY + YX = rv + *3u3 = [xi, *2, *з].
If the measure of (ui, v) is θ radians then б, г are polar coordinates of Y.
The numbers б, г, *з determine the point X; they are called cylindrical
coordinates of X, and X is called the point (б, г, *з). The set
of points for which r = constant ^ 0 and Θ, x% are
arbitrary is a right circular cylinder intersecting the plane
*з = 0 in a circle and having the vector из as an axis.
Hence the name "cylindrical coordinates." The set of
points for which θ = constant and г, *з are arbitrary is a
plane containing из and intersecting the plane *з = 0
in a line through the origin. The plane is divided into
halves by the line containing из, and if we restrict r to be
non-negative (as is usually done), then the set of points
θ = constant consists of only one of these halves. The
set of points for which *з = constant is a plane parallel
to *з = 0. The point (б, г, *з), where r ^ 0, is thus the
intersection of a half-plane, a cylinder, and a plane.
Figure 70.1 shows the point (π/Ъ, 2, 3) at the intersection of the half-plane
θ = π/3, г ^ 0, the cylinder r = 2, and the plane *з = 3. In drawing the
intersections of these surfaces with one another and with the coordinate
planes, recall the instructions in Section 37. To draw the vector ν = [cos
π/3, sin π/3, 0], note that 2v = [1, \/3, 0] and that the tip of 2v lies on the
circle r = 2, x3 = 0. and also on the line χλ = 1,*з = 0.
Spherical coordinates are described as follows: Let r, ν be defined as in
cylindrical coordinates; let ρ (Greek letter rho) be the non-negative scalar
Figure 70.1

Sec. 70] CYLINDRICAL AND SPHERICAL COORDINATES 273
and w the unit vector such that OX = pw. Then
X = (4M = (f?.2)
OX = pw = rv + *зиз>
where r ^ 0 and б again determines v. The vector w lies in the half-plane
containing U3, v, and U3, ν is an orthonormal system. If the measure of
(из, w) is φ (Greek letter phi), then φ determines the vector w in this half-
plane, and hence θ, φ, ρ (theta, phi, rho) determine X. These three numbers
are called the spherical coordinates of X,
and X is called the point (θ, φ, ρ). The
points for which ρ = constant ^ 0 are
at distance ρ from the origin; i.e., this
set of points is the sphere with center at
the origin and radius p. Hence the
name "spherical coordinates." The set
of points for which θ = constant, r
^ 0 is a half-plane. The set of points
for which φ = constant, ρ ^ 0 is one
nappe of a right circular cone with axis
U3. This cone intersects the sphere in a
circle that forms a base of the cone.
The point (θ, φ, ρ) is the intersection of
a half-plane, a cone, and a sphere.
Figure 70.2 shows the point (π/3,
π/6, 2) at the intersection of the half-
plane θ = π/3, г ^ 0, the cone φ = π/6, ρ ^ 0, and the sphere ρ = 2.
The sphere intersects the plane θ = constant in a circle. At the point where
this circle intersects the line containing u3, the tangent to the circle is parallel
to v, and at the point where the circle intersects the line containing v, the
tangent is parallel to U3. Draw short tangents at these points before drawing
the isometric projection of the circle; erase the tangents afterward. To draw
the vector w = U3 cos φ + ν sin φ, note that 2w = U3 s/b + ν and that the
tip of 2w lies on the intersection of the circle θ = π/Ъ, ρ = 2 with the line
θ = π/3, r = 1.
The number θ is called the longitude and φ the colatitude (complement of
the latitude). The measure π/2 — φ of (w, v) is called the latitude. On the
surface of the earth, latitude and longitude are measured in degrees and
latitude is divided into east and west latitude measured east and west from
the meridian half-circle (Θ = 0, r ^ 0, ρ = the radius of the earth) through
Greenwich. In locating a star, the origin is taken at the point from which
the star is being observed, and a left-handed coordinate system is used in
Figure 70.2

274 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
which Vi points northward, V2 eastward, and V3 upward. The longitude
converted into degrees is called the azimuth and the latitude converted into
degrees is called the altitude. This system is used in astronomy and navigation.
In astronomy the distance of the star from the observer is also important,
and this distance is the coordinate p.
If a point has the cylindrical coordinates (0, r, x3) and the spherical
coordinates (0, φ, ρ), then
*3 = ρ cos φ, r = ρ sin φ, ρ = Vxj2 + r2,
and φ is such that
— = cos φ, - = sin φ.
Ρ Ρ
These are the relations between cylindrical and spherical coordinates. The
cartesian coordinates are related to the cylindrical and spherical coordinates
as follows:
x\ = r cos θ = ρ cos θ sin φ,
*2 = r sin θ = ρ sin θ sin φ,
*з = ρ cos φ,
Ρ = λΛι2 + *22 + *з2.
EXAMPLE 70.1
The equation θ = π/3 (spherical coordinates) is converted into cartesian
coordinates as follows:
*i
*2
*2
*i
*2
π _ ρ sin φ
ρ cos — sin φ = >
3 2
sin φ
ρ sin — sin ώ = >
3 2
sin ф)/2 ^ ^^
(p sin ф)/2
= *,л/з.

Sec. 70] CYLINDRICAL AND SPHERICAL COORDINATES 275
EXAMPLE 70.2
The equation φ = π/4 is converted into cartesian coordinates as follows:
7Г 7Г
xi2 + *22 = p2 cos2 0 sin2 - + p2 sin2 0 sin2 -
4 4
2
4 2 '
7Γ ρ
= ρ2 (cos2 0 + sin2 0) sin2 - = -
and hence
or
2 2 2 "" P
V = p-'cos - = —■
4 2
*3Z p72
*i2 + *22 P2/2
= 1,
2 2 ι 2
*3 = *1 + *2 ·
EXAMPLE 70.3
The equation ρ = 2 is equivalent to
p2 = «i2 + x22 + *32 = 4.
Questions
1. Let Ζ be a point in three-dimensional space. Show that there exist unit vectors
v, w and scalars хз, г, р such that
OX = pw = OK + ΚΛ-, OK = rv, ГЛ- = ^зиз,
where К is a point of the plane of ui, иг. 2. Let the measures of (щ, ν), (из, w) be
respectively 0, φ. Show that the cylindrical coordinates 0, г, дгз determine X and that
the spherical coordinates 0, φ, ρ also determine ΛΓ. 3. Describe the surface r —
constant ^ 0, the surface 0 = constant, r ^ 0, the surface дгз = constant, and state
how X is related to such surfaces. 4. Describe the surfaces ρ = constant ^ 0 and φ —
constant. 5. How are the spherical coordinates 0, ρ related to longitude and latitude
and how are they related to azimuth and altitude? 6. What is the interpretation of ρ
in astronomy? 7. Find equations relating cylindrical coordinates to spherical
coordinates. 8. Find equations relating cartesian coordinates to both cylindrical
coordinates and spherical coordinates.

276 ALTERNATIVE COORDINATE SYSTEMS (Ch. 6
Problems
1. This problem is concerned with the location of the point (π/6, 2, 4) (cylindrical
coordinates). For this point ν = [V3/2, Vi, 0]. Draw the intersections of the
plane *з = 4 with the coordinate planes x\ — 0, χι — 0. Draw the intersections
of the cylinder r — 2 with the planes *3 = 0, *з = 4, by first drawing the
squares containing these circles, next drawing the circles, and then erasing
the squares. Note that the tip of the vector 2v = [λ/3, 1,0] is at the
intersection of the circle r — 2, *3 = 0, with the line χι — 1, *3 = 0. Locate this
tip and then draw the intersections of the half-plane θ — 7г/6, г ^ 0, with the
planes дгз = 0, дг3 = 4 and the cylinder r — 2. Locate the point (π/6, 2, 4).
2. This problem is concerned with the location of the point (π/ό, π/4, 2) (spherical
coordinates). Draw the intersections of the sphere ρ — 2 with the coordinate
planes by first drawing the three squares containing these circles and then
drawing the circles. Draw the intersections of the cone φ — 7г/4, with the planes
xi — 0, ДГ2 = 0. Note that these lines of intersection pass through corners of the
above squares. Now erase the squares. Draw the intersection of the cone
φ — 7г/4, with the sphere ρ = 2, by first drawing the square containing this
circle, next drawing the circle, and then erasing the square. Draw the
intersection of the half-plane θ — 7г/6, г ^ 0, with the plane *3 = 0. See Problem 1.
Draw the intersection of the half-plane θ — π/6, г ^ 0, with the sphere ρ = 2,
by first drawing the rectangle containing this half-circle, next drawing the
half-circle, and then erasing the rectangle. Note that two sides of the rectangle
are parallel to the line θ — π/6, χ3 — 0. Draw the intersection of the half-
plane θ — π/6, г ^ 0, with the cone φ = π/4. Locate the point (π/6, π/4, 2).
3. Find the cylindrical coordinates and cartesian coordinates of the point
(тг/6, тг/4, 2).
4. Show that the plane θ — π/4 has the equation x\ — хг — 0.
5. Show that the cone φ — π/6 has the equation *i2 + дг22 = *з2/3·

COMPLEX
NUMBERS
7
71. ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS
If i is a number such that i2 = — 1 and αχ, a2 are real numbers, then
a = ai + ia2
is called a complex number, αγ is called the real part of a, and a2 is called the
imaginary part. If a2 = 0, then a = αγ is a real number, and if αϊ =0 but
Д2 И 0, then a = ia2 is called a />urf imaginary number. The sum of two
complex numbers is defined in a natural way. Thus, if b = b\ + ib2 is another
complex number, then
a + b = ai + ia2 + bx + ib2 = (αλ + bi) + i(a2 + b2) = b + a.

278
COMPLEX NUMBERS
lCh.7
This formula resembles the formula for the sum of two vectors. Thus, if
a = k, a2], b = [bu b2], then
a + b = [au a2] + [bu b2] = [αϊ + bu a2 + b2] = b + a.
Recall that two vectors are equal if and only if their components are both
equal, and that a vector is zero if and only if both of its components are zero.
Similarly, two complex numbers are equal if and only if their real and
imaginary parts are equal, and a complex number is zero if and only if its real and
imaginary parts are both zero. The zero complex number is denoted by 0.
A complex number can thus be represented geometrically as a vector. The
numbers a, b are represented respectively by a, b, and the sum a + b of
the numbers is represented by the vector sum a + b. The negative of a
complex number and the difference between complex numbers are
represented respectively by the negative of the representative vector and the
difference between the representative vectors. Thus
-a = (-αϊ) + i(-a2) = -αϊ - ia2,
b - a = (bi - αϊ) + i(b2 — a2),
are represented by — a and b — a, respectively. Note that
α + (b + c) = (αϊ + ii + Cl) + i(a2 + b2 + c2) = (a + b) + с
The product of two complex numbers is also defined in a natural way by
the formula
ab = (αϊ + ia2)(b\ + ib2) = a\b\ + ia\b2 + ia2b\ + i2a2b2
= {a\bi — a2b2) + (a\b2 + 02^1)г = ba.
if a is real, i.e., a2 = 0, a = a\, then ab = a\bi + ia\b2 = abi + iab2. With a
little computation one can check that (see Problem 2)
a(bc) = (ab)c.
If β = βι + ia2, then
ά = a\ — ia2
is called the complex conjugate of a. Note that
α + ά = 2αι and αά = a\ + a2
are both real. The absolute value of a complex number is denoted by \a\
and defined to be the magnitude a of the representative vector a. Thus

Sec. 71] ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS 279
Note that when a is real (i.e., when a2 = 0 and a = αϊ), then
\a\ = | «ι ί = Va?
and that this agrees with the formula for the absolute value of a real number.
Also note that a = 0 if and only if
|«| = Vai2 + a22 = 0.
The quotient χ = a/b (when b И 0) is defined to be the number χ such
that a = bx. This equation implies that
al· = bbx,
and since bl· and \/bl· are real, it also implies that
1 1
— al· = — bbx = x;
bl· bl·
i.e.,
1 1
χ = — al· = — — ((aibi + a2b2) + i(a2bi - aib2))
bl· bi2 + b22
a\b\ + 02^2 .02^1 — 01^2
ьх2 + ь22 +l bx2 + b22 '
Conversely, if χ is so defined, then
1 1
bx = b — al· = — Λα = a:
bl· bl·
a al· a\b\ + 02^2 .α2^ι — Q\b2
I ^x = Vl· = ь,2 + ь22 +l ь,2 + ь22 '
i.e.,
Questions
1. What is a complex number? 2. What is the real part of a complex number?
3. What is a pure imaginary number? 4. Define the sum of two complex numbers.
5. Define the representative vector of a complex number. 6. Show that the
representative of the sum is the sum of the representatives, the representative of the negative is
the negative of the representatives, and the representative of the difference is the
difference of the representatives. 7. Define the zero complex number and show that a
complex number is zero if and only if its representative is the zero vector. 8. Define
the product of two complex numbers. 9. Define the conjugate of a complex number.
10. Define the absolute value of a complex number. 11. Show that a = 0 if and only if
| «a | = 0. 12. Define the quotient of two complex numbers and show how to obtain its
real and imaginary parts.

280
COMPLEX NUMBERS
(Ch. 7
Problems
1. Find the real and imaginary parts of (3 — 2i) + ( — 7 + t), (3 — 2i) —
(-7 + 0, and (3 - 2t)(-7 + t).
2. Let a — a\ + i<4, b — b\ + ibt, с — c\ + tC2· Show that
(ai)c = {a\b\c\ — αη}>4£\ — аф&г — агЬ\сг)
+ i(aib\C2 + а\Ь&\ + 02*1^1 — 02*2^2)
= a(bc).
3. Show that ab — ~ab.
4. Show that \ab\ = VablTb = |α|-|*Ι·
5. Show that if χ — a/b, then i = a/b.
6. Use the triangle inequality to show that
\a + b\ й \а\ + \b\.
Does this inequality hold if a, b are real? Test for the case a — 3, b — 4 and
the case a — 3, b — —4.
7. Find real and imaginary parts of (3 — 2t)/( —7 + t).
72. MULTIPLICATION AND ROTATION
Let υ be a complex number such that | o| =1 and let Vi, V2 be respectively
the representative vectors of» and iv. Let the measure of (ui, vi) be a (alpha)
radians. Then
Vi = [cos a, sin α], υ = cos a + i sin a,
iv = — sin a + i cos a, V2 = [ — sin a, cos a];
i.e., vi, V2 is a right-handed orthonormal system. If χ = [xly x2] is the
representative vector of χ = xi + ix2, then
vx = v(xi + ix2) = Mi + (iv)x2,
and hence the representative vector of ox is
vi*i + v2*2 = R(x).
where R = [vi, V2] is the rotation through a radians. Similarly, if w is such
that I w I =1, the representative vectors of w, iw are respectively wj, W2, and
S = [wi, W2] is the rotation through /3 (beta) radians, then the representative

Sec.72] MULTIPLICATION AND ROTATION 281
vector of wx is S(x) and the representative vector of wvx is
S(R(x)) = SR(x).
Thus the multiplication of χ by wo corresponds to the product rotation SR
through a + β radians applied to the representative x. As a check on this
result, note that
vw = (cos a + i sin a) (cos β + i sin β)
= (cos a cos β — sin a sin β) + (sin a cos β + cos a sin β)
= cos (a + β) + i sin (a + 0).
The complex number i is such that |i| = 1, and its representative vector is
иг- A multiplication by i corresponds to the rotation J = [u2, — щ] through
π/2 radians. The number 1 has the representative vector ub and a
multiplication by 1 corresponds to the identity rotation I = [ui, иг] through zero
radians.
The representative vector [χι, x2] of the number χ = X\ + ix2 has its tip at
the point (*i, хг). We can choose polar coordinates Θ, r of the tip so that
r= VXl2 + ,22 = \x\Z0.
Then
X\ = r cos θ = \x\ cos Θ, *2 = r sin θ = | χ| sin Θ,
and hence
χ = Χι + ix2 = \x\ cos θ + i\ χ\ sin θ
= \x\ (cos θ + i sin Θ) = \x\v,
where |x| is real and non-negative and ν = cos θ + i sin θ is such that
\v\ =1.
If a, b are any two complex numbers, they can be written in the form
a = | a | (cos a + i sin a), b — \ b \ (cos β + i sin β),
and hence
ab = \a\ (cos a + i sin a) \ b \ (cos β + i sin β)
= \a\ - | 6 | (сое (a + β) + г sin (α + β)),
where
\a\-\b\ = \ab\ Ш0
and
| cos (a + β) + i sin (a + β) \ = 1.

282
COMPLEX NUMBERS
(Ch. 7
Questions
1. Let ν be a complex number such that |»| =1. Define a rotation R such that if
χ is the representative vector of x, then R(x) is the representative vector of vx. 2. Define
a corresponding rotation S for a complex number w for which \w\ = 1, and show
that SR(x) is the representative vector of wvx. 3. Let R, S be defined as in Questions
1 and 2; let R be the rotation through a radians and S be the rotation through β
radians. Show that
iiw — cos (a + β) + t'(sin (a + β).
4. Show that J = [иг, — ui] is the rotation associated with the complex number t,
and I = [ui, иг] is the rotation associated with the complex number 1. 5. Explain
the formula
χ — \x\ (cos θ + t sin Θ).
6. Show that if
a — I a I (cos a + i sin β), b = | b \ (cos β + i sin β),
then
ab = |e|-|*|(cos(a + β) + t sin (a + β)).
Problems
1. Show that if |d| = 1, then \/v — v.
2. Show that if \/v = v, then |d| = 1.
3. Show that if R is the rotation associated with v, where |v| =1, then R' is the
rotation associated with \/v.
4. Let χ — — 1 + «'. Draw the representative vector of χ and find θ such that
χ = |*| (cos θ + t sin Θ).
5. Let дг = — 1 — tv3. Draw the representative vector and find θ such that
χ = |*| (cos6 + t'sin0).
6. Find θ such that — t = \—i\ (cos б + t sin Θ).
7. Show that
1 = cos 2π + t sin 2π = cos 6π + t sin όπ.
8. Show that if * = cos θ + i sin Θ, then l/x = cos б — t sin 6. Show that
χ -\- χ — 2 cos 6.
9. Let a — a\ + 102 and let a, a* be the representative vectors of α, ία,
respectively. Then A = [a, a*] is called the associated matrix of the complex number a.
Write A in the form displaying its components.
10. Let A, B be the associated matrices of the complex numbers a, b, respectively.
Show that AB is the associated matrix of ab. See Problem 3 of Section 49.

Sec. 73] DE MOIVRE'S THEOREM 283
73. DE MOIVRE'S THEOREM
We have seen that if
a = | a | (cos a + i sin a), b = \b\ (cos β + i sin /3),
then
ab = \a\-\b\ (cos (a + β) + i sin (a + β)).
In particular if a = b = χ, α = β = θ, then
χ2 = |*|2(cos (θ + θ) + ι sin (θ + Θ))
= |*|2(cos20 + г sin 26).
Furthermore
x3 = x2x = | x21 ■ | χ| (cos 2Θ + г sin 20)(cos θ + i sin Θ)
= |*|3(cos30 + ί sin 30),
and in general, we have T73.1.
T73.1. If χ = \x\ (cos 0 + г sin 0) am/ η is a positive integer, then
xn = |*|n(cosn0 + г'зтя0).
This result is called De Moiore's theorem.
EXAMPLE 73.1
Find three roots of the equation x3 = 1. Since cos 2π = 1, sin 2π = 0,
we can write 1 in the form
1 = cos 2π + г sin 2π,
and if χ = |*| (cos 0 + г sin 0), then *3 = |*|3(cos 30 + г sin 30) and the
equation x3 = 1 becomes
| χ | 3(cos 30 + i sin 30) = 1 (cos 2тт + i sin 2тг).
This equation is satisfied if
| jc |3 = 1 or |*|=1, 30 = 2тг or 0 = —;

284
COMPLEX NUMBERS
ICh. 7
i.e., if
2π 2π
χ = cos (- i sin —
3 3
/ Λ . . / Λ
= COS ^7Г - -J + I Sin \^Г - -J =
7Г 7Г
■ cos —(- ι sin —
3 3
ι 1V3
= -2+-Γ·
To find another root, we note that
(*2)3 = (*3)2 = l2 = 1,
and hence
is another root. Obviously
6π
4π 4π
χ = cos (- ί sin —
3 3
6π
д: = cos (- ι sin — = cos 2π + г sin 2π = 1
is also a root. We cannot obtain any additional roots by forming χ4, χ , etc.,
since
x4 = Λ = χ, χ5 = x3x2 = x2.
The roots x, x2, x3 correspond respectively to rotations through 2π/3, 47τ/3, 2π
radians. If the tips of the vectors representing these roots are connected by
straight-line segments, the resulting figure is an equilateral triangle inscribed
21, . . 2*
cos "y + i sin -j- — x
z = cos -j- +i sin -5-
Figure 73.1
in a circle of unit radius (see Fig. 73.1). It is instructive to check that χ is a
root, by the following direct computation:

Sec. 73]
DE MOIVRE'S THEOREM
285
„ / 1 гЛ/з\2 1 3 / л/з л/з\
ι i-v/з
2 2 '
—(-ΐ^Χ-ΐ-ΐ)
1 3
= - + - = 1.
4 4
EXAMPLE 73.2
Solve the equation y* = 1. If j^ = |_y | (cos 6 + г sin 6), then jy4 = |jy|4
(cos 40 + г sin 46), and the equation jy4 = 1 becomes
\y | 4(cos 46 + i sin 46) = 1 (cos 2π + ι sin 2π).
This equation is satisfied if
\y\A = 1 or |^|=1, 46 = 2тг or θ = —,
i.e., if
7Γ 7Γ
ν = cos h ι: sin — = i.
2 2
The remaining roots are
у = cos π + i sin π = — 1,
y3 = cos 3π/2 + i sin 3π/2 = — i
y* = cos 2π + г sin 2π = 1,
since
(/)4 = (/)2 = i, (У5)4 = (/)8 = i.
The roots correspond respectively to rotations through π/2, π, 3π/2, 2π
radians. If the tips of the vectors representing these roots are connected by
straight-line segments, the resulting figure is a square inscribed in a circle of
unit radius (see Fig. 73.1). Similarly, the roots of the equation z5 = 1 are
ζ = cos 2π/5 + i sin 2π/5, ζ2, ζ3, ζ4, ζ5 = 1. If the tips of the vectors
representing these roots are connected by straight-line segments, the resulting
figure is the regular pentagon of Fig. 73.1.

286 COMPLEX NUMBERS (Ch. 7
EXAMPLE 73.3
Solve the equation x3 = a, where a = — 1 + i. The vector a = [ —1, 1]
represents the number a and \a\ = y/\ + 1 = y/2. By drawing the vector
a, it is easily seen that the angle (ui, a) has degree measure 135 and radian
measure 3π/4 Thus the equation x3 = a is equivalent to
x3 = | χ |3(cos 30 + i sin 30) = λ/2 (cos \- i sin — J.
and this equation is satisfied if
I jc |3 = V2 or \x\=y/l, 30 = — or 0 = -;
4 4
i.e., if
, = ^(cos^ + isin-;) = ^(^ + ^)
= V2(1+0-
The remaining roots are obtained by utilizing the results of Example 73.1.
Thus, since
</2 /-ι iVlw* /<Гг \3/-i iVb\3
+ —)
V2
(-1 +0-1
/- \ Δ Δ / / \ /- /\2 2/
V2 V2
and similarly
it follows that
y/l /-1 гЛ/з\ \/2 r
= 2^т= ((-1 - V3) + ,-(-i + V3)),
y/l /-1 г"л/з\ -^2 /- r-
are the remaining two roots.

Sec. 73]
DE MOIVRE'S THEOREM
287
EXAMPLE 73.4
Solve the equation x* = b, where b = — 1 + iy/b- The vector b =
[ —1, \/3] represents the number b, and |ό| = y/\ + 3 = 2. The angle
(ui, b) has radian measure 2π/3. Hence the equation x* = b is equivalent to
/ 2тг 2τγ\
| χ |4(cos 46 + i sin 40) = 2 f cos h i sin — J ·
This equation is satisfied if
(Г 2ТГ 7Г
I jc |4 = 2 or |*| =V2, 40 = — or 0 = -;
3 6
i.e., if
* = -V^fcos- + г sin- ) = (л/з +
Ονο 6/2
Since
(-^ (V3 + ,>■) = ^ (V3 + ο) (04 = (-i + «V3)-i,
it follows that another root is
>^2 /- \fl r-
(V3 + г>' = (-1 + »'V3).
2 2
The remaining roots are
— (V3 + 0(-i) = — (-V3 - o,
>^2 r- \fl r
— (V3 + 0(-0 = — (1 - iV 3).
In Problem 7 you are asked to show that
cos 30 = 4 cos3 θ — 3 cos 0, sin 30 = 3 sin 0 — 4 sin3 0.
If χ = cos 0, then cos 0 is a root of the equation
4*3 - Ъх - cos 30 = 0.
If we can trisect a given angle of 30 radians, then we can construct (with
ruler and compass) an angle of 0 radians, and hence we can construct a

288
COMPLEX NUMBERS
(Ch. 7
segment of length cos 0 relative to a given segment chosen to represent unit
length. That is, χ = cos б is a constructible root of the above cubic. Obviously
we can construct the angle of 30 = 2π/3 radians. If we can trisect this angle,
then since cos 2π/3 = — И, it follows that χ = cos 0 = cos 2π/9 must be a
constructible root of the equation
4x3 - Ъх + $ = 0 or 8*3 - 6x + 1 = 0.
The candidates for rational roots of the equation
8*3 - 6x + 1 = 0
are 1, У2, }^i, M> a"d their negatives. Since none of these numbers satisfies
the equation, this cubic does not have a rational root and hence does not
have a constructible root. That is, we cannot trisect the angle of 2π/3 radians.
Questions
1. Show that if χ — \x\ (cos 0 + i sin 0), then
xk = |*|4(cos40 -1- tsin 40),
1. State De Moivre's theorem. 3. Show how to find three roots of the equation x3 — 1
and three roots of the equation хг — —1 + i. 4. Discuss the possibility of trisecting
the angle of 2π/3 radians. Assume the formula cos 36 = 4 cos3 θ — 3 cos Θ.
Problems
1. Solve the equation x3 — 1 = 0 by the method of Section 53 and check with
the results of Example 73.1.
2. Find one root of the equation *5 = —l/y/2 — i/Vl.
3. Solve the equation *2 = i (2 roots).
4. Solve the equation хг — i (3 roots).
5. Solve the equation *8 = 1.
6. Solve the equation дг2 = 5 + tv3/2.
7. Show that
(cos θ + i sin 0)3 = cos3 θ - 3 cos θ sin2 θ + t'(3 cos2 θ sin θ - sin3 Θ)
= 4 cos3 θ - 3 cos θ + «'(3 sin θ - 4 sin3 θ).
Use the formula cos2 θ + sin2 0=1. Use this result to show that
cos 30 = 4 cos3 0 - 3 cos 0,
sin 30 = 3 sin 0 - 4 sin3 0.

Sec. 73]
DE MOIVRE'S THEOREM
289
8. Set cos 6 — χ, sin 6 — у, 36 = 7г/2 and solve
4*3 - Ъх - cos3 6 = 0, 4y3 - Ъу + sin 36 = О,
and use your results to find cos π/6, sin π/6.
9. By a similar procedure (with 36 = π) find cos 6/3, sin 6/3.
10. Show that 1 is a root of the equation x3 — 1 =0 and that the remaining
roots satisfy the equation χ2 + χ + 1 =0. Divide both members of this
quadratic equation by χ and show that it is equivalent to
*+l+-=*+*+l =0.
χ
Show that the fact that χ — cos 2π/3 + i sin 2π/3 is a root of хг — 1 = 0
implies that (see Problem 8 of Section 72)
2π
χ + χ + 1 = 2 cos — + 1 =0.
Solve for cos 2π/3.
11. Show that 1 is a root of the equation *5 — 1 =0 and that the remaining
roots satisfy the equation *4 + χ3 + χ2 + χ + 1 =0. Divide both members
of this quartic by x2 and show that it is equivalent to
x2 + x + \+- + -2 = x2 + ? + x + x~+\
X X*
= (* + x)2 + (x + x) - 1 =0.
Show that the fact that χ = cos 2π/5 + i sin 2π/5 is a root of *5 — 1 =0
implies that
J-- 9 27Γ
jy = дс + дс == 2 cos ■
is a root of
and hence that
У _
2
y2+y-\ ■
-ι ± νΊ
4
5
= 0
2тг
= cos —
Show that the plus sign must be chosen and hence that
2π _„0 У/1- 1
cos-— = cos 72 =
5 4
2тг . 700 / /у/5-Ц2 y/\Q + 2V5

290 COMPLEX NUMBERS (Ch. 7
Figure 73.2
Use the result of Problem 9 of Section 57 to obtain exact values of cos 3°, sin
3°. Figure 73.2 indicates how the results of this problem can be used to
produce a ruler-and-compass construction of a regular pentagon.

APPENDIX

Table 1
THREE-PLACE TRIGONOMETRIC FUNCTIONS
0°
1°
2°
3°
4°
5°
6°
7°
8°
9°
10°
11°
12°
13°
14°
15°
16°
17°
18°
19°
20°
21°
22°
23°
24°
25°
26°
27°
28°
29°
30°
31°
32°
33°
34°
35°
36°
37°
38°
39°
40°
41°
42°
43°
44°
45°
value
0.000
0.017
0.035
0.052
0.070
0.087
0.105
0.122
0.139
0.156
0.174
0.191
0.208
0.225
0.242
0.259
0.276
0.292
0.309
0.326
0.342
0.358
0.375
0.391
0.407
0.423
0.438
0.454
0.469
0.485
0.500
0.515
0.530
0.545
0.559
0.574
0.588
0.602
0.616
0.629
0.643
0.656
0.669
0.682
0.695
0.707
sin
log
0.242-2
0.543-2
0.719-2
0.844-2
0.940-2
0.019-1
0.086-1
0.144-1
0.194-1
0.240-1
0.281-1
0.318-1
0.352-1
0.384-1
0.413-1
0.440-1
0.466-1
0.490-1
0.513-1
0.534-1
0.554-1
0.574-1
0.592-1
0.609-1
0.626-1
0.642-1
0.657-1
0.672-1
0.686-1
0.699-1
0.712-1
0.724-1
0.736-1
0.748-1
0.759-1
0.769-1
0.779-1
0.789-1
0.799-1
0.808-1
0.817-1
0.826-1
0.834-1
0.842-1
0.849-1
cos
tan
value
0.000
0.017
0.035
0.052
0.070
0.087
0.105
0.123
0.141
0.158
0.176
0.194
0.213
0.231
0.249
0.268
0.287
0.306
0.325
0.344
0.364
0.384
0.404
0.424
0.445
0.466
0.488
0.510
0.532
0.554
0.577
0.601
0.625
0.649
0.675
0.700
0.727
0.754
0.781
0.810
0.839
0.869
0.900
0.933
0.966
1.000
log
0.242-2
0.543-2
0.719-2
0.845-2
0.942-2
0.022-1
0.089-1
0.148-1
0.200-1
0.246-1
0.289-1
0.327-1
0.363-1
0.397-1
0.428-1
0.457-1
0.485-1
0.512-1
0.537-1
0.561-1
0.584-1
0.606-1
0.628-1
0.649-1
0.669-1
0.688-1
0.707-1
0.726-1
0.744-1
0.761-1
0.779-1
0.796-1
0.813-1
0.829-1
0.845-1
0.861-1
0.877-1
0.893-1
0.908-1
0.924-1
0.939-1
0.954-1
0.970-1
0.985-1
0.000
ctn
sec
value
1.00
1.00
1.00
1.00
1.00
1.00
1.01
1.01
1.01
1.01
1.02
1.02
1.02
1.03
1.03
1.04
1.04
1.05
1.05
1.06
1.06
1.07
1.08
1.09
1.10
1.10
1.11
1.12
1.13
1.14
1.15
1.17
1.18
1.19
1.21
1.22
1.24
1.25
1.27
1.29
1.31
1.33
1.35
1.37
1.39
1.41
CSC
CSC
value
57.30
28.65
19.11
14.34
11.47
9.57
8.21
7.19
6.39
5.76
5.24
4.81
4.45
4.13
3.86
3.63
3.42
3.24
3.07
2.92
2.79
2.67
2.56
2.46
2.37
2.28
2.20
2.13
2.06
2.00
1.94
1.89
1.84
1.79
1.74
1.70
1.66
1.62
1.59
1.56
1.52
1.49
1.47
1.44
1.41
sec
ctn
value
57^29
28.64
19.08
14.30
11.43
9.51
8.14
7.12
6.31
5.67
5.14
4.70
4.33
4.01
3.73
3.49
3.27
3.08
2.90
2.75
2.61
2.48
2.36
2.25
2.14
2.05
1.96
1.88
1.80
1.73
1.66
1.60
1.54
1.48
1.43
1.38
1.33
1.28
1.23
1.19
1.15
1.11
1.07
1.04
1.00
log
1J58
1.457
1.281
1.155
1.058
0.978
0.911
0.852
0.800
0.754
0.711
0.673
0.637
0.603
0.572
0.543
0.515
0.488
0.463
0.439
0.416
0.394
0.372
0.351
0.331
0.312
0.293
0.274
0.256
0.239
0.221
0.204
0.187
0.171
0.155
0.139
0.123
0.107
0.092
0.076
0.061
0.046
0.030
0.015
0.000
tan
cos
value
1.000
1.000
0.999
0.999
0.998
0.996
0.995
0.993
0.990
0.988
0.985
0.982
0.978
0.974
0.970
0.966
0.961
0.956
0.951
0.946
0.940
0.934
0.927
0.921
0.914
0.906
0.899
0.891
0.883
0.875
0.866
0.857
0.848
0.839
0.829
0.819
0.809
0.799
0.788
0.777
0.766
0.755
0.743
0.731
0.719
0.707
log
0.000
0.000
0.000
0.999-1
0.999-1
0.998-1
0.998-1
0.997-1
0.996-1
0.995-1
0.993-1
0.992-1
0.990-1
0.989-1
0.987-1
0.985-1
0.983-1
0.981-1
0.978-1
0.976-1
0.973-1
0.970-1
0.967-1
0.964-1
0.961-1
0.957-1
0.954-1
0.950-1
0.946-1
0.942-1
0.938-1
0.933-1
0.928-1
0.924-1
0.919-1
0.913-1
0.908-1
0.902-1
0.897-1
0.891-1
0.884-1
0.878-1
0.871-1
0.864-1
0.857-1
0.849-1
sin
90°
89°
88°
87°
86°
85°
84°
83°
82°
81°
80°
79°
78°
77°
76°
75°
74°
73°
72°
71°
70°
69°
68°
67°
66°
65°
64°
63°
62°
61°
60°
59°
58°
57°
56°
55°
54°
53°
52°
51°
50°
49°
48°
47°
46°
45°

294 APPENDIX
Table 2
THREE-PLACE COMMON LOGARITHMS
N
1
2
3
4
5
6
7
8
9
0
000
301
477
602
699
778
845
903
954
1
041
322
491
613
708
785
851
908
959
2
079
342
505
623
716
792
857
914
964
3
114
362
519
633
724
799
863
919
968
4
146
380
531
643
732
806
869
924
973
5
176
398
544
653
740
813
875
929
978
6
204
415
556
663
748
820
881
934
982
7
230
431
568
672
756
826
886
940
987
8
255
447
580
681
763
833
892
944
991
9
279
462
591
690
771
839
898
949
996

INDEX
Abelian group, 29, 221
Absolute value, 12, 71
Addition of vectors, 6
Altitude, 274
Angle, 215, 225
Angle sum, 231
Angular velocity vector, 7
Area of parallelogram, 45
Area of triangle, 45
Associative law, 6
Asymptote, 84
Azimuth, 274
Barycentric coordinates, 87
Base of logarithm, 251
Bisector of an angle, 74
Cartesian coordinates, 87
Characteristic equation, 187, 201
Characteristic of logarithm, 246
Circle, 55
center and radius of, 177
Colatitude, 273
Commutative group, 29
Commutative law, 6, 21
Completing the square, 177, 178
Component of a vector, 15
Complex numbers, 277, 278
Cone, 144,211,273
Conic section, 77, 211
Conjugate complex, 278
Consistency of postulates, 149
Constant term, 168

296
INDEX
Coordinates, 12, 18, 27, 34, 35
barycentric, 89
cartesian, 87
cylindrical, 272
polar, 265
spherical, 272
Coordinate plane, 128
Cosecant, 233
Cosine, 216
inverse, 262
Cosines, law of, 230, 231
Cotangent, 233
Cross product, 113
Cross product term, 185
Cubic equation, 207
Cyclic order, 119
Cylinder, 144, 272
Cylindrical coordinates, 272
Degree, 168
De Moivre's theorem, 283
Determinant, second order, 50
third order, 91
higher order, 153
Diagonal, principle, 52
secondary, 52
Direction, 1
Direction perpendicular to given direction,
41
Directrix, 77
Discriminant, 178
Displacement, 17
Distance between two points, 30, 35, 88
Distance formula, 35, 88
Distance from point to line, 42, 47, 108
Distance from point to plane, 109, 122
Distributive law, 21, 118
Dividing a segment in a given ratio, 39, 91
(problem 7)
Dot product, 21
Eccentricity, 77
Eigen-value, 187, 201
Eigen-vector, 187, 201
Ellipse, 75, 77, 79
Ellipsoid, 136, 137
Elliptic paraboloid, 140
Empty set, 61
Equation of a line, 47, 88, 164
Equation of locus, 73
Equations, linear, 54, 99
Equations, simultaneous, 160
Equilibrium of forces, 10
Equivalent vectors, 2
Expansion of determinant, 52, 93, 154, 156
Exponents, 248
Extrapolation, 241
Focus, 77
Force, 7
Function, linear vector, 47
inverse trigonometric, 261
scalar, 216
trigonometric, 216, 233
Graphs of trigonometric functions, 238, 239,
240
Group, 29, 221
additive, 221
multiplicative, 221
Higher degree equation, 207
Hyperbola, 77, 82
conjugate hyperbolas, 85
Hyperbolic paraboloid, 141
Hyperboloid, 137, 138, 139
Identity, 256
Imaginary numbers, 277
Inconsistent, 149
Inequality, Schwarz, 108
triangle, 109
Intercept, 67
Intercept formula, 68
Interpolation, 240
Isometric projection, 26
Intersection of lines, 59
Intersection of planes, 126, 130
Intersection of sets, 61
Inverse, 221
Latitude. 273
Latus rectum, 77
Law of cosines, 230, 231
Law of sines, 230
Left handed, 71, 106
Line, 36, 88, 150
Line parallel to a given line, 66
Line perpendicular to a given line, 66
Linear dependence, 62, 119
Linear equations, 54, 65, 122, 160
geometric properties of, 65
Linear independence, 62, 110
Linear vector function, 183, 186
Locus, equation of, 73, 136
Logarithm, 244
base of, 244, 251
Longitude, 273

INDEX
297
Magnitude of a vector, 1, 22, 30
Mantissa, 246
Matrix, 183, 186, 189, 199, 204
Minor of a determinant, 52, 93, 94, 95, 154
Models of postulates, 147
Multiplication of a vector by a number, 11
η-dimensional space, 31, 147
Negative of a vector, 9
Normal vector to a line, 47
to a plane, 123
Obvious solution, 63
Opposite sense, 2
Orientation, 71, 106
Oriented angle, 215, 227
Origin, 34
of a ray, 214
of a vector, 3
Orthonormal system, 18, 25, 31
left handed, 71, 106
orientation of, 71, 106
right handed, 71, 106
Parabola, 77
Parabolic cylinder, 143
Paraboloid, 136, 140, 141
Parallelepiped, volume of, 111
Parallelogram, area of, 45
Parameter, 37, 88
Parametric equations, 37, 88
Perpendicular, 16, 22
to a line, 47
to a plane, 123
Point-slope formula, 68
two-point formula, 68
Plane, 90
Plane analytic geometry, 18, 34
Polar coordinates, 265
Postulates, 5, 29
Product of two determinants, 69, 104
Product, scalar (or dot), 21
scalar triple, 116
vector (or cross), 113
vector triple, 121
Projection, 15
Pure imaginary, 277
Pythagorean theorem, 23, 28
Quadratic equation, 169, 173, 199
Quadratic formula, 177, 178
Quadric surface, 137
Radian measure, 235
Ray, 214
Real, 277
Reduction formula, 223
Right handed, 71, 106
Rotation, 169, 180, 182, 194, 280
through an angle, 218
Ruled surface, 142
Ruler and compass constructions, 210, 287,
288, 289
Ruling, 144
Scalar, 22
Scalar product, 21
Scalar triple product, 116
Schwarz inequality, 108
Secant, 233
Sense, 1
Set, 31, 35
empty, 61
Sets, intersection of, 61
union of, 74
Signed minor, 51, 92, 153
Sine, 216
inverse, 262
Sines, law of, 230
Slide rule, 242
Slope, 67
Slope-intercept formula, 68
Space, 31, 147
Sphere, 88, 273
Spherical coordinates, 272
Standard form of conic section, 85
Sum of vectors, 6
Surface, quadric, 137
ruled, 142
Symmetric determinant, 176
Symmetric matrix, 190, 204
Symmetry, 77, 80, 83
Tangent, 233
Terminal point, 3
Tetrahedron, volume of, 111
Three dimensional space, 25, 31, 147
Tip, 3
Translation, 169, 193
Transpose of a determinant, 53, 97, 156
of a matrix, 190, 204
Triangle, area of, 45
solution of, 232, 254
Triangle inequality, 109
Trigonometry, 214
Two-point formula, 68

298
INDEX
Union of sets, 74 Velocity, 2
Unit vector, 3 Velocity vector, 2
Unoriented angle, 215, 226, 227 Vertex of a cone, 144
Vector, 2
Vector function, 183
Vector product, 113
Vector space, 31
Work, 17, 20
Zero vector, 9