Vector Calculus

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Vector Calculus

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Vector Calculus

4

th EDITION

Susan Jane Colley Oberlin College

Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal ´ Toronto Delhi Mexico City Sao ˜ Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

Editor in Chief: Deirdre Lynch Senior Acquisitions Editor: William Hoffman Sponsoring Editor: Caroline Celano Editorial Assistant: Brandon Rawnsley Senior Managing Editor: Karen Wernholm Production Project Manager: Beth Houston Executive Marketing Manager: Jeff Weidenaar Marketing Assistant: Caitlin Crain Senior Author Support/Technology Specialist: Joe Vetere Rights and Permissions Advisor: Michael Joyce Manufacturing Buyer: Debbie Rossi Design Manager: Andrea Nix Senior Designer: Beth Paquin Production Coordination and Composition: Aptara, Inc. Cover Designer: Suzanne Duda Cover Image: Alessandro Della Bella/AP Images Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and Pearson Education was aware of a trademark claim, the designations have been printed in initial caps or all caps. Library of Congress Cataloging-in-Publication Data Colley, Susan Jane. Vector calculus / Susan Jane Colley. – 4th ed. p. cm. Includes index. ISBN-13: 978-0-321-78065-2 ISBN-10: 0-321-78065-5 1. Vector analysis. I. Title. QA433.C635 2012 515’.63–dc23 2011022433 c 2012, 2006, 2002 Pearson Education, Inc. Copyright  All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. For information on obtaining permission for use of material in this work, please submit a written request to Pearson Education, Inc., Rights and Contracts Department, 501 Boylston Street, Suite 900, Boston, MA 02116, fax your request to 617-671-3447, or e-mail at http://www.pearsoned.com/legal/permissions.htm. 1 2 3 4 5 6 7 8 9 10—EB—15 14 13 12 11

www.pearsonhighered.com

ISBN 13: 978-0-321-78065-2 ISBN 10: 0-321-78065-5

To Will and Diane, with love

John Seyfried

About the Author Susan Jane Colley

Susan Colley is the Andrew and Pauline Delaney Professor of Mathematics at Oberlin College and currently Chair of the Department, having also previously served as Chair. She received S.B. and Ph.D. degrees in mathematics from the Massachusetts Institute of Technology prior to joining the faculty at Oberlin in 1983. Her research focuses on enumerative problems in algebraic geometry, particularly concerning multiple-point singularities and higher-order contact of plane curves. Professor Colley has published papers on algebraic geometry and commutative algebra, as well as articles on other mathematical subjects. She has lectured internationally on her research and has taught a wide range of subjects in undergraduate mathematics. Professor Colley is a member of several professional and honorary societies, including the American Mathematical Society, the Mathematical Association of America, Phi Beta Kappa, and Sigma Xi.

Contents Preface

ix

To the Student: Some Preliminary Notation

xv

1

Vectors

1

1.1

Vectors in Two and Three Dimensions

1

1.2

More About Vectors

8

1.3

The Dot Product

18

1.4

The Cross Product

27

1.5

Equations for Planes; Distance Problems

40

1.6

Some n-dimensional Geometry

48

1.7

New Coordinate Systems

62

True/False Exercises for Chapter 1

75

Miscellaneous Exercises for Chapter 1

75

2

Differentiation in Several Variables

82

2.1

Functions of Several Variables; Graphing Surfaces

82

2.2

Limits

2.3

The Derivative

116

2.4

Properties; Higher-order Partial Derivatives

134

2.5

The Chain Rule

142

2.6

Directional Derivatives and the Gradient

158

2.7

Newton’s Method (optional)

176

True/False Exercises for Chapter 2

182

Miscellaneous Exercises for Chapter 2

183

3

Vector-Valued Functions

189

3.1

Parametrized Curves and Kepler’s Laws

189

3.2

Arclength and Differential Geometry

202

3.3

Vector Fields: An Introduction

221

3.4

Gradient, Divergence, Curl, and the Del Operator

227

True/False Exercises for Chapter 3

237

Miscellaneous Exercises for Chapter 3

237

4

Maxima and Minima in Several Variables

244

4.1

Differentials and Taylor’s Theorem

244

4.2

Extrema of Functions

263

4.3

Lagrange Multipliers

278

4.4

Some Applications of Extrema

293

True/False Exercises for Chapter 4

305

Miscellaneous Exercises for Chapter 4

306

97

viii

Contents

5

Multiple Integration

310

5.1

Introduction: Areas and Volumes

310

5.2

Double Integrals

314

5.3

Changing the Order of Integration

334

5.4

Triple Integrals

337

5.5

Change of Variables

349

5.6

Applications of Integration

373

5.7

Numerical Approximations of Multiple Integrals (optional)

388

True/False Exercises for Chapter 5

401

Miscellaneous Exercises for Chapter 5

403

6

Line Integrals

408

6.1

Scalar and Vector Line Integrals

408

6.2

Green’s Theorem

429

6.3

Conservative Vector Fields

439

True/False Exercises for Chapter 6

450

Miscellaneous Exercises for Chapter 6

451

7

Surface Integrals and Vector Analysis

455

7.1

Parametrized Surfaces

455

7.2

Surface Integrals

469

7.3

Stokes’s and Gauss’s Theorems

490

7.4

Further Vector Analysis; Maxwell’s Equations

510

True/False Exercises for Chapter 7

522

Miscellaneous Exercises for Chapter 7

523

8

Vector Analysis in Higher Dimensions

530

8.1

An Introduction to Differential Forms

530

8.2

Manifolds and Integrals of k-forms

536

8.3

The Generalized Stokes’s Theorem

553

True/False Exercises for Chapter 8

561

Miscellaneous Exercises for Chapter 8

561

Suggestions for Further Reading

563

Answers to Selected Exercises

565

Index

599

Preface Physical and natural phenomena depend on a complex array of factors. The sociologist or psychologist who studies group behavior, the economist who endeavors to understand the vagaries of a nation’s employment cycles, the physicist who observes the trajectory of a particle or planet, or indeed anyone who seeks to understand geometry in two, three, or more dimensions recognizes the need to analyze changing quantities that depend on more than a single variable. Vector calculus is the essential mathematical tool for such analysis. Moreover, it is an exciting and beautiful subject in its own right, a true adventure in many dimensions. The only technical prerequisite for this text, which is intended for a sophomore-level course in multivariable calculus, is a standard course in the calculus of functions of one variable. In particular, the necessary matrix arithmetic and algebra (not linear algebra) are developed as needed. Although the mathematical background assumed is not exceptional, the reader will still be challenged in places. My own objectives in writing the book are simple ones: to develop in students a sound conceptual grasp of vector calculus and to help them begin the transition from first-year calculus to more advanced technical mathematics. I maintain that the first goal can be met, at least in part, through the use of vector and matrix notation, so that many results, especially those of differential calculus, can be stated with reasonable levels of clarity and generality. Properly described, results in the calculus of several variables can look quite similar to those of the calculus of one variable. Reasoning by analogy will thus be an important pedagogical tool. I also believe that a conceptual understanding of mathematics can be obtained through the development of a good geometric intuition. Although I state many results in the case of n variables (where n is arbitrary), I recognize that the most important and motivational examples usually arise for functions of two and three variables, so these concrete and visual situations are emphasized to explicate the general theory. Vector calculus is in many ways an ideal subject for students to begin exploration of the interrelations among analysis, geometry, and matrix algebra. Multivariable calculus, for many students, represents the beginning of significant mathematical maturation. Consequently, I have written a rather expansive text so that they can see that there is a story behind the results, techniques, and examples—that the subject coheres and that this coherence is important for problem solving. To indicate some of the power of the methods introduced, a number of topics, not always discussed very fully in a first multivariable calculus course, are treated here in some detail: • an early introduction of cylindrical and spherical coordinates (§1.7); • the use of vector techniques to derive Kepler’s laws of planetary motion (§3.1); • the elementary differential geometry of curves in R3 , including discussion of curvature, torsion, and the Frenet–Serret formulas for the moving frame (§3.2); • Taylor’s formula for functions of several variables (§4.1);

x

Preface

• the use of the Hessian matrix to determine the nature (as local extrema) of critical points of functions of n variables (§4.2 and §4.3); • an extended discussion of the change of variables formula in double and triple integrals (§5.5); • applications of vector analysis to physics (§7.4); • an introduction to differential forms and the generalized Stokes’s theorem (Chapter 8). Included are a number of proofs of important results. The more technical proofs are collected as addenda at the ends of the appropriate sections so as not to disrupt the main conceptual flow and to allow for greater flexibility of use by the instructor and student. Nonetheless, some proofs (or sketches of proofs) embody such central ideas that they are included in the main body of the text.

New in the Fourth Edition I have retained the overall structure and tone of prior editions. New features in this edition include the following: • 210 additional exercises, at all levels; • a new, optional section (§5.7) on numerical methods for approximating multiple integrals; • reorganization of the material on Newton’s method for approximating solutions to systems of n equations in n unknowns to its own (optional) section (§2.7); • new proofs in Chapter 2 of limit properties (in §2.2) and of the general multivariable chain rule (Theorem 5.3 in §2.5); • proofs of both single-variable and multivariable versions of Taylor’s theorem in §4.1; • various additional refinements and clarifications throughout the text, including many new and revised examples and explanations; R R R PowerPoint files and Wolfram Mathematica notebooks • new Microsoft that coordinate with the text and that instructors may use in their teaching (see “Ancillary Materials” below).

How to Use This Book There is more material in this book than can be covered comfortably during a single semester. Hence, the instructor will wish to eliminate some topics or subtopics—or to abbreviate the rather leisurely presentations of limits and differentiability. Since I frequently find myself without the time to treat surface integrals in detail, I have separated all material concerning parametrized surfaces, surface integrals, and Stokes’s and Gauss’s theorems (Chapter 7), from that concerning line integrals and Green’s theorem (Chapter 6). In particular, in a one-semester course for students having little or no experience with vectors or matrices, instructors can probably expect to cover most of the material in Chapters 1–6, although no doubt it will be necessary to omit some of the optional subsections and to downplay

Preface

xi

many of the proofs of results. A rough outline for such a course, allowing for some instructor discretion, could be the following: Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6

8–9 lectures 9 lectures 4–5 lectures 5–6 lectures 8 lectures 4 lectures 38–41 lectures

If students have a richer background (so that much of the material in Chapter 1 can be left largely to them to read on their own), then it should be possible to treat a good portion of Chapter 7 as well. For a two-quarter or two-semester course, it should be possible to work through the entire book with reasonable care and rigor, although coverage of Chapter 8 should depend on students’ exposure to introductory linear algebra, as somewhat more sophistication is assumed there. The exercises vary from relatively routine computations to more challenging and provocative problems, generally (but not invariably) increasing in difficulty within each section. In a number of instances, groups of problems serve to introduce supplementary topics or new applications. Each chapter concludes with a set of miscellaneous exercises that both review and extend the ideas introduced in the chapter. A word about the use of technology. The text was written without reference to any particular computer software or graphing calculator. Most of the exercises can be solved by hand, although there is no reason not to turn over some of the more tedious calculations to a computer. Those exercises that require a computer for computational or graphical purposes are marked with the symbol T and should be amenable to software such as Mathematica® , Maple® , or MATLAB.

◆

Ancillary Materials In addition to this text a Student Solutions Manual is available. An Instructor’s Solutions Manual, containing complete solutions to all of the exercises, is available to course instructors from the Pearson Instructor Resource Center R R (www.pearsonhighered.com/irc), as are many Microsoft PowerPoint files and R  Wolfram Mathematica notebooks that can be adapted for classroom use. The reader can find errata for the text and accompanying solutions manuals at the following address: www.oberlin.edu/math/faculty/colley/VCErrata.html

Acknowledgments I am very grateful to many individuals for sharing with me their thoughts and ideas about multivariable calculus. I would like to express particular appreciation to my Oberlin colleagues (past and present) Bob Geitz, Kevin Hartshorn, Michael Henle (who, among other things, carefully read the draft of Chapter 8), Gary Kennedy, Dan King, Greg Quenell, Michael Raney, Daniel Steinberg, Daniel Styer, Richard Vale, Jim Walsh, and Elizabeth Wilmer for their conversations with me. I am also grateful to John Alongi, Northwestern University; Matthew Conner, University of California, Davis; Henry C. King, University of Maryland; Stephen B. Maurer,

xii

Preface

Swarthmore College; Karen Saxe, Macalester College; David Singer, Case Western Reserve University; and Mark R. Treuden, University of Wisconsin at Stevens Point, for their helpful comments. Several colleagues reviewed various versions of the manuscript, and I am happy to acknowledge their efforts and many fine suggestions. In particular, for the first three editions, I thank the following reviewers: Raymond J. Cannon, Baylor University; Richard D. Carmichael, Wake Forest University; Stanley Chang, Wellesley College; Marcel A. F. D´eruaz, University of Ottawa (now emeritus); Krzysztof Galicki, University of New Mexico (deceased); Dmitry Gokhman, University of Texas at San Antonio; Isom H. Herron, Rensselaer Polytechnic Institute; Ashwani K. Kapila, Rensselaer Polytechnic Institute; Christopher C. Leary, State University of New York, College at Geneseo; David C. Minda, University of Cincinnati; Jeffrey Morgan, University of Houston; Monika Nitsche, University of New Mexico; Jeffrey L. Nunemacher, Ohio Wesleyan University; Gabriel Prajitura, State University of New York, College at Brockport; Florin Pop, Wagner College; John T. Scheick, The Ohio State University (now emeritus); Mark Schwartz, Ohio Wesleyan University; Leonard M. Smiley, University of Alaska, Anchorage; Theodore B. Stanford, New Mexico State University; James Stasheff, University of North Carolina at Chapel Hill (now emeritus); Saleem Watson,California State University, Long Beach; Floyd L. Williams, University of Massachusetts, Amherst (now emeritus). For the fourth edition, I thank: Justin Corvino, Lafayette College; Carrie Finch, Washington and Lee University; Soomin Kim, Johns Hopkins University; Tanya Leise, Amherst College; Bryan Mosher, University of Minnesota. Many people at Oberlin College have been of invaluable assistance throughout the production of all the editions of Vector Calculus. I would especially like to thank Ben Miller for his hard work establishing the format for the initial drafts and Stephen Kasperick-Postellon for his manifold contributions to the typesetting, indexing, proofreading, and friendly critiquing of the original manuscript. I am very grateful to Linda Miller and Michael Bastedo for their numerous typographical contributions and to Catherine Murillo for her help with any number of tasks. Thanks also go to Joshua Davis and Joaquin Espinoza Goodman for their assistance with proofreading. Without the efforts of these individuals, this project might never have come to fruition. The various editorial and production staff members have been most kind and helpful to me. For the first three editions, I would like to express my appreciation to my editor, George Lobell, and his editorial assistants Gale Epps, Melanie Van Benthuysen, and Jennifer Urban; to production editors Nicholas Romanelli, Barbara Mack, and Debbie Ryan at Prentice Hall, and Lori Hazzard at Interactive Composition Corporation; to Ron Weickart and the staff at Network Graphics

Preface

xiii

for their fine rendering of the figures, and to Tom Benfatti of Prentice Hall for additional efforts with the figures; and to Dennis Kletzing for his careful and enthusiastic composition work. For this edition, it is a pleasure to acknowledge my upbeat editor, Caroline Celano, and her assistant, Brandon Rawnsley; they have made this new edition fun to do. In addition, I am most grateful to Beth Houston, my production manager at Pearson, Jogender Taneja and the staff at Aptara, Inc., Donna Mulder, Roger Lipsett, and Thomas Wegleitner. Finally, I thank the many Oberlin students who had the patience to listen to me lecture and who inspired me to write and improve this volume. SJC [email protected]

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To the Student: Some Preliminary Notation Here are the ideas that you need to keep in mind as you read this book and learn vector calculus. Given two sets A and B, I assume that you are familiar with the notation A ∪ B for the union of A and B—those elements that are in either A or B (or both): A ∪ B = {x | x ∈ A or x ∈ B}. Similarly, A ∩ B is used to denote the intersection of A and B—those elements that are in both A and B: A ∩ B = {x | x ∈ A and x ∈ B}. The notation A ⊆ B, or A ⊂ B, indicates that A is a subset of B (possibly empty or equal to B). x

−3 −2 −1

0

1

2

3

Figure 1 The coordinate line R.

y

y0

(x0, y0)

1 x 1

x0

Figure 2 The coordinate plane R2 .

One-dimensional space (also called the real line or R) is just a straight line. We put real number coordinates on this line by placing negative numbers on the left and positive numbers on the right. (See Figure 1.) Two-dimensional space, denoted R2 , is the familiar Cartesian plane. If we construct two perpendicular lines (the x- and y-coordinate axes), set the origin as the point of intersection of the axes, and establish numerical scales on these lines, then we may locate a point in R2 by giving an ordered pair of numbers (x, y), the coordinates of the point. Note that the coordinate axes divide the plane into four quadrants. (See Figure 2.) Three-dimensional space, denoted R3 , requires three mutually perpendicular coordinate axes (called the x-, y- and z-axes) that meet in a single point (called the origin) in order to locate an arbitrary point. Analogous to the case of R2 , if we establish scales on the axes, then we can locate a point in R3 by giving an ordered triple of numbers (x, y, z). The coordinate axes divide three-dimensional space into eight octants. It takes some practice to get your sense of perspective correct when sketching points in R3 . (See Figure 3.) Sometimes we draw the coordinate axes in R3 in different orientations in order to get a better view of things. However, we always maintain the axes in a right-handed configuration. This means that if you curl the fingers of your right hand from the positive x-axis to the positive y-axis, then your thumb will point along the positive z-axis. (See Figure 4.) Although you need to recall particular techniques and methods from the calculus you have already learned, here are some of the more important concepts to keep in mind: Given a function f (x), the derivative f  (x) is the limit (if it exists) of the difference quotient of the function: f  (x) = lim

h→0

f (x + h) − f (x) . h

xvi

To the Student: Some Preliminary Notation

z (−1, −2, 2) 2 −1

(2, 4, 5) z

−2

−1

1

1

x

5 y

2

y

x

4 x

y

y

z

Figure 3 Three-dimensional

Figure 4 The x-, y-, and z-axes in R3 are always

space R3 . Selected points are graphed.

drawn in a right-handed configuration.

The significance of the derivative f  (x0 ) is that it measures the slope of the line tangent to the graph of f at the point (x0 , f (x0 )). (See Figure 5.) The derivative may also be considered to give the instantaneous rate of change of f at x = x0 . We also denote the derivative f  (x) by d f /d x. b The definite integral a f (x) d x of f on the closed interval [a, b] is the limit (provided it exists) of the so-called Riemann sums of f :

(x0, f (x0))

x

 Figure 5 The derivative f  (x 0 ) is

the slope of the tangent line to y = f (x) at (x0 , f (x0 )).

b

f (x) d x =

a

n 

lim

all xi →0

f (xi∗ )xi .

i=1

Here a = x0 < x1 < x2 < · · · < xn = b denotes a partition of [a, b] into subintervals [xi−1 , xi ], the symbol xi = xi − xi−1 (the length of the subinterval), and xi∗ denotes any point in [xi−1 , xi ]. If f (x) ≥ 0 on [a, b], then each term f (xi∗ )xi in the Riemannsum is the area of a rectangle related to the graph of f . The n Riemann sum i=1 f (xi∗ )xi thus approximates the total area under the graph of f between x = a and x = b. (See Figure 6.) y

… … a

…

x1 x2 x3 … … xi − 1

xi … xn − 1 b

x

x*i Figure 6 If f (x) ≥ 0 on [a, b], then the Riemann sum approximates the area under y = f (x) by giving the sum of areas of rectangles.

To the Student: Some Preliminary Notation

xvii

y

y = f (x)

a

b

x

Figure 7 The area under the graph of y = f (x) is

b a

f (x) d x.

b The definite integral a f (x) d x, if it exists, is taken to represent the area under y = f (x) between x = a and x = b. (See Figure 7.) The derivative and the definite integral are connected by an elegant result known as the fundamental theorem of calculus. Let f (x) be a continuous function of one variable, and let F(x) be such that F  (x) = f (x). (The function F is called an antiderivative of f .) Then 

b

f (x) d x = F(b) − F(a);  x d f (t) dt = f (x). 2. dx a 1.

a

Finally, the end of an example is denoted by the symbol ◆ and the end of a proof by the symbol ■.

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Vector Calculus

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1

Vectors

1.1

Vectors in Two and Three Dimensions

1.1 Vectors in Two and Three Dimensions

1.2

More About Vectors

1.3

The Dot Product

1.4

The Cross Product

1.5

Equations for Planes; Distance Problems

For your study of the calculus of several variables, the notion of a vector is fundamental. As is the case for many of the concepts we shall explore, there are both algebraic and geometric points of view. You should become comfortable with both perspectives in order to solve problems effectively and to build on your basic understanding of the subject.

1.6

Some n-dimensional Geometry

1.7

New Coordinate Systems True/False Exercises for Chapter 1 Miscellaneous Exercises for Chapter 1

Vectors in R2 and R3 : The Algebraic Notion A vector in R2 is simply an ordered pair of real numbers. That is, a vector in R2 may be written as (a1 , a2 ) (e.g., (1, 2) or (π, 17)). Similarly, a vector in R3 is simply an ordered triple of real numbers. That is, a vector in R3 may be written as √ (a1 , a2 , a3 ) (e.g., (π, e, 2)). DEFINITION 1.1

To emphasize that we want to consider the pair or triple of numbers as a single unit, we will use boldface letters; hence a = (a1 , a2 ) or a = (a1 , a2 , a3 ) will be our standard notation for vectors in R2 or R3 . Whether we mean that a is a vector in R2 or in R3 will be clear from context (or else won’t be important to the discussion). When doing handwritten work, it is difficult to “boldface” anything, so you’ll want to put an arrow over the letter. Thus, a will mean the same thing as a. Whatever notation you decide to use, it’s important that you distinguish the vector a (or a ) from the single real number a. To contrast them with vectors, we will also refer to single real numbers as scalars. In order to do anything interesting with vectors, it’s necessary to develop some arithmetic operations for working with them. Before doing this, however, we need to know when two vectors are equal. DEFINITION 1.2 Two vectors a = (a1 , a2 ) and b = (b1 , b2 ) in R2 are

equal if their corresponding components are equal, that is, if a1 = b1 and a2 = b2 . The same definition holds for vectors in R3 : a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) are equal if their corresponding components are equal, that is, if a1 = b1 , a2 = b2 , and a3 = b3 .

2

Chapter 1

Vectors

EXAMPLE 1 The vectors a = (1, 2) and b = (1, 2, 3) and d = (2, 3, 1) are not equal in R3 .

3 3

 , 63 are equal in R2 , but c =

◆

Next, we discuss the operations of vector addition and scalar multiplication. We’ll do this by considering vectors in R3 only; exactly the same remarks will hold for vectors in R2 if we simply ignore the last component. DEFINITION 1.3 (VECTOR ADDITION) Let a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) be two vectors in R3 . Then the vector sum a + b is the vector in R3 obtained via componentwise addition: a + b = (a1 + b1 , a2 + b2 , a3 + b3 ).

EXAMPLE 2 We have (0, 1, 3) + (7, −2, 10) = (7, −1, 13) and (in R2 ): √ √ (1, 1) + (π, 2) = (1 + π, 1 + 2).

◆

Properties of vector addition. We have 1. a + b = b + a for all a, b in R3 (commutativity); 2. a + (b + c) = (a + b) + c for all a, b, c in R3 (associativity); 3. a special vector, denoted 0 (and called the zero vector), with the property that a + 0 = a for all a in R3 . These three properties require proofs, which, like most facts involving the algebra of vectors, can be obtained by explicitly writing out the vector components. For example, for property 1, we have that if a = (a1 , a2 , a3 )

and

b = (b1 , b2 , b3 ),

then a + b = (a1 + b1 , a2 + b2 , a3 + b3 ) = (b1 + a1 , b2 + a2 , b3 + a3 ) = b + a, since real number addition is commutative. For property 3, the “special vector” is just the vector whose components are all zero: 0 = (0, 0, 0). It’s then easy to check that property 3 holds by writing out components. Similarly for property 2, so we leave the details as exercises. (SCALAR MULTIPLICATION) Let a = (a1 , a2 , a3 ) be a vector in R3 and let k ∈ R be a scalar (real number). Then the scalar product ka is the vector in R3 given by multiplying each component of a by k: ka = (ka1 , ka2 , ka3 ). DEFINITION 1.4

EXAMPLE 3 If a = (2, 0,

√ √ 2) and k = 7, then ka = (14, 0, 7 2).

◆

The results that follow are not difficult to check—just write out the vector components.

1.1

3

Vectors in Two and Three Dimensions

Properties of scalar multiplication. For all vectors a and b in R3 (or R2 ) and scalars k and l in R, we have 1. (k + l)a = ka + la (distributivity); 2. k(a + b) = ka + kb (distributivity); 3. k(la) = (kl)a = l(ka). It is worth remarking that none of these definitions or properties really depends on dimension, that is, on the number of components. Therefore we could have introduced the algebraic concept of a vector in Rn as an ordered n-tuple (a1 , a2 , . . . , an ) of real numbers and defined addition and scalar multiplication in a way analogous to what we did for R2 and R3 . Think about what such a generalization means. We will discuss some of the technicalities involved in §1.6.

y

(a1, a2)

x

Figure 1.1 A vector a ∈ R2

corresponds to a point in R2 . z

(a1, a2, a3)

y

x Figure 1.2 A vector a ∈ R3

corresponds to a point in R3 .

Vectors in R2 and R3 : The Geometric Notion Although the algebra of vectors is certainly important and you should become adept at working algebraically, the formal definitions and properties tend to present a rather sterile picture of vectors. A better motivation for the definitions just given comes from geometry. We explore this geometry now. First of all, the fact that a vector a in R2 is a pair of real numbers (a1 , a2 ) should make you think of the coordinates of a point in R2 . (See Figure 1.1.) Similarly, if a ∈ R3 , then a may be written as (a1 , a2 , a3 ), and this triple of numbers may be thought of as the coordinates of a point in R3 . (See Figure 1.2.) All of this is fine, but the results of performing vector addition or scalar multiplication don’t have very interesting or meaningful geometric interpretations in terms of points. As we shall see, it is better to visualize a vector in R2 or R3 as an arrow that begins at the origin and ends at the point. (See Figure 1.3.) Such a depiction is often referred to as the position vector of the point (a1 , a2 ) or (a1 , a2 , a3 ). If you’ve studied vectors in physics, you have heard them described as objects having “magnitude and direction.” Figure 1.3 demonstrates this concept, provided that we take “magnitude” to mean “length of the arrow” and “direction” to be the orientation or sense of the arrow. (Note: There is an exception to this approach, namely, the zero vector. The zero vector just sits at the origin, like a point, and has no magnitude and, therefore, an indeterminate direction. This exception will not pose much difficulty.) However, in physics, one doesn’t demand that all vectors In R2

In R3

y

z (a1, a2, a3) (a1, a2) a a y

x x 2

3

Figure 1.3 A vector a in R or R is represented by an arrow from the

origin to a.

4

Chapter 1

Vectors

be represented by arrows having their tails bound to the origin. One is free to “parallel translate” vectors throughout R2 and R3 . That is, one may represent the vector a = (a1 , a2 , a3 ) by an arrow with its tail at the origin (and its head at (a1 , a2 , a3 )) or with its tail at any other point, so long as the length and sense of the arrow are not disturbed. (See Figure 1.4.) For example, if we wish to represent a by an arrow with its tail at the point (x1 , x2 , x3 ), then the head of the arrow would be at the point (x1 + a1 , x2 + a2 , x3 + a3 ). (See Figure 1.5.) z

a

a

a (a1, a2, a3) a

z (x1 + a1, x2 + a2, x3 + a3)

a

a (x1, x2, x3)

y x

y

a x

Figure 1.4 Each arrow is a parallel translate of the position vector of the point (a1 , a2 , a3 ) and represents the same vector.

a+b

b

a Figure 1.6 The vector

a + b may be represented by an arrow whose tail is at the tail of a and whose head is at the head of b.

Figure 1.5 The vector a = (a1 , a2 , a3 ) represented by an arrow with tail at the point (x1 , x2 , x3 ).

With this geometric description of vectors, vector addition can be visualized in two ways. The first is often referred to as the “head-to-tail” method for adding vectors. Draw the two vectors a and b to be added so that the tail of one of the vectors, say b, is at the head of the other. Then the vector sum a + b may be represented by an arrow whose tail is at the tail of a and whose head is at the head of b. (See Figure 1.6.) Note that it is not immediately obvious that a + b = b + a from this construction! The second way to visualize vector addition is according to the so-called parallelogram law: If a and b are nonparallel vectors drawn with their tails emanating from the same point, then a + b may be represented by the arrow (with its tail at the common initial point of a and b) that runs along a diagonal of the parallelogram determined by a and b (Figure 1.7). The parallelogram law is completely consistent with the head-to-tail method. To see why, just parallel translate b to the opposite side of the parallelogram. Then the diagonal just described is the result of adding a and (the translate of) b, using the head-to-tail method. (See Figure 1.8.) We still should check that these geometric constructions agree with our algebraic definition. For simplicity, we’ll work in R2 . Let a = (a1 , a2 ) and b = (b1 , b2 ) as usual. Then the arrow obtained from the parallelogram law addition of a and b is the one whose tail is at the origin O and whose head is at the point P in Figure 1.9. If we parallel translate b so that its tail is at the head of a, then it is immediate that the coordinates of P must be (a1 + b1 , a2 + b2 ), as desired. Scalar multiplication is easier to visualize: The vector ka may be represented by an arrow whose length is |k| times the length of a and whose direction is the same as that of a when k > 0 and the opposite when k < 0. (See Figure 1.10.) It is now a simple matter to obtain a geometric depiction of the difference between two vectors. (See Figure 1.11.) The difference a − b is nothing more

1.1

Vectors in Two and Three Dimensions

a+b

b

a+b

b

a

5

b (translated)

a

Figure 1.7 The vector a + b may be represented by the arrow that runs along the diagonal of the parallelogram determined by a and b.

Figure 1.8 The equivalence of the parallelogram law and the head-to-tail methods of vector addition.

y P B b2

b

a2

a

a

A

a1

x b1

Figure 1.9 The point P has coordinates

(a1 + b1 , a2 + b2 ).

c=a−b

b

2a

b

−3a 2 Figure 1.10 Visualization of

scalar multiplication.

than a + (−b) (where −b means the scalar −1 times the vector b). The vector a − b may be represented by an arrow pointing from the head of b toward the head of a; such an arrow is also a diagonal of the parallelogram determined by a and b. (As we have seen, the other diagonal can be used to represent a + b.) Here is a construction that will be useful to us from time to time.

a Figure 1.11 The

geometry of vector subtraction. The vector c is such that b + c = a. Hence, c = a − b.

Given two points P1 (x1 , y1 , z 1 ) and P2 (x2 , y2 , z 2 ) in R3 , the displacement vector from P1 to P2 is −−→ P1 P2 = (x2 − x1 , y2 − y1 , z 2 − z 1 ). DEFINITION 1.5

z

P2 P1 O

x Figure 1.12 The displacement −−→ vector P1 P2 , represented by the arrow from P1 to P2 , is the difference between the position vectors of these two points.

y

This construction is not hard to understand if we consider Figure 1.12. Given −−→ −−→ the points P1 and P2 , draw the corresponding position vectors O P1 and O P2 . −−→ −−→ −−→ Then we see that P1 P2 is precisely O P2 − O P1 . An analogous definition may be made for R2 . In your study of the calculus of one variable, you no doubt used the notions of derivatives and integrals to look at such physical concepts as velocity, acceleration, force, etc. The main drawback of the work you did was that the techniques involved allowed you to study only rectilinear, or straight-line, activity. Intuitively, we all understand that motion in the plane or in space is more complicated than straightline motion. Because vectors possess direction as well as magnitude, they are ideally suited for two- and three-dimensional dynamical problems.

6

Vectors

Chapter 1

For example, suppose a particle in space is at the point (a1 , a2 , a3 ) (with respect to some appropriate coordinate system). Then it has position vector a = (a1 , a2 , a3 ). If the particle travels with constant velocity v = (v1 , v2 , v3 ) for t seconds, then the particle’s displacement from its original position is tv, and its new coordinate position is a + tv. (See Figure 1.13.)

z

tv v a

(a1, a2, a3) y

EXAMPLE 4 If a spaceship is at position (100, 3, 700) and is traveling with velocity (7, −10, 25) (meaning that the ship travels 7 mi/sec in the positive x-direction, 10 mi/sec in the negative y-direction, and 25 mi/sec in the positive z-direction), then after 20 seconds, the ship will be at position (100, 3, 700) + 20(7, −10, 25) = (240, −197, 1200), and the displacement from the initial position is (140, −200, 500).

x Figure 1.13 After t seconds, the point starting at a, with velocity v, moves to a + tv.

y

x v2 current

v1 ship (with respect to still water)

◆

EXAMPLE 5 The S.S. Calculus is cruising due south at a rate of 15 knots (nautical miles √ per hour) with respect to still water. However, there is also a current of 5 2 knots southeast. What is the total velocity of the ship? If the ship is initially at the origin and a lobster pot is at position (20, −79), will the ship collide with the lobster pot? Since velocities are vectors, the total velocity of the ship is v1 + v2 , where v1 is the velocity of the ship with respect to still water and v2 is the southeast-pointing velocity of the current. Figure 1.14 makes it fairly straightforward to compute these velocities. We have that v1 = (0, −15). Since v2 points southeastward, its direction must be along the line y = −x. Therefore, v2 can be written as v2 = (v, −v), where v√is a positive real number. By the Pythagorean theorem, if the √ length of v2 is 5 2, then we must have v 2 + (−v)2 = (5 2)2 or 2v 2 = 50, so that v = 5. Thus, v2 = (5, −5), and, hence, the net velocity is (0, −15) + (5, −5) = (5, −20).

Net velocity

Figure 1.14 The length of √ v1 is

15, and the length of v2 is 5 2.

After 4 hours, therefore, the ship will be at position (0, 0) + 4(5, −20) = (20, −80) ◆

and thus will miss the lobster pot.

EXAMPLE 6 The theory behind the venerable martial art of judo is an excellent example of vector addition. If two people, one relatively strong and the other relatively weak, have a shoving match, it is clear who will prevail. For example, someone pushing one way with 200 lb of force will certainly succeed in overpowering another pushing the opposite way with 100 lb of force. Indeed, as Figure 1.15 shows, the net force will be 100 lb in the direction in which the stronger person is pushing. 100 lb

200 lb

=

100 lb

Figure 1.15 A relatively strong person pushing with a

force of 200 lb can quickly subdue a relatively weak one pushing with only 100 lb of force.

> 200 lb 100 lb

200 lb

Figure 1.16 Vector addition in

judo.

Dr. Jigoro Kano, the founder of judo, realized (though he never expressed his idea in these terms) that this sort of vector addition favors the strong over the weak. However, if the weaker participant applies his or her 100 lb of force in a direction only slightly different from that of the stronger, he or she will effect a vector sum of length large enough to surprise the opponent. (See Figure 1.16.)

1.1

7

Exercises

This is the basis for essentially all of the throws of judo and why judo is described as the art of “using a person’s strength against himself or herself.” In fact, the word “judo” means “the giving way.” One “gives in” to the strength of another by ◆ attempting only to redirect his or her force rather than to oppose it.

1.1 Exercises 

12. Sketch the vectors a = (2, −7, 8) and b = − 1,

1. Sketch the following vectors in R2 :

(a) (2, 1)

(b) (3, 3)

(c) (−1, 2)

2. Sketch the following vectors in R3 :

(a) (1, 2, 3)

(b) (−2, 0, 2)

(c) (2, −3, 1)

3. Perform the indicated algebraic operations. Express

your answers in the form of a single vector a = (a1 , a2 ) in R2 . (a) (3, 1) + (−1, 7) (b) −2(8, 12) (c) (8, 9) + 3(−1, 2) (d) (1, 1) + 5(2, 6) − 3(10, 2) (e) (8, 10) + 3 ((8, −2) − 2(4, 5))

4. Perform the indicated algebraic operations. Express

your answers in the form of a single vector a = (a1 , a2 , a3 ) in R3 . (a) (2, 1, 2) + (−3, 9, 7) (b) 12 (8, 4, 1) + 2(5, −7, 14 )   (c) −2 (2, 0, 1) − 6( 12 , −4, 1)

5. Graph the vectors a = (1, 2), b = (−2, 5), and a +

b = (1, 2) + (−2, 5), using both the parallelogram law and the head-to-tail method.

6. Graph the vectors a = (3, 2) and b = (−1, 1). Also

calculate and graph a − b, 12 a, and a + 2b.

7. Let A be the point with coordinates (1, 0, 2), let B be

the point with coordinates (−3, 3, 1), and let C be the point with coordinates (2, 1, 5). −→ −→ (a) Describe the vectors AB and B A. −→ −→ −→ −→ (b) Describe the vectors AC, BC, and AC + C B. −→ −→ −→ (c) Explain, with pictures, why AC + C B = AB. 8. Graph (1, 2, 1) and (0, −2, 3), and calculate and graph

(1, 2, 1) + (0, −2, 3), −1(1, 2, 1), and 4(1, 2, 1).

9. If (−12, 9, z) + (x, 7, −3) = (2, y, 5), what are x, y,

and z? 10. What is the length (magnitude) of the vector (3, 1)?

(Hint: A diagram will help.) 11. Sketch the vectors a = (1, 2) and b = (5, 10). Explain

why a and b point in the same direction.

7 , −4 2



. Explain why a and b point in opposite directions. 13. How would you add the vectors (1, 2, 3, 4) and

(5, −1, 2, 0) in R4 ? What should 2(7, 6, −3, 1) be? In general, suppose that a = (a1 , a2 , . . . , an )

and

b = (b1 , b2 , . . . , bn )

are two vectors in Rn and k ∈ R is a scalar. Then how would you define a + b and ka? 14. Find the displacement vectors from P1 to P2 , where P1

−−→ and P2 are the points given. Sketch P1 , P2 , and P1 P2 . (a) P1 (1, 0, 2), P2 (2, 1, 7) (b) P1 (1, 6, −1), P2 (0, 4, 2) (c) P1 (0, 4, 2), P2 (1, 6, −1) (d) P1 (3, 1), P2 (2, −1)

15. Let P1 (2, 5, −1, 6) and P2 (3, 1, −2, 7) be two points

in R4 . How would you define and calculate the displacement vector from P1 to P2 ? (See Exercise 13.)

16. If A is the point in R3 with coordinates (2, 5, −6) and

the displacement vector from A to a second point B is (12, −3, 7), what are the coordinates of B? 17. Suppose that you and your friend are in New York talk-

ing on cellular phones. You inform each other of your own displacement vectors from the Empire State Building to your current position. Explain how you can use this information to determine the displacement vector from you to your friend. 18. Give the details of the proofs of properties 2 and 3 of

vector addition given in this section. 19. Prove the properties of scalar multiplication given in

this section. 20. (a) If a is a vector in R2 or R3 , what is 0a? Prove your

answer. (b) If a is a vector in R2 or R3 , what is 1a? Prove your answer. 21. (a) Let a = (2, 0) and b = (1, 1). For 0 ≤ s ≤ 1 and

0 ≤ t ≤ 1, consider the vector x = sa + tb. Explain why the vector x lies in the parallelogram

8

Chapter 1

Vectors

determined by a and b. (Hint: It may help to draw a picture.) (b) Now suppose that a = (2, 2, 1) and b = (0, 3, 2). Describe the set of vectors {x = sa + tb | 0 ≤ s ≤ 1, 0 ≤ t ≤ 1}. 22. Let a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) be two nonzero

vectors such that b = ka. Use vectors to describe the set of points inside the parallelogram with vertex P0 (x0 , y0 , z 0 ) and whose adjacent sides are parallel to a and b and have the same lengths as a and b. (See Figure 1.17.) (Hint: If P(x, y, z) is a point in the par−→ allelogram, describe O P, the position vector of P.)

24. A plane takes off from an airport with velocity vector

(50, 100, 4). Assume that the units are miles per hour, that the positive x-axis points east, and that the positive y-axis points north. (a) How fast is the plane climbing vertically at takeoff? (b) Suppose the airport is located at the origin and a skyscraper is located 5 miles east and 10 miles north of the airport. The skyscraper is 1,250 feet tall. When will the plane be directly over the building? (c) When the plane is over the building, how much vertical clearance is there? 25. As mentioned in the text, physical forces (e.g., gravity)

are quantities possessing both magnitude and direction and therefore can be represented by vectors. If an object has more than one force acting on it, then the resultant (or net) force can be represented by the sum of the individual force vectors. Suppose that two forces, F1 = (2, 7, −1) and F2 = (3, −2, 5), act on an object. (a) What is the resultant force of F1 and F2 ? (b) What force F3 is needed to counteract these forces (i.e., so that no net force results and the object remains at rest)?

z

b P0

P a

O

y

26. A 50 lb sandbag is suspended by two ropes. Suppose

x Figure 1.17 Figure for Exercise 22.

23. A flea falls onto marked graph paper at the point (3, 2).

She begins moving from that point with velocity vector v = (−1, −2) (i.e., she moves 1 graph paper unit per minute in the negative x-direction and 2 graph paper units per minute in the negative y-direction). (a) What is the speed of the flea? (b) Where is the flea after 3 minutes? (c) How long does it take the flea to get to the point (−4, −12)? (d) Does the flea reach the point (−13, −27)? Why or why not?

1.2

that a three-dimensional coordinate system is introduced so that the sandbag is at the origin and the ropes are anchored at the points (0, −2, 1) and (0, 2, 1). (a) Assuming that the force due to gravity points parallel to the vector (0, 0,−1), give a vector F that describes this gravitational force. (b) Now, use vectors to describe the forces along each of the two ropes. Use symmetry considerations and draw a figure of the situation. 27. A 10 lb weight is suspended in equilibrium by two

ropes. Assume that the weight is at the point (1, 2, 3) in a three-dimensional coordinate system, where the positive z-axis points straight up, perpendicular to the ground, and that the ropes are anchored at the points (3, 0, 4) and (0, 3, 5). Give vectors F1 and F2 that describe the forces along the ropes.

More About Vectors

The Standard Basis Vectors In R2 , the vectors i = (1, 0) and j = (0, 1) play a special notational role. Any vector a = (a1 , a2 ) may be written in terms of i and j via vector addition and scalar multiplication: (a1 , a2 ) = (a1 , 0) + (0, a2 ) = a1 (1, 0) + a2 (0, 1) = a1 i + a2 j. (It may be easier to follow this argument by reading it in reverse.) Insofar as notation goes, the preceding work simply establishes that one can write either (a1 , a2 )

1.2

More About Vectors

9

y

y

a = a1i + a2 j a2 j j x

i

x

a1i

Figure 1.18 Any vector in R2 can be written in terms of i and j.

z

z

k

a3k

a

j

y

a2 j

i

y a1i

x

x 3

Figure 1.19 Any vector in R can be written in terms of i, j, and k.

y

y=3

x

Figure 1.20 In R2 , the equation

y = 3 describes a line. z

EXAMPLE 1 We may write the vector (1, −2) as i − 2j and the vector ◆ (7, π, −3) as 7i + πj − 3k. y=3 y

x Figure 1.21 In R3 , the equation

y = 3 describes a plane.

or a1 i + a2 j to denote the vector a. It’s your choice which notation to use (as long as you’re consistent), but the ij-notation is generally useful for emphasizing the “vector” nature of a, while the coordinate notation is more useful for emphasizing the “point” nature of a (in the sense of a’s role as a possible position vector of a point). Geometrically, the significance of the standard basis vectors i and j is that an arbitrary vector a ∈ R2 can be decomposed pictorially into appropriate vector components along the x- and y-axes, as shown in Figure 1.18. Exactly the same situation occurs in R3 , except that we need three vectors, i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1), to form the standard basis. (See Figure 1.19.) The same argument as the one just given can be used to show that any vector a = (a1 , a2 , a3 ) may also be written as a1 i + a2 j + a3 k. We shall use both coordinate and standard basis notation throughout this text.

Parametric Equations of Lines In R2 , we know that equations of the form y = mx + b or Ax + By = C describe straight lines. (See Figure 1.20.) Consequently, one might expect the same sort of equation to define a line in R3 as well. Consideration of a simple example or two (such as in Figure 1.21) should convince you that a single such linear equation describes a plane, not a line. A pair of simultaneous equations in x, y, and z is required to define a line. We postpone discussing the derivation of equations for planes until §1.5 and concentrate here on using vectors to give sets of parametric equations for lines in R2 or R3 (or even Rn ).

10

Chapter 1

Vectors

y

First, we remark that a curve in the plane may be described analytically by points (x, y), where x and y are given as functions of a third variable (the parameter) t. These functions give rise to parametric equations for the curve:  x = f (t) . y = g(t)

t = π /2

t=π

t=0

x

EXAMPLE 2 The set of equations  x = 2 cos t y = 2 sin t

t = 3π /2

0 ≤ t < 2π

describes a circle of radius 2, since we may check that Figure 1.22 The graph of the

x 2 + y 2 = (2 cos t)2 + (2 sin t)2 = 4.

parametric equations x = 2 cos t, y = 2 sin t, 0 ≤ t < 2π.

Parametric equations may be used as readily to describe curves in R3 ; a curve in R3 is the set of points (x, y, z) whose coordinates x, y, and z are each given by a function of t: ⎧ ⎨x = f (t) y = g(t) . ⎩z = h(t)

z l P0

a a y

x Figure 1.23 The line l is the unique line passing through P0 and parallel to the vector a.

P

a

The advantages of using parametric equations are twofold. First, they offer a uniform way of describing curves in any number of dimensions. (How would you define parametric equations for a curve in R4 ? In R128 ?) Second, they allow you to get a dynamic sense of a curve if you consider the parameter variable t to represent time and imagine that a particle is traveling along the curve with time according to the given parametric equations. You can represent this geometrically by assigning a “direction” to the curve to signify increasing t. Notice the arrow in Figure 1.22. Now, we see how to provide equations for lines. First, convince yourself that a line in R2 or R3 is uniquely determined by two pieces of geometric information: (1) a vector whose direction is parallel to that of the line and (2) any particular point lying on the line—see Figure 1.23. In Figure 1.24, we seek the vector −→ r = OP between the origin O and an arbitrary point P on the line l (i.e., the position −→ vector of P(x, y, z)). O P is the vector sum of the position vector b of the given −−→ point P0 (i.e., O P0 ) and a vector parallel to a. Any vector parallel to a must be a scalar multiple of a. Letting this scalar be the parameter variable t, we have −→ −−→ r = O P = O P0 + ta,

z

ta P0

◆

(See Figure 1.22.)

r = OP

b = OP0

and we have established the following proposition: O

y

PROPOSITION 2.1 The vector parametric equation for the line through the point

x Figure 1.24 The graph of a line

in R3 .

−−→ P0 (b1 , b2 , b3 ), whose position vector is O P0 = b = b1 i + b2 j + b3 k, and parallel to a = a1 i + a2 j + a3 k is r(t) = b + ta.

(1)

1.2

More About Vectors

11

Expanding formula (1), −→ r(t) = O P = b1 i + b2 j + b3 k + t(a1 i + a2 j + a3 k) = (a1 t + b1 )i + (a2 t + b2 )j + (a3 t + b3 )k. −→ Next, write O P as xi + yj + zk so that P has coordinates (x, y, z). Then, extracting components, we see that the coordinates of P are (a1 t + b1 , a2 t + b2 , a3 t + b3 ) and our parametric equations are ⎧ ⎨x = a1 t + b1 y = a2 t + b2 , (2) ⎩ z = a3 t + b3 where t is any real number. These parametric equations work just as well in R2 (if we ignore the zcomponent) or in Rn where n is arbitrary. In Rn , formula (1) remains valid, where we take a = (a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ). The resulting parametric equations are ⎧ x 1 = a 1 t + b1 ⎪ ⎪ ⎪ ⎨ x 2 = a 2 t + b2 . .. ⎪ ⎪ . ⎪ ⎩ xn = an t + bn EXAMPLE 3 To find the parametric equations of the line through (1, −2, 3) and parallel to the vector πi − 3j + k, we have a = π i − 3j + k and b = i − 2j + 3k so that formula (1) yields r(t) = i − 2j + 3k + t(π i − 3j + k) = (1 + πt)i + (−2 − 3t)j + (3 + t)k. The parametric equations may be read as ⎧ ⎪ ⎨ x = πt + 1 y = −3t − 2 . ⎪ ⎩z = t + 3 z

◆

EXAMPLE 4 From Euclidean geometry, two distinct points determine a unique line in R2 or R3 . Let’s find the parametric equations of the line through the points P0 (1, −2, 3) and P1 (0, 5, −1). The situation is suggested by Figure 1.25. To use formula (1), we need to find a vector a parallel to the desired line. The vector with tail at P0 and head at P1 is such a vector. That is, we may use for a the vector −−→ P0 P1 = (0 − 1, 5 − (−2), −1 − 3) = −i + 7j − 4k.

P0

y P1 x Figure 1.25 Finding equations for a line through two points in Example 4.

For b, the position vector of a particular point on the line, we have the choice of taking either b = i − 2j + 3k or b = 5j − k. Hence, the equations in (2) yield parametric equations ⎧ ⎧ ⎪ ⎪ ⎨x = 1 − t ⎨ x = −t or y = −2 + 7t y = 5 + 7t . ⎪ ⎪ ⎩ z = 3 − 4t ⎩ z = −1 − 4t ◆

12

Chapter 1

Vectors

In general, given two arbitrary points P0 (a1 , a2 , a3 )

and

P1 (b1 , b2 , b3 ),

the line joining them has vector parametric equation −−→ −−→ r(t) = O P0 + t P0 P1 . Equation (3) gives parametric equations ⎧ ⎪ ⎨ x = a1 + (b1 − a1 )t y = a2 + (b2 − a2 )t . ⎪ ⎩ z = a + (b − a )t 3 3 3

(3)

(4)

Alternatively, in place of equation (3), we could use the vector equation −−→ −−→ r(t) = O P1 + t P0 P1 ,

(5)

−−→ −−→ r(t) = O P1 + t P1 P0 ,

(6)

or perhaps

each of which gives rise to somewhat different sets of parametric equations. Again, we refer you to Figure 1.25 for an understanding of the vector geometry involved. Example 4 brings up an important point, namely, that parametric equations for a line (or, more generally, for any curve) are never unique. In fact, the two sets of equations calculated in Example 4 are by no means the only ones; we −−→ could have taken a = P1 P0 = i − 7j + 4k or any nonzero scalar multiple of −−→ P0 P1 for a. If parametric equations are not determined uniquely, then how can you check your work? In general, this is not so easy to do, but in the case of lines, there are two approaches to take. One is to produce two points that lie on the line specified by the first set of parametric equations and see that these points lie on the line given by the second set of parametric equations. The other approach is to use the parametric equations to find what is called the symmetric form of a line in R3 . From the equations in (2), assuming that each ai is nonzero, one can eliminate the parameter variable t in each equation to obtain: ⎧ x − b1 ⎪ t= ⎪ ⎪ ⎪ a1 ⎪ ⎪ ⎪ ⎨ y − b2 . t= ⎪ a2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ t = z − b3 a3 The symmetric form is y − b2 z − b3 x − b1 = = . a1 a2 a3

(7)

1.2

More About Vectors

13

In Example 4, the two sets of parametric equations give rise to corresponding symmetric forms x −1 y+2 z−3 x y−5 z+1 = = and = = . −1 7 −4 −1 7 −4 It’s not difficult to see that adding 1 to each “side” of the second symmetric form yields the first one. In general, symmetric forms for lines can differ only by a constant term or constant scalar multiples (or both). The symmetric form is really a set of two simultaneous equations in R3 . For example, the information in (7) can also be written as ⎧x −b y − b2 1 ⎪ = ⎪ ⎨ a1 a2 . ⎪ z − b x − b 1 3 ⎪ ⎩ = a1 a3 This illustrates that we require two “scalar” equations in x, y, and z to describe a line in R3 , although a single vector parametric equation, formula (1), is sufficient. The next two examples illustrate how to use parametric equations for lines to identify the intersection of a line and a plane or of two lines. EXAMPLE 5 We find where the line with parametric equations ⎧ ⎪ ⎨x = t + 5 y = −2t − 4 ⎪ ⎩ z = 3t + 7 intersects the plane 3x + 2y − 7z = 2. To locate the point of intersection, we must find what value of the parameter t gives a point on the line that also lies in the plane. This is readily accomplished by substituting the parametric values for x, y, and z from the line into the equation for the plane 3(t + 5) + 2(−2t − 4) − 7(3t + 7) = 2.

(8)

Solving equation (8) for t, we find that t = −2. Setting t equal to −2 in the parametric equations for the line yields the point (3, 0, 1), which, indeed, lies in ◆ the plane as well. EXAMPLE 6 We determine whether and where the two lines ⎧ ⎧ ⎪ ⎪ ⎨x = t + 1 ⎨x = 3t − 3 y = 5t + 6 and y=t ⎪ ⎪ ⎩ z = −2t ⎩z = t + 1 intersect. The lines intersect provided that there is a specific value t1 for the parameter of the first line and a value t2 for the parameter of the second line that generate the same point. In other words, we must be able to find t1 and t2 so that, by equating the respective parametric expressions for x, y, and z, we have ⎧ ⎪ ⎨t1 + 1 = 3t2 − 3 5t1 + 6 = t2 . (9) ⎪ ⎩−2t = t + 1 1 2

14

Chapter 1

Vectors

The last two equations of (9) yield t2 = 5t1 + 6 = −2t1 − 1

⇒

t1 = −1.

Using t1 = −1 in the second equation of (9), we find that t2 = 1. Note that the values t1 = −1 and t2 = 1 also satisfy the first equation of (9); therefore, we have solved the system. Setting t = −1 in the set of parametric equations for the first ◆ line gives the desired intersection point, namely, (0, 1, 2).

Parametric Equations in General Vector geometry makes it relatively easy to find parametric equations for a variety of curves. We provide two examples. EXAMPLE 7 If a wheel rolls along a flat surface without slipping, a point on the rim of the wheel traces a curve called a cycloid, as shown in Figure 1.26. y

x

y Figure 1.26 The graph of a cycloid.

P t

A x

O Figure 1.27 The result of the wheel in Figure 1.26 rolling through a central angle of t.

P 3π /2 − t

t

A

Suppose that the wheel has radius a and that coordinates in R2 are chosen so that the point of interest on the wheel is initially at the origin. After the wheel has rolled through a central angle of t radians, the situation is as shown in Figure 1.27. −→ We seek the vector O P, the position vector of P, in terms of the parameter t. −→ −→ −→ Evidently, O P = O A + A P, where the point A is the center of the wheel. The −→ vector O A is not difficult to determine. Its j-component must be a, since the center of the wheel does not vary vertically. Its i-component must equal the distance the wheel has rolled; if t is measured in radians, then this distance is at, the length −→ of the arc of the circle having central angle t. Hence, O A = ati + aj. −→ The value of vector methods becomes apparent when we determine A P. −→ Parallel translate the picture so that A P has its tail at the origin, as in Figure 1.28. From the parametric equations of a circle of radius a,



3π 3π −→ − t i + a sin − t j = −a sin t i − a cos t j, A P = a cos 2 2 from the addition formulas for sine and cosine. We conclude that

−→ Figure 1.28 A P with its tail at the origin.

−→ −→ −→ O P = O A + A P = (ati + aj) + (−a sin ti − a cos tj) = a(t − sin t)i + a(1 − cos t)j,

1.2

so the parametric equations are 

x = a(t − sin t) y = a(1 − cos t)

15

More About Vectors

.

◆

EXAMPLE 8 If you unwind adhesive tape from a nonrotating circular tape dispenser so that the unwound tape is held taut and tangent to the dispenser roll, then the end of the tape traces a curve called the involute of the circle. Let’s find the parametric equations for this curve, assuming that the dispensing roll has constant radius a and is centered at the origin. (As more and more tape is unwound, the radius of the roll will, of course, decrease. We’ll assume that little enough tape is unwound so that the radius of the roll remains constant.) −→ Considering Figure 1.29, we see that the position vector O P of the desired −→ −→ −→ −→ point P is the vector sum O B + B P. To determine O B and B P, we use the angle −→ θ between the positive x-axis and O B as our parameter. Since B is a point on the circle, −→ O B = a cos θ i + a sin θ j. y

y Unwound tape

B

θ

O

P

in Example 8. The point P describes a curve known as the involute of the circle.

y

x

P θ − π /2

x

−→

Figure 1.30 The vector B P must

make an angle of θ − π/2 with the positive x-axis.

−→ To find the vector B P, parallel translate it so that its tail is at the origin. Figure 1.30 −→ shows that B P’s length must be aθ, the amount of unwound tape, and its direction must be such that it makes an angle of θ − π/2 with the positive x-axis. From our experience with circular geometry and, perhaps, polar coordinates, we see that −→ B P is described by π π −→ i + aθ sin θ − j = aθ sin θ i − aθ cos θ j. B P = aθ cos θ − 2 2 Hence, −→ −→ −→ O P = O B + B P = a(cos θ + θ sin θ) i + a(sin θ − θ cos θ ) j. So

Figure 1.31 The involute.

θ

aθ

x

(a, 0)

Figure 1.29 Unwinding tape, as

Generating circle

a

Involute

B



x = a(cos θ + θ sin θ) y = a(sin θ − θ cos θ)

are the parametric equations of the involute, whose graph is pictured in ◆ Figure 1.31.

16

Chapter 1

Vectors

1.2 Exercises In Exercises 1–5, write the given vector by using the standard basis vectors for R2 and R3 . 1. (2, 4)

2. (9, −6)

4. (−1, 2, 5)

5. (2, 4, 0)

3. (3, π, −7)

In Exercises 6–10, write the given vector without using the standard basis notation. 6. i + j − 3k √ 7. 9i − 2j + 2 k 8. −3(2i − 7k) 9. πi − j (Consider this to be a vector in R2 .) 10. πi − j (Consider this to be a vector in R3 .) 11. Let a1 = (1, 1) and a2 = (1, −1).

(a) Write the vector b = (3, 1) as c1 a1 + c2 a2 , where c1 and c2 are appropriate scalars. (b) Repeat part (a) for the vector b = (3, −5). (c) Show that any vector b = (b1 , b2 ) in R2 may be written in the form c1 a1 + c2 a2 for appropriate choices of the scalars c1 , c2 . (This shows that a1 and a2 form a basis for R2 that can be used instead of i and j.) 12. Let a1 = (1, 0, −1), a2 = (0, 1, 0), and a3 = (1, 1, −1). (a) Find scalars c1 , c2 , c3 , so as to write the vector b = (5, 6, −5) as c1 a1 + c2 a2 + c3 a3 . (b) Try to repeat part (a) for the vector b = (2, 3, 4). What happens? (c) Can the vectors a1 , a2 , a3 be used as a basis for R3 , instead of i, j, k? Why or why not? In Exercises 13–18, give a set of parametric equations for the lines so described. 13. The line in R3 through the point (2, −1, 5) that is parallel to the vector i + 3j − 6k. 14. The line in R3 through the point (12, −2, 0) that is

parallel to the vector 5i − 12j + k.

15. The line in R2 through the point (2, −1) that is parallel

to the vector i − 7j.

16. The line in R3 through the points (2, 1, 2) and

(3, −1, 5).

5 R √ through the points (9, π, −1, 5, 2) and (−1, 1, 2, 7, 1).

21. (a) Write a set of parametric equations for the line in

R3 through the point (−1, 7, 3) and parallel to the vector 2i − j + 5k. (b) Write a set of parametric equations for the line through the points (5, −3, 4) and (0, 1, 9). (c) Write different (but equally correct) sets of equations for parts (a) and (b). (d) Find the symmetric forms of your answers in (a)–(c). 22. Give a symmetric form for the line having parametric

equations x = 5 − 2t, y = 3t + 1, z = 6t − 4.

23. Give a symmetric form for the line having parametric

equations x = t + 7, y = 3t − 9, z = 6 − 8t.

24. A certain line in R3 has symmetric form

x −2 y−3 z+1 = = . 5 −2 4 Write a set of parametric equations for this line. 25. Give a set of parametric equations for the line with

symmetric form x +5 y−1 z + 10 = = . 3 7 −2 26. Are the two lines with symmetric forms

x −1 y+2 z+1 = = 5 −3 4 and x −4 y−1 z+5 = = 10 −5 8 the same? Why or why not? 27. Show that the two sets of equations

x −2 y−1 z x +1 y+6 z+5 = = and = = 3 7 5 −6 −14 −10 actually represent the same line in R3 . 28. Determine whether the two lines l1 and l2 defined by

3

17. The line in R

(2, 4, −1).

20. Write a set of parametric equations for the line in

through the points (1, 4, 5) and

18. The line in R2 through the points (8, 5) and (1, 7). 19. Write a set of parametric equations for the line in R4

through the point (1, 2, 0, 4) and parallel to the vector (−2, 5, 3, 7).

the sets of parametric equations l1 : x = 2t − 5, y = 3t + 2, z = 1 − 6t, and l2 : x = 1 − 2t, y = 11 − 3t, z = 6t − 17 are the same. (Hint: First find two points on l1 and then see if those points lie on l2 .)

29. Do the parametric equations l1 : x = 3t + 2, y =

t − 7, z = 5t + 1, and l2 : x = 6t − 1, y = 2t − 8, z = 10t − 3 describe the same line? Why or why not?

1.2

30. Do the parametric equations x = 3t 3 + 7, y = 2 − t 3 ,

z = 5t 3 + 1 determine a line? Why or why not?

31. Do the parametric equations x = 5t − 1, y = 2t + 2

2

3, z = 1 − t 2 determine a line? Explain.

32. A bird is flying along the straight-line path x = 2t + 7,

y = t − 2, z = 1 − 3t, where t is measured in minutes. (a) Where is the bird initially (at t = 0)? Where is the bird 3 minutes later? (b) Give a vector that is parallel to the bird’s path.   (c) When does the bird reach the point 34 , 1 , − 11 ? 3 6 2 (d) Does the bird reach (17, 4, −14)?

17

Exercises

44. (a) Find the distance from the point (−2, 1, 5) to any

point on the line x = 3t − 5, y = 1 − t, z = 4t + 7. (Your answer should be in terms of the parameter t.) (b) Now find the distance between the point (−2, 1, 5) and the line x = 3t − 5, y = 1 − t, z = 4t + 7. (The distance between a point and a line is the distance between the given point and the closest point on the line.)

45. (a) Describe the curve given parametrically by



33. Find where the line x = 3t − 5, y = 2 − t, z = 6t in-

x = 2 cos 3t y = 2 sin 3t

0≤t
0

Figure 1.65 If the angle between

So, again, it follows that (ka) × b = k(a × b).

■

a and b is θ , then the angle between ka and b is either θ (if k > 0) or π − θ (if k < 0).

1.4 Exercises Evaluate the determinants in Exercises 1–4.     2  0 4  5  1.  2.   −1 1 3 6     1  −2 3 5  0      2 7  3.  0 4.  3 6  −1  0 3   4 −8

18. Find the volume of the parallelepiped determined by

a = 3i − j, b = −2i + k, and c = i − 2j + 4k.

   

19. What is the volume of the parallelepiped with vertices

 1  2   −1   2 

In Exercises 5–7, calculate the indicated cross products, using both formulas (2) and (3). 5. (1, 3, −2) × (−1, 5, 7) 6. (3i − 2j + k) × (i + j + k) 7. (i + j) × (−3i + 2j) 8. Prove property 3 of cross products, using properties 1

and 2. 9. If a × b = 3i − 7j − 2k, what is (a + b) × (a − b)?

(3, 0, −1), (4, 2, −1), (−1, 1, 0), (4, 3, 5), (−1, 2, 6), and (0, 4, 6)?   a1 a2  20. Verify that (a × b) · c =  b1 b2  c1 c2

(3, 1, 5), (0, 3, 0), a3 b3 c3

   .  

21. Show that (a × b) · c = a · (b × c) using Exercise 20. 22. Use

geometry |b · (a × c)|.

to

show

that

|(a × b) · c| =

23. (a) Show that the area of the triangle with vertices

P1 (x1 , y1 ), P2 (x2 , y2 ), and P3 (x3 , y3 ) is given by the absolute value of the expression     1  1 1 1  x1 x2 x3 . 2  y y y  1 2 3

10. Calculate the area of the parallelogram having vertices

(1, 1), (3, 2), (1, 3), and (−1, 2). 11. Calculate the area of the parallelogram having vertices

(1, 2, 3), (4, −2, 1), (−3, 1, 0), and (0, −3, −2).

12. Find a unit vector that is perpendicular to both 2i +

j − 3k and i + k.

13. If (a × b) · c = 0, what can you say about the geomet-

(b) Use part (a) to find the area of the triangle with vertices (1, 2), (2, 3), and (−4, −4). 24. Suppose that a, b, and c are noncoplanar vectors in R3 ,

so that they determine a tetrahedron as in Figure 1.66. c

ric relation between a, b, and c? Compute the area of the triangles described in Exercises 14–17. 14. The triangle determined by the vectors a = i + j and

b

b = 2i − j

15. The triangle determined by the vectors a = i − 2j +

6k and b = 4i + 3j − k

16. The triangle having vertices (1, 1), (−1, 2), and

(−2, −1)

17. The triangle having vertices (1, 0, 1), (0, 2, 3), and

(−1, 5, −2)

a Figure 1.66 The tetrahedron of

Exercise 24.

Give a formula for the surface area of the tetrahedron in terms of a, b, and c. (Note: More than one formula is possible.)

Exercises

1.4

39

25. Suppose that you are given nonzero vectors a, b, and c

in R3 . Use dot and cross products to give expressions for vectors satisfying the following geometric descriptions: (a) A vector orthogonal to a and b (b) A vector of length 2 orthogonal to a and b (c) The vector projection of b onto a (d) A vector with the length of b and the direction of a (e) A vector orthogonal to a and b × c (f) A vector in the plane determined by a and b and perpendicular to c. 26. Suppose a, b, c, and d are vectors in R3 . Indicate which

of the following expressions are vectors, which are scalars, and which are nonsense (i.e., neither a vector nor a scalar). (b) (a · b) · c (a) (a × b) × c (c) (a · b) × (c · d) (d) (a × b) · c (e) (a · b) × (c × d) (f) a × [(b · c)d] (g) (a × b) · (c × d) (h) (a · b)c − (a × b) Exercises 27–32 concern several identities for vectors a, b, c, and d in R3 . Each of them can be verified by hand by writing the vectors in terms of their components and by using formula (2) for the cross product and Definition 3.1 for the dot product. However, this is quite tedious to do. Instead, use a computer algebra system to define the vectors a, b, c, and d in general and to verify the identities.

◆ T 28. a · (b × c) = b · (c × a) = c · (a × b) ◆ = −a · (c × b) = −c · (b × a) T 27. (a × b) × c = (a · c)b − (b · c)a

= −b · (a × c) b) · (c × d) = (a · c)(b · d) − (a · d)(b · c) T 29. (a × ◆    a·c a·d  = 

 b·c b·d 

T 30. (a × b) × c + (b × c) × a + (c × a) × b = 0 (this is ◆ known as the Jacobi identity). T 31. (a × b) × (c × d) = [a · (c × d)]b − [b · (c × d)]a ◆ T 32. (a × b) · (b × c) × (c × a) = [a · (b × c)] ◆ 2

45

F F = 20 lb O 4 ft

P

Figure 1.67 Figure for Exercise 34.

40 lb 30

Figure 1.68 The configuration for

Exercise 35.

that it makes an angle of 30◦ with the horizontal. (See Figure 1.68.) Gertrude exerts 40 lb of force straight down to turn the bolt. (a) If the length of the arm of the wrench is 1 ft, how much torque does Gertrude impart to the bolt? (b) What if she has a second tire iron whose length is 18 in? 36. Egbert is trying to open a jar of grape jelly. The ra-

dius of the lid of the jar is 2 in. If Egbert imparts 15 lb of force tangent to the edge of the lid to open the jar, how many ft-lb, and in what direction, is the resulting torque? 37. A 50 lb child is sitting on one end of a seesaw, 3 ft

from the center fulcrum. (See Figure 1.69.) When she is

33. Establish the identity

(a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c) of Exercise 29 without resorting to a computer algebra system by using the results of Exercises 27 and 28. 34. Egbert applies a 20 lb force at the edge of a 4 ft

wide door that is half-open in order to close it. (See Figure 1.67.) Assume that the direction of force is perpendicular to the plane of the doorway. What is the torque about the hinge on the door?

3 ft 1.5 ft

35. Gertrude is changing a flat tire with a tire iron. The tire

iron is positioned on one of the bolts of the wheel so

Figure 1.69 The seesaw of Exercise 37.

40

Chapter 1

Vectors

1.5 ft above the horizontal position, what is the amount of torque she exerts on the seesaw? 38. For this problem, note that the radius of the earth is

approximately 3960 miles. (a) Suppose that you are standing at 45◦ north latitude. Given that the earth spins about its axis, how fast are you moving? (b) How fast would you be traveling if, instead, you were standing at a point on the equator? 39. Archie, the cockroach, and Annie, the ant, are on an

LP record. Archie is at the edge of the record (approximately 6 in from the center) and Annie is 2 in closer to the center of the record. How much faster is Archie traveling than Annie? (Note: A record playing on a turntable spins at a rate of 33 13 revolutions per minute.) 40. A top is spinning with a constant angular speed of 12

radians/sec. Suppose that the top spins about its axis

1.5

of symmetry and we orient things so that this axis is the z-axis and the top spins counterclockwise about it. (a) If, at a certain instant, a point P in the top has coordinates (2, −1, 3), what is the velocity of the point at that instant? (b) What are the (approximate) coordinates of P one second later? 41. There is a difficulty involved with our definition of

the angular velocity vector ω, namely, that we cannot properly consider this vector to be “free” in the sense of being able to parallel translate it at will. Consider the rotations of a rigid body about each of two parallel axes. Then the corresponding angular velocity vectors ω1 and ω2 are parallel. Explain, perhaps with a figure, that even if ω1 and ω2 are equal as “free vectors,” the corresponding rotational motions that result must be different. (Therefore, when considering more than one angular velocity, we should always assume that the axes of rotation pass through a common point.)

Equations for Planes; Distance Problems

In this section, we use vectors to derive analytic descriptions of planes in R3 . We also show how to solve a variety of distance problems involving “flat objects” (i.e., points, lines, and planes). z n P Π

P0

Coordinate Equations of Planes A plane in R3 is determined uniquely by the following geometric information: a particular point P0 (x0 , y0 , z 0 ) in the plane and a particular vector n = Ai + Bj + Ck that is normal (perpendicular) to the plane. In other words, is the −−→ set of all points P(x, y, z) in space such that P0 P is perpendicular to n. (See Figure 1.70.) This means that is defined by the vector equation

y

−−→ n · P0 P = 0.

(1)

x Figure 1.70 The plane in R3

through the point P0 and perpendicular to the vector n.

−−→ Since P0 P = (x − x0 )i + (y − y0 )j + (z − z 0 )k, equation (1) may be rewritten as (Ai + Bj + Ck) · ((x − x0 )i + (y − y0 )j + (z − z 0 )k) = 0 or A(x − x0 ) + B(y − y0 ) + C(z − z 0 ) = 0. This is equivalent to Ax + By + C z = D, where D = Ax0 + By0 + C z 0 .

(2)

1.5

Equations for Planes; Distance Problems

41

EXAMPLE 1 The plane through the point (3, 2, 1) with normal vector 2i − j + 4k has equation (2i − j + 4k) · ((x − 3)i + (y − 2)j + (z − 1)k) = 0 ⇐⇒ 2(x − 3) − (y − 2) + 4(z − 1) = 0 ⇐⇒ 2x − y + 4z = 8.

◆

Not only does a plane in R3 have an equation of the form given by equation (2), but, conversely, any equation of this form must describe a plane. Moreover, it is easy to read off the components of a vector normal to the plane from such an equation: They are just the coefficients of x, y, and z. EXAMPLE 2 Given the plane with equation 7x + 2y − 3z = 1, find a normal vector to the plane and identify three points that lie on that plane. A possible normal vector is n = 7i + 2j − 3k. However, any nonzero scalar multiple of n will do just as well. Algebraically, the effect of using a scalar multiple of n as normal is to multiply equation (2) by such a scalar. Finding three points in the plane is not difficult. First, let y = z = 0 in the defining equation and solve for x: 7x + 2 · 0 − 3 · 0 = 1 ⇐⇒ 7x = 1 ⇐⇒ x = 17 .   Thus 17 , 0, 0 is a point on the plane. Next, let x = z = 0 and solve for y: 7 · 0 + 2y − 3 · 0 = 1

⇐⇒

y = 12 .

  So 0, 12 , 0 is another point on the plane. Finally, let x = y = 0 and solve for z.   ◆ You should find that 0, 0, − 13 lies on the plane. EXAMPLE 3 Put coordinate axes on R3 so that the z-axis points vertically. Then a plane in R3 is vertical if its normal vector n is horizontal (i.e., if n is parallel to the x y-plane). This means that n has no k-component, so n can be written in the form Ai + Bj. It follows from equation (2) that a vertical plane has an equation of the form A(x − x0 ) + B(y − y0 ) = 0. Hence, a nonvertical plane has an equation of the form A(x − x0 ) + B(y − y0 ) + C(z − z 0 ) = 0, where C = 0.

◆

EXAMPLE 4 From high school geometry, you may recall that a plane is determined by three (noncollinear) points. Let’s find an equation of the plane that contains the points P0 (1, 2, 0), P1 (3, 1, 2), and P2 (0, 1, 1). There are two ways to solve this problem. The first approach is algebraic and rather uninspired. From the aforementioned remarks, any plane must have an equation of the form Ax + By + C z = D for suitable constants A, B, C, and D. Thus, we need only to substitute the coordinates of P0 , P1 , and P2 into this equation and solve for A, B, C, and D. We have that • substitution of P0 gives A + 2B = D; • substitution of P1 gives 3A + B + 2C = D; and • substitution of P2 gives B + C = D.

42

Chapter 1

Vectors

Hence, we must solve a system of three equations in four unknowns: ⎧ ⎪ =D ⎨ A + 2B 3A + B + 2C = D . ⎪ ⎩ B+ C=D

(3)

In general, such a system has either no solution or else infinitely many solutions. We must be in the latter case, since we know that the three points P0 , P1 , and P2 lie on some plane (i.e., that some set of constants A, B, C, and D must exist). Furthermore, the existence of infinitely many solutions corresponds to the fact that any particular equation for a plane may be multiplied by a nonzero constant without altering the plane defined. In other words, we can choose a value for one of A, B, C, or D, and then the other values will be determined. So let’s multiply the first equation given in (3) by 3, and subtract it from the second equation. We obtain ⎧ ⎪ = D ⎨ A + 2B (4) −5B + 2C = −2D . ⎪ ⎩ B+ C= D Now, multiply the third equation in (4) by 5 and add it to the second: ⎧ ⎪ = D ⎨ A + 2B 7C = 3D . ⎪ ⎩ B+ C= D

(5)

Multiply the third equation appearing in (5) by 2 and subtract it from the first: ⎧ ⎪ −2C = −D ⎨A (6) 7C = 3D . ⎪ ⎩ B+ C= D By adding appropriate multiples of the second equation to both the first and third equations of (6), we find that ⎧ ⎪ = − 17 D ⎨A (7) 7C = 3D . ⎪ ⎩ 4 B = 7D Thus, if in (7) we take D = −7 (for example), then A = 1, B = −4, C = −3, and the equation of the desired plane is x − 4y − 3z = −7. z

P1

P2 P0

x

n

Figure 1.71 The plane determined by the points P0 , P1 , and P2 in Example 4.

y

The second method of solution is cleaner and more geometric. The idea is to make use of equation (1). Therefore, we need to know the coordinates of a particular point on the plane (no problem—we are given three such points) and −−→ −−→ a vector n normal to the plane. The vectors P0 P1 and P0 P2 both lie in the plane. (See Figure 1.71.) In particular, the normal vector n must be perpendicular to them both. Consequently, the cross product provides just what we need. That is, we may take −−→ −−→ n = P0 P1 × P0 P2 = (2i − j + 2k) × (−i − j + k)    i j k   2  = i − 4j − 3k. =  2 −1  −1 −1 1

1.5

Equations for Planes; Distance Problems

43

If we take P0 (1, 2, 0) to be the particular point in equation (1), we find that the equation we desire is (i − 4j − 3k) · ((x − 1)i + (y − 2)j + zk) = 0 or (x − 1) − 4(y − 2) − 3z = 0. This is the same equation as the one given by the first method. z x − 2y + z = 4

2x + y + 3z = −7

x

y

Figure 1.72 The line of intersection of the planes x − 2y + z = 4 and 2x + y + 3z = −7 in Example 5.

◆

EXAMPLE 5 Consider the two planes having equations x − 2y + z = 4 and 2x + y + 3z = −7. We determine a set of parametric equations for their line of intersection. (See Figure 1.72.) We use Proposition 2.1. Thus, we need to find a point on the line and a vector parallel to the line. To find the point on the line, we note that the coordinates (x, y, z) of any such point must satisfy the system of simultaneous equations given by the two planes  x − 2y + z = 4 . (8) 2x + y + 3z = −7 From the equations given in (8), it is not too difficult to produce a single solution (x, y, z). For example, if we let z = 0 in (8), we obtain the simpler system  x − 2y = 4 . (9) 2x + y = −7 The solution to the system of equations (9) is readily calculated to be x = −2, y = −3. Thus, (−2, −3, 0) are the coordinates of a point on the line. To find a vector parallel to the line of intersection, note that such a vector must be perpendicular to the two normal vectors to the planes. The normal vectors to the planes are i − 2j + k and 2i + j + 3k. Therefore, a vector parallel to the line of intersection is given by (i − 2j + k) × (2i + j + 3k) = −7i − j + 5k. Hence, Proposition 2.1 implies that a vector parametric equation for the line is r(t) = (−2i − 3j) + t(−7i − j + 5k), and a standard set of parametric equations is ⎧ ⎪ ⎨ x = −7t − 2 y = −t − 3 . ⎪ ⎩ z = 5t

◆

Parametric Equations of Planes Another way to describe a plane in R3 is by a set of parametric equations. First, suppose that a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) are two nonzero, nonparallel vectors in R3 . Then a and b determine a plane in R3 that passes through the origin. (See Figure 1.73.) To find the coordinates of a point P(x, y, z) in this plane, draw a parallelogram whose sides are parallel to a and b and that has two opposite vertices at the origin and at P, as shown in Figure 1.74. Then there must −→ exist scalars s and t so that the position vector of P is O P = sa + tb. The plane

44

Chapter 1

Vectors

z

z

tb b

y

b y

a x

a

sa

P

x Figure 1.73 The plane through

the origin determined by the vectors a and b.

may be described as

Figure 1.74 For the point P in

−→ the plane shown, O P = sa + tb for appropriate scalars s and t.



 x ∈ R3 | x = sa + tb; s, t ∈ R . Now, suppose that we seek to describe a general plane (i.e., one that does not necessarily pass through the origin). Let −−→ c = (c1 , c2 , c3 ) = O P0 denote the position vector of a particular point P0 in and let a and b be two (nonzero, nonparallel) vectors that determine the plane through the origin parallel to . By parallel translating a and b so that their tails are at the head of c (as in Figure 1.75), we adapt the preceding discussion to see that the position vector of any point P(x, y, z) in may be described as −→ O P = sa + tb + c. To summarize, we have shown the following:

z

Π

P0 b P c a y

x Figure 1.75 The plane passing through P0 (c1 , c2 , c3 ) and parallel to a and b.

PROPOSITION 5.1 A vector parametric equation for the plane containing

−−→ the point P0 (c1 , c2 , c3 ) (whose position vector is O P0 = c) and parallel to the nonzero, nonparallel vectors a and b is x(s, t) = sa + tb + c.

(10)

By taking components in formula (10), we readily obtain a set of parametric equations for : ⎧ ⎪ ⎨ x = sa1 + tb1 + c1 y = sa2 + tb2 + c2 . ⎪ ⎩ z = sa + tb + c 3 3 3

(11)

Compare formula (10) with that of equation (1) in Proposition 2.1. We need to use two parameters s and t to describe a plane (instead of a single parameter t that appears in the vector parametric equation for a line) because a plane is a two-dimensional object.

1.5

Equations for Planes; Distance Problems

45

EXAMPLE 6 We find a set of parametric equations for the plane that passes through the point (1, 0, −1) and is parallel to the vectors 3i − k and 2i + 5j + 2k. From formula (10), any point on the plane is specified by x(s, t) = s(3i − k) + t(2i + 5j + 2k) + (i − k) = (3s + 2t + 1)i + 5tj + (2t − s − 1)k. The individual parametric equation may be read off as ⎧ ⎪ ⎨ x = 3s + 2t + 1 . y = 5t ⎪ ⎩ z = 2t − s − 1

Distance Problems Cross products and vector projections provide convenient ways to understand a range of distance problems involving lines and planes: Several examples follow. What is important about these examples are the vector techniques for solving geometric problems that they exhibit, not the general formulas that may be derived from them.

P0

BP0 − proja BP0

B

◆

a

Figure 1.76 A general configuration for finding the distance between a point and a line, using vector projections.

P0

EXAMPLE 7 (Distance between a point and a line) We find the distance between the point P0 (2, 1, 3) and the line l(t) = t(−1, 1, −2) + (2, 3, −2) in two ways. METHOD 1. From the vector parametric equations for the given line, we read off a point B on the line—namely, (2, 3, −2)—and a vector a parallel to the line—namely, a = (−1, 1, −2). Using Figure 1.76, the length of the vector −−→ −−→ B P0 − proja B P0 provides the desired distance between P0 and the line. Thus, we calculate that −−→ B P0 = (2, 1, 3) − (2, 3, −2) = (0, −2, 5);  −−→  a · B P0 −−→ a proja B P0 = a·a

(−1, 1, −2) · (0, −2, 5) (−1, 1, −2) = (−1, 1, −2) · (−1, 1, −2) = (2, −2, 4).

D

METHOD 2. In this case, we use a little trigonometry. If θ denotes the angle −−→ between the vectors a and B P0 as in Figure 1.77, then

θ

B

The desired distance is √ −−→ −−→

B P0 − proja B P0 = (0, −2, 5) − (2, −2, 4) = (−2, 0, 1) = 5.

a

Figure 1.77 Another general configuration for finding the distance between a point and a line.

D sin θ = −−→ ,

B P0

where D denotes the distance between P0 and the line. Hence, −−→ −−→

a × B P0

a B P0 sin θ −−→ = . D = B P0 sin θ =

a

a

46

Chapter 1

Vectors

Therefore, we calculate

   i j k  −−→  1 −2  = i + 5j + 2k, a × B P0 =  −1  0 −2 5

so that the distance sought is

√

i + 5j + 2k

30 √ = √ = 5, D=

−i + j − 2k

6

which agrees with the answer obtained by Method 1.

◆

EXAMPLE 8 (Distance between parallel planes) The planes P2

Π2

D

n

Π1

P1

Figure 1.78 The general configuration for finding the distance D between two parallel planes.

1 : 2x − 2y + z = 5

and

2 : 2x − 2y + z = 20

are parallel. (Why?) We see how to compute the distance between them. Using Figure 1.78 as a guide, we see that the desired distance D is given by −−→

projn P1 P2 , where P1 is a point on 1 , P2 is a point on 2 , and n is a vector normal to both planes. First, the vector n that is normal to both planes may be read directly from the equation for either 1 or 2 as n = 2i − 2j + k. It is not hard to find a point P1 on 1 : the point P1 (0, 0, 5) will do. Similarly, take P2 (0, 0, 20) for a point on

2 . Then −−→ P1 P2 = (0, 0, 15), and calculate −−→ projn P1 P2 =





−−→  (2, −2, 1) · (0, 0, 15) n · P1 P2 n= (2, −2, 1) n·n (2, −2, 1) · (2, −2, 1) = − 15 (2, −2, 1) 9 = − 53 (2, −2, 1).

Hence, the distance D that we seek is √ −−→ D = projn P1 P2 = 53 9 = 5.

◆

EXAMPLE 9 (Distance between two skew lines) Find the distance between the two skew lines B1 l1

l2

B2

a1

a2

Figure 1.79 Configuration for determining the distance between two skew lines in Example 9.

l1 (t) = t(2, 1, 3) + (0, 5, −1) 3

and

l2 (t) = t(1, −1, 0) + (−1, 2, 0).

(Two lines in R are said to be skew if they are neither intersecting nor parallel. It follows that the lines must lie in parallel planes and that the distance between the lines is equal to the distance between the planes.) −−→ To solve this problem, we need to find projn B1 B2 , the length of the projection of the vector between a point on each line onto a vector n that is perpendicular to both lines, hence, also perpendicular to the parallel planes that contain the lines. (See Figure 1.79.) From the vector parametric equations for the lines, we read that the point B1 (0, 5, −1) is on the first line and B2 (−1, 2, 0) is on the second. Hence, −−→ B1 B2 = (−1, 2, 0) − (0, 5, −1) = (−1, −3, 1).

1.5

Exercises

47

For a vector n that is perpendicular to both lines, we may use n = a1 × a2 , where a1 = (2, 1, 3) is a vector parallel to the first line and a2 = (1, −1, 0) is parallel to the second. (We may read these vectors from the parametric equations.) Thus,    i  j k    1 3  = 3i + 3j − 3k, n = a1 × a2 =  2  1 −1 0 and so, −−→ projn B1 B2 =





−−→  (−1, −3, 1) · (3, 3, −3) n · B1 B2 n= (3, 3, −3) n·n (3, 3, −3) · (3, 3, −3) = − 15 (3, 3, −3) 27

= − 53 (1, 1, −1). √ −−→ The desired distance is projn B1 B2 = 53 3.

◆

1.5 Exercises 1. Calculate an equation for the plane containing the point

(3, −1, 2) and perpendicular to i − j + 2k.

2. Find an equation for the plane containing the point

(9, 5, −1) and perpendicular to i − 2k.

3. Find an equation for the plane containing the points

(3, −1, 2), (2, 0, 5), and (1, −2, 4).

4. Find an equation for the plane containing the points

(A, 0, 0), (0, B, 0), and (0, 0, C). Assume that at least two of A, B, and C are nonzero. 5. Give an equation for the plane that is parallel to the

plane 5x − 4y + z = 1 and that passes through the point (2, −1, −2).

6. Give an equation for the plane parallel to the plane 2x −

3y + z = 5 that passes through the point (−1, 1, 2).

7. Find an equation for the plane parallel to the plane x −

y + 7z = 10 that passes through the point (−2, 0, 1).

8. Give an equation for the plane parallel to the plane

2x + 2y + z = 5 and that contains the line with parametric equations x = 2 − t, y = 2t + 1, z = 3 − 2t.

9. Explain why there is no plane parallel to the plane

5x − 3y + 2z = 10 that contains the line with parametric equations x = t + 4, y = 3t − 2, z = 5 − 2t.

10. Find an equation for the plane that contains the line x =

2t − 1, y = 3t + 4, z = 7 − t and the point (2, 5, 0).

11. Find an equation for the plane that is perpendicular

to the line x = 3t − 5, y = 7 − 2t, z = 8 − t and that passes through the point (1, −1, 2).

12. Find an equation for the plane that contains the two

lines l1 : x = t + 2, y = 3t − 5, z = 5t + 1 and l2 : x = 5 − t, y = 3t − 10, z = 9 − 2t.

13. Give a set of parametric equations for the line of inter-

section of the planes x + 2y − 3z = 5 and 5x + 5y − z = 1.

14. Give a set of parametric equations for the line through

(5, 0, 6) that is perpendicular to the plane 2x − 3y + 5z = −1.

15. Find a value for A so that the planes 8x − 6y + 9Az =

6 and Ax + y + 2z = 3 are parallel.

16. Find values for A so that the planes Ax − y + z = 1

and 3Ax + Ay − 2z = 5 are perpendicular.

Give a set of parametric equations for each of the planes described in Exercises 17–22. 17. The plane that passes through the point (−1, 2, 7) and

is parallel to the vectors 2i − 3j + k and i − 5k

18. The plane that passes through the point (2, 9, −4)

and is parallel to the vectors −8i + 2j + 5k and 3i − 4j − 2k

19. The plane that contains the lines l1 : x = 2t + 5, y =

−3t − 6, z = 4t + 10 and l2 : x = 5t − 1, y = 10t + 3, z = 7t − 2

20. The plane that passes through the three points (0, 2, 1),

(7, −1, 5), and (−1, 3, 0)

21. The plane that contains the line l: x = 3t − 5, y =

10 − 3t, z = 2t + 9 and the point (−2, 4, 7)

48

Chapter 1

Vectors

22. The plane determined by the equation 2x − 3y +

5z = 30

D1 and Ax + By + C z = D2 is d= √

23. Find a single equation of the form Ax + By + C z = D

that describes the plane given parametrically as x = 3s − t + 2, y = 4s + t, z = s + 5t + 3. (Hint: Begin by writing the parametric equations in vector form and then find a vector normal to the plane.)

24. Find the distance between the point (1, −2, 3) and the

line l: x = 2t − 5, y = 3 − t, z = 4.

|D1 − D2 | A2 + B 2 + C 2

.

34. Two planes are given parametrically by the vector

equations x1 (s, t) = (−3, 4, −9) + s(9, −5, 9) + t(3, −2, 3) x2 (s, t) = (5, 0, 3) + s(−9, 2, −9) + t(−4, 7, −4).

25. Find the distance between the point (2, −1) and the

(a) Give a convincing explanation for why these planes are parallel. (b) Find the distance between the planes.

26. Find the distance between the point (−11, 10, 20) and

35. Write equations for the planes that are parallel to

line l: x = 3t + 7, y = 5t − 3.

the line l: x = 5 − t, y = 3, z = 7t + 8.

27. Determine the distance between the two lines l1 (t) =

t(8, −1, 0) + (−1, 3, 5) (0, 3, 4).

and

l2 (t) = t(0, 3, 1) +

28. Compute the distance between the two lines

l1 (t) = (t − 7)i + (5t + 1)j + (3 − 2t)k and l2 (t) = 4ti + (2 − t)j + (8t + 1)k.

29. (a) Find the distance between the two lines l1 (t) =

t(3, 1, 2) + (4, 0, 2) (2, 1, 3).

and

l2 (t) = t(1, 2, 3) +

(b) What does your answer in part (a) tell you about the relative positions of the lines? 30. (a) The lines l1 (t) = t(1, −1, 5) + (2, 0, −4) and

l2 (t) = t(1, −1, 5) + (1, 3, −5) are parallel. Explain why the method of Example 9 cannot be used to calculate the distance between the lines. (b) Find another way to calculate the distance. (Hint: Try using some calculus.)

31. Find the distance between the two planes given by the

equations x − 3y + 2z = 1 and x − 3y + 2z = 8.

32. Calculate the distance between the two planes

5x − 2y + 2z = 12 and

− 10x + 4y − 4z = 8.

33. Show that the distance d between the two parallel

planes determined by the equations Ax + By + C z =

1.6

x + 3y − 5z = 2 and lie three units from it.

36. Suppose that l1 (t) = ta + b1 and l2 (t) = ta + b2 are

parallel lines in either R2 or R3 . Show that the distance D between them is given by D=

a × (b2 − b1 )

.

a

(Hint: Consider Example 7.) 37. Let be the plane in R3 with normal vector n that

passes through the point A with position vector a. If b is the position vector of a point B in R3 , show that the distance D between B and is given by D=

|n · (b − a)| .

n

38. Show that the distance D between parallel planes with

normal vector n is given by |n · (x2 − x1 )| ,

n

where x1 is the position vector of a point on one of the planes, and x2 is the position vector of a point on the other plane. 39. Suppose that l1 (t) = ta1 + b1 and l2 (t) = ta2 + b2 are

skew lines in R3 . Use the geometric reasoning of Example 9 to show that the distance D between these lines is given by D=

|(a1 × a2 ) · (b2 − b1 )| .

a1 × a2

Some n-dimensional Geometry

Vectors in Rn The algebraic idea of a vector in R2 or R3 is defined in §1.1, in which we asked you to consider what would be involved in generalizing the operations of vector addition, scalar multiplication, etc., to n-dimensional vectors, where n can be arbitrary. We explore some of the details of such a generalization next.

1.6

Some n-dimensional Geometry

49

A vector in Rn is an ordered n-tuple of real numbers. We use a = (a1 , a2 , . . . , an ) as our standard notation for a vector in Rn . DEFINITION 6.1

EXAMPLE 1 The 5-tuple (2, 4, 6, 8, 10) is a vector in R5 . The (n + 1)-tuple (2n, 2n − 2, 2n − 4, . . . , 2, 0) is a vector in Rn+1 , where n is arbitrary. ◆ Exactly as is the case in R2 or R3 , we call two vectors a = (a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ) equal just in case ai = bi for i = 1, 2, . . . , n. Vector addition and scalar multiplication are defined in complete analogy with Definitions 1.3 and 1.4: If a = (a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ) are two vectors in Rn and k ∈ R is any scalar, then a + b = (a1 + b1 , a2 + b2 , . . . , an + bn ) and ka = (ka1 , ka2 , . . . , kan ). The properties of vector addition and scalar multiplication given in §1.1 hold (with proofs that are no different from those in the two- and three-dimensional cases). Similarly, the dot product of two vectors in Rn is readily defined: a · b = a1 b1 + a2 b2 + · · · + an bn . The dot product properties given in §1.3 continue to hold in n dimensions; we leave it to you to check that this is so. What we cannot do in dimensions larger than three is to develop a pictorial representation for vectors as arrows. Nonetheless, the power of our algebra and analogy does allow us to define a number of geometric ideas. We define the length of a vector in a ∈ Rn by using the dot product: √

a = a · a. The distance between two vectors a and b in Rn is Distance between a and b = a − b . We can even define the angle between two nonzero vectors by using a generalized version of equation (4) of §1.3: θ = cos−1

a·b .

a b

Here a, b ∈ Rn and θ is taken so that 0 ≤ θ ≤ π. (Note: At this point in our discussion, it is not clear that we have −1 ≤

a·b ≤ 1,

a b

which is a necessary condition if our definition of the angle θ is to make sense. Fortunately, the Cauchy–Schwarz inequality—formula (1) that follows—takes care of this issue.) Thus, even though we are not able to draw pictures of vectors in Rn , we can nonetheless talk about what it means to say that two vectors are perpendicular or parallel, or how far apart two vectors may be. (Be careful about this business. We are defining notions of length, distance, and angle entirely in

50

Chapter 1

Vectors

terms of the dot product. Results like Theorem 3.3 have no meaning in Rn , since the ideas of angles between vectors and dot products are not independent.) There is no simple generalization of the cross product. However, see Exercises 39–42 at the end of this section for the best we can do by way of analogy. We can create a standard basis of vectors in Rn that generalize the i, j, k-basis in R3 . Let e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), .. . en = (0, 0, . . . , 0, 1). Then it is not difficult to see (check for yourself ) that a = (a1 , a2 , . . . , an ) = a1 e1 + a2 e2 + · · · + an en . Here are two famous (and often handy) inequalities: Cauchy–Schwarz inequality. For all vectors a and b in Rn , we have |a · b| ≤ a b .

(1)

PROOF If n = 2 or 3, this result is virtually immediate in view of Theorem 3.3.

However, in dimensions larger than three, we do not have independent notions of inner products and angles, so a different proof is required. First note that the inequality holds if either a or b is 0. So assume that a and b are nonzero. Then we may define the projection of b onto a just as in §1.3:

a·b proja b = a = ka. a·a Here k is, of course, the scalar a · b/a · a. Let c = b − ka (so that b = ka + c). Then we have a · c = 0, since a · c = a · (b − ka) = a · b − ka · a

a·b a·a = a·b − a·a = a·b − a·b = 0. We leave it to you to check that the “Pythagorean theorem” holds, namely, that the following equation is true:

b 2 = k 2 a 2 + c 2 . Multiply this equation by a 2 = a · a. We obtain

a 2 b 2 = a 2 k 2 a 2 + a 2 c 2

2 2 a·b

a 2 + a 2 c 2 = a

a·a

1.6



Some n-dimensional Geometry

a·b = (a · a) a·a b

c

51

2 (a · a) + a 2 c 2

= (a · b)2 + a 2 c 2 . Now, the quantity a 2 c 2 is nonnegative. Hence,

a 2 b 2 ≥ (a · b)2 . ■

Taking square roots in this last inequality yields the result desired.

a ka Figure 1.80 The geometry behind the proof of the Cauchy–Schwarz inequality.

The geometric motivation for this proof of the Cauchy–Schwarz inequality comes from Figure 1.80.1 The triangle inequality. For all vectors a, b ∈ Rn we have

a + b ≤ a + b .

(2)

PROOF Strategic use of the Cauchy–Schwarz inequality yields

a + b 2 = (a + b) · (a + b) = a · a + 2a · b + b · b ≤ a · a + 2 a b + b · b

by (1)

= a + 2 a b + b

2

2

= ( a + b )2 . b

Thus, the result desired holds by taking square roots, since the quantities on both ■ sides of the inequality are nonnegative.

a a+b Figure 1.81 The triangle inequality visualized.

In two or three dimensions the triangle inequality has the following obvious proof from which the inequality gets its name: Since a , b , and a + b can be viewed as the lengths of the sides of a triangle, inequality (2) says nothing more than that the sum of the lengths of two sides of a triangle must be at least as large as the length of the third side, as demonstrated by Figure 1.81.

Matrices We had a brief glance at matrices and determinants in §1.4 in connection with the computation of cross products. Now it’s time for another look. A matrix is defined in §1.4 as a rectangular array of numbers. To extend our discussion, we need a good notation for matrices and their individual entries. We used the uppercase Latin alphabet to denote entire matrices and will continue to do so. We shall also adopt the standard convention and use the lowercase Latin alphabet and two sets of indices (one set for rows, the other for columns) to identify matrix entries. Thus, the general m × n matrix can be written as ⎡ ⎤ a11 a12 · · · a1n ⎢ a21 a22 · · · a2n ⎥ A=⎢ . ⎥ = (shorthand) (ai j ). .. . . ⎣ .. . .. ⎦ . . am1 am2 · · · amn

1

See J. W. Cannon, Amer. Math. Monthly 96 (1989), no. 7, 630–631.

52

Chapter 1

Vectors

The first index always will represent the row position and the second index, the column position. Vectors in Rn can also be thought of as matrices. We shall have occasion to write the vector a = (a1 , a2 , . . . , an ) either as a row vector (a 1 × n matrix),   a = a1 a2 · · · an , or, more typically, as a column vector (an n × 1 matrix), ⎤ ⎡ a1 ⎢ a2 ⎥ ⎥ a=⎢ ⎣ .. ⎦ . . an We did not use double indices since there is only a single row or column present. It will be clear from context (or else indicated explicitly) in which form a vector a will be viewed. An m × n matrix A can be thought of as a “vector of vectors” in two ways: (1) as m row vectors in Rn , ⎡   ⎤ a11 a12 · · · a1n ⎢   ⎥ ⎥ ⎢ a21 a22 · · · a2n ⎥, ⎢ A=⎢ ⎥ .. ⎦ ⎣ .   am1 am2 · · · amn or (2) as n column vectors in Rm , ⎤ ⎡⎡ a11 ⎢ ⎢ a21 ⎥ ⎥ ⎢ A=⎢ ⎣ ⎣ .. ⎦ . am1

⎡

⎤ a12 ⎢ a22 ⎥ ⎢ . ⎥ ··· ⎣ . ⎦ .

⎡

⎤⎤ a1n ⎢ a2n ⎥ ⎥ ⎢ . ⎥ ⎥. ⎣ . ⎦⎦ .

am2

amn

We now define the basic matrix operations. Matrix addition and scalar multiplication are really no different from the corresponding operations on vectors (and, moreover, they satisfy essentially the same properties). DEFINITION 6.2 (MATRIX ADDITION) Let A and B be two m × n matrices. Then their matrix sum A + B is the m × n matrix obtained by adding corresponding entries. That is, the entry in the ith row and jth column of A + B is ai j + bi j , where ai j and bi j are the i jth entries of A and B, respectively.

EXAMPLE 2 If  1 A= 4

2 5

3 6



then

 and 

A+B =

8 2

B=

2 10

2 6

7 −2

0 −1 5 0

 ,

 .

 7 1 , then A + B is not defined, since B does not have However, if B = 5 3 the same dimensions as A. ◆ 

1.6

Some n-dimensional Geometry

53

Properties of matrix addition. For all m × n matrices A, B, and C we have 1. A + B = B + A (commutativity); 2. A + (B + C) = (A + B) + C (associativity); 3. An m × n matrix O (the zero matrix) with the property that A + O = A for all m × n matrices A.

DEFINITION 6.3 (SCALAR MULTIPLICATION) If A is an m × n matrix and k ∈ R is any scalar, then the product k A of the scalar k and the matrix A is obtained by multiplying every entry in A by k. That is, the i jth entry of k A is kai j (where ai j is the i jth entry of A).

 EXAMPLE 3 If A =

1 4

2 5

  3 3 , then 3A = 12 6

6 15

 9 . 18

◆

Properties of scalar multiplication. If A and B are any m × n matrices and k and l are any scalars, then 1. (k + l)A = k A + l A (distributivity); 2. k(A + B) = k A + k B (distributivity); 3. k(l A) = (kl)A = l(k A). We leave it to you to supply proofs of these addition and scalar multiplication properties if you wish. Just as defining products of vectors needed to be “unexpected” in order to be useful, so it is with defining products of matrices. To a degree, matrix multiplication is a generalization of the dot product of two vectors. DEFINITION 6.4 (MATRIX MULTIPLICATION) Let A be an m × n matrix and B an n × p matrix. Then the matrix product AB is the m × p matrix whose i jth entry is the dot product of the ith row of A and the jth column of B (considered as vectors in Rn ). That is, the i jth entry of ⎡ ⎤ a11 a12 · · · a1n ⎡ ⎡ ⎤ ⎤ b1 j .. ⎥ b11 · · · ⎢ .. · · · b1 p . ⎥ ⎢ b21 ⎢ . b2 p ⎥ ⎢ b2 j ⎥ ⎢ ⎥ ⎢ . ⎥ ⎢[ai1 ai2 · · · ain ]⎥ ⎢ .. .. ⎥ ⎣ ⎣ ⎦ . ⎢ . ⎥ . . . ⎦ .. ⎦ ⎣ .. · · · · · · . bn1 bn j bnp am1 am2 · · · amn

is ai1 b1 j + ai2 b2 j + · · · + ain bn j = (more compactly)

n  k=1

aik bk j .

54

Chapter 1

Vectors

EXAMPLE 4 If  A=

1 4

2 5

3 6

⎡

 and

0 B=⎣ 7 2

⎤ 1 0 ⎦, 4

then the (2, 1) entry of AB is the dot product of the second row of A and the first column of B: ⎡ ⎤ 0   5 6 · ⎣ 7 ⎦ = (4)(0) + (5)(7) + (6)(2) = 47. (2, 1) entry = 4 2 The full product AB is the 2 × 2 matrix   20 13 . 47 28 On the other hand, B A is the 3 × 3 matrix ⎡ ⎤ 4 5 6 ⎣ 7 14 21 ⎦ . 18 24 30

◆

Order matters in matrix multiplication. To multiply two matrices we must have Number of columns of left matrix = number of rows of right matrix. In Example 4, the products AB and B A are matrices of different dimensions; hence, they could not possibly be the same. A worse situation occurs when the matrix product is defined in one order and not the other. For example, if A is 2 × 3 and B is 3 × 3, then AB is defined (and is a 2 × 3 matrix), but B A is not. However, even if both products AB and B A are defined and of the same dimensions (as is the case if A and B are both n × n, for example), it is in general still true that AB = B A. Despite this negative news, matrix multiplication does behave well in a number of respects, as the following results indicate: Properties of matrix multiplication. Suppose A, B, and C are matrices of appropriate dimensions (meaning that the expressions that follow are all defined) and that k is a scalar. Then 1. 2. 3. 4.

A(BC) = (AB)C; k(AB) = (k A)B = A(k B); A(B + C) = AB + AC; (A + B)C = AC + BC.

The proofs of these properties involve little more than Definition 6.4, although the notation can become somewhat involved, as in the proof of property 1. One simple operation on matrices that has no analogue in the real number system is the transpose. The transpose of an m × n matrix A is the n × m matrix

1.6

Some n-dimensional Geometry

55

A T obtained by writing the rows of A as columns. For example, if ⎡ ⎤   1 4 1 2 3 5 ⎦. , then A T = ⎣ 2 A= 4 5 6 3 6 More abstractly, the i jth entry of A T is a ji , the jith entry of A. The transpose operation turns row vectors into column vectors and vice versa. We also have the following results: (A T )T = A,

for any matrix A.

(3)

(AB)T = B T A T ,

where A is m × n and B is n × p.

(4)

The transpose will largely function as a notational convenience for us. For example, consider a, b ∈ Rn to be column vectors. Then the dot product a · b can be written in matrix form as ⎡ ⎤ b1   ⎢ b2 ⎥ T ⎥ a2 · · · an ⎢ a · b = a1 b1 + a2 b2 + · · · + an bn = a1 ⎣ ... ⎦ = a b. bn EXAMPLE 5 Matrix multiplication is defined the way it is so that, roughly speaking, working with vectors or quantities involving several variables can be made to look as much as possible like working with a single variable. This idea will become clearer throughout the text, but we can provide an important example now. A linear function in a single variable is a function of the form f (x) = ax where a is a constant. The natural generalization of this to higher dimensions is a linear mapping F: Rn → Rm , F(x) = Ax, where A is a (constant) m × n matrix and x ∈ Rn . More explicitly, F is a function that takes a vector in Rn (written as a column vector) and returns a vector in Rm (also written as a column). That is, ⎡ ⎤⎡ ⎤ x1 a11 a12 · · · a1n ⎢ a21 a22 · · · a2n ⎥ ⎢ x2 ⎥ F(x) = Ax = ⎢ . ⎥⎢ . ⎥. .. . . ⎣ ... . .. ⎦ ⎣ .. ⎦ . am1 am2 · · · amn

xn

The function F has the properties that F(x + y) = F(x) + F(y) for all x, y ∈ Rn and F(kx) = kF(x) for all x ∈ Rn , k ∈ R. These properties are also satisfied by f (x) = ax, of course. Perhaps more important, however, is the fact that linear mappings behave nicely with respect to composition. Suppose F is as just defined and G: Rm → R p is another linear mapping defined by G(x) = Bx, where B is a p × m matrix. Then there is a composite function G ◦ F: Rn → R p defined by G ◦ F(x) = G(F(x)) = G(Ax) = B(Ax) = (B A)x by the associativity property of matrix multiplication. Note that B A is defined and is a p × n matrix. Hence, we see that the composition of two linear mappings is again a linear mapping. Part of the reason we defined matrix multiplication the ◆ way we did is so that this is the case. EXAMPLE 6 We saw that by interpreting equation (1) in §1.2 in n dimensions, we obtain parametric equations of a line in Rn . Equation (2) of §1.5, the equation

56

Chapter 1

Vectors

for a plane in R3 through a given point (x0 , y0 , z 0 ) with given normal vector n = Ai + Bj + Ck, can also be generalized to n dimensions: A1 (x1 − b1 ) + A2 (x2 − b2 ) + · · · + An (xn − bn ) = 0. If we let A = (A1 , A2 , . . . , An ), b = (b1 , b2 , . . . , bn ) (“constant” vectors), and x = (x1 , x2 , . . . , xn ) (a “variable” vector), then the aforementioned equation can be rewritten as A · (x − b) = 0 or, considering A, b, and x as n × 1 matrices, as AT (x − b) = 0. This is the equation for a hyperplane in Rn through the point b with normal vector A. The points x that satisfy this equation fill out an (n − 1)-dimensional subset of Rn . ◆ At this point, it is easy to think that matrix arithmetic and the vector geometry of Rn , although elegant, are so abstract and formal as to be of little practical use. However, the next example, from the field of economics,2 shows that this is not the case. EXAMPLE 7 Suppose that we have n commodities. If the price per unit of the ith commodity is pi , then the cost of purchasing xi (> 0) units of commodity i is pi xi . If p = ( p1 , . . . , pn ) is the price vector of all the commodities and x = (x1 , . . . , xn ) is the commodity bundle vector, then p · x = p1 x 1 + p2 x 2 + · · · + pn x n represents the total cost of the commodity bundle. Now suppose that we have an exchange economy, so that we may buy and sell items. If you have an endowment vector w = (w1 , . . . , wn ), where wi is the amount of commodity i that you can sell (trade), then, with prices given by the price vector p, you can afford any commodity bundle x where p · x ≤ p · w. We may rewrite this last equation as p · (x − w) ≤ 0. In other words, you can afford any commodity bundle x in the budget set {x | p · (x − w) ≤ 0}. The equation p · (x − w) = 0 defines a budget hyperplane ◆ passing through w with normal vector p.

Determinants We have already defined determinants of 2 × 2 and 3 × 3 matrices. (See §1.4.) Now we define the determinant of any n × n (square) matrix in terms of determinants of (n − 1) × (n − 1) matrices. By “iterating the definition,” we can calculate any determinant. 2

See D. Saari, “Mathematical complexity of simple economics,” Notices of the American Mathematical Society 42 (1995), no. 2, 222–230.

Some n-dimensional Geometry

1.6

57

Let A = (ai j ) be an n × n matrix. The determinant of A is the real number given by

DEFINITION 6.5

|A| = (−1)1+1 a11 |A11 | + (−1)1+2 a12 |A12 | + · · · + (−1)1+n a1n |A1n |, where Ai j is the (n − 1) × (n − 1) submatrix of A obtained by deleting the ith row and jth column of A. ⎡

1 2 1 1 0 ⎢ −2 EXAMPLE 8 If A = ⎣ 4 2 −1 3 −2 1 ⎡ 1 2 1 1 0 ⎢ −2 A12 = ⎣ 4 2 −1 3 −2 1

⎤ 3 5⎥ , then 0⎦ 1 ⎤ ⎡ 3 −2 0 5⎥ ⎣ 4 −1 = ⎦ 0 3 1 1

⎤ 5 0 ⎦. 1

According to Definition 6.5, ⎤ ⎡ ⎤ ⎡ 1 2 1 3 1 0 5 1 0 5⎥ ⎢ −2 0⎦ = (−1)1+1 (1) det ⎣ 2 −1 det ⎣ 4 2 −1 0⎦ −2 1 1 3 −2 1 1 ⎡ ⎤ −2 0 5 0⎦ + (−1)1+2 (2) det ⎣ 4 −1 3 1 1 ⎡

−2 1 2 + (−1)1+3 (1) det ⎣ 4 3 −2

⎤ 5 0⎦ 1

⎡

⎤ −2 1 0 2 −1 ⎦ + (−1)1+4 (3) det ⎣ 4 3 −2 1 = (1)(1)(−1) + (−1)(2)(37) + (1)(1)(−78) + (−1)(3)(−7) = −132.

◆

The determinant of the submatrix Ai j of A is called the i jth minor of A, and the quantity (−1)i+ j |Ai j | is called the i jth cofactor. Definition 6.5 is known as cofactor expansion of the determinant along the first row, since det A is written as the sum of the products of each entry of the first row and the corresponding cofactor (i.e., the sum of the terms a1 j times (−1)i+ j |Ai j |). It is natural to ask if one can compute determinants by cofactor expansion along other rows or columns of A. Happily, the answer is yes (although we shall not prove this).

58

Chapter 1

Vectors

Convenient Fact. The determinant of A can be computed by cofactor expansion along any row or column. That is, |A| = (−1)i+1 ai1 |Ai1 | + (−1)i+2 ai2 |Ai2 | + · · · + (−1)i+n ain |Ain | (expansion along the ith row), |A| = (−1)1+ j a1 j |A1 j | + (−1)2+ j a2 j |A2 j | + · · · + (−1)n+ j an j |An j | (expansion along the jth column). EXAMPLE 9 To compute the determinant of ⎡ 1 2 0 4 0 0 9 ⎢2 ⎢ 5 1 −1 ⎢7 ⎣0 2 0 0 3 1 0 0

5 0 0 2 0

⎤ ⎥ ⎥ ⎥, ⎦

expansion along the first row involves more calculation than necessary. In particular, one would need to calculate four 4 × 4 determinants on the way to finding the desired 5 × 5 determinant. (To make matters worse, these 4 × 4 determinants would, in turn, need to be expanded also.) However, if we expand along the third column, we find that det A = (−1)1+3 (0) det A13 + (−1)2+3 (0) det A23 + (−1)3+3 (1) det A33 + (−1)4+3 (0) det A43 + (−1)5+3 (0) det A53 = det A33  1 2  0 2 = 2 0 3 1

4 9 0 0

5 0 2 0

    .  

There are several good ways to evaluate this 4 × 4 determinant. We’ll expand about the bottom row:       

1 2 0 3

2 0 2 1

4 9 0 0

5 0 2 0

   2    4+1  = (−1) (3)  0  2 

4 9 0

5 0 2

  1     + (−1)4+2 (1)  2   0 

4 9 0

5 0 2

     

= (−1)(3)(−54) + (1)(1)(2) = 164.

◆

Of course, not all matrices contain well-distributed zeros as in Example 9, so there is by no means always an obvious choice for an expansion that avoids much calculation. Indeed, one does not compute determinants of large matrices by means of cofactor expansion. Instead, certain properties of determinants are used to make hand computations feasible. Since we shall rarely need to consider determinants larger than 3 × 3, we leave such properties and their significance to the exercises. (See, in particular, Exercises 26 and 27.)

59

Exercises

1.6

1.6 Exercises 1. Rewrite in terms of the standard basis for Rn :

(a) (1, 2, 3, . . . , n) (b) (1, 0, −1, 1, 0, −1, . . . , 1, 0, −1) (Assume that n is a multiple of 3.) In Exercises 2– 4 write the given vectors without recourse to standard basis notation. 2. e1 + e2 + · · · + en 3. e1 − 2e2 + 3e3 − 4e4 + · · · + (−1)

15. Suppose that you run a grain farm that produces six n+1

nen

4. e1 + en 5. Calculate the following, where a = (1, 3, 5, . . . ,

2n − 1) and b = (2, −4, 6, . . . , (−1)n+1 2n): (a) a + b

(b) a − b

(d) a

(e) a · b

of 30 caps that can be sold for $10 each, 16 caps that can be sold for $10 each, 20 caps that can be sold for $12 each, and 28 caps that can be sold for $15 each. You suggest swapping half your inventory of each type of T-shirt for half his inventory of each type of baseball cap. Is your friend likely to accept your offer? Why or why not?

(c) −3a

6. Let n be an even number. Verify the triangle in-

equality in Rn for a = (1, 0, 1, 0, . . . , 0) and b = (0, 1, 0, 1, . . . , 1).

7. Verify that the Cauchy–Schwarz inequality holds for

the vectors a = (1, 2, . . . , n) and b = (1, 1, . . . , 1).

8. If a = (1, −1, 7, 3, 2) and b = (2, 5, 0, 9, −1), calcu-

late the projection proja b. 9. Show, for all vectors a, b, c ∈ Rn , that

a − b ≤ a − c + c − b . 10. Prove the Pythagorean theorem. That is, if a, b, and c are vectors in Rn such that a + b = c and a · b = 0, then

a 2 + b 2 = c 2 . Why is this called the Pythagorean theorem? 11. Let a and b be vectors in Rn . Show that if a + b =

a − b , then a and b are orthogonal.

12. Let a and b be vectors in Rn . Show that if a − b >

a + b , then the angle between a and b is obtuse (i.e., more than π/2).

13. Describe “geometrically” the set of points in R5 satis-

fying the equation 2(x1 − 1) + 3(x2 + 2) − 7x3 + x4 − 4 − 5(x5 + 1) = 0. 14. To make some extra money, you decide to print four

types of silk-screened T-shirts that you sell at various prices. You have an inventory of 20 shirts that you can sell for $8 each, 30 shirts that you sell for $10 each, 24 shirts that you sell for $12 each, and 20 shirts that you sell for $15 each. A friend of yours runs a side business selling embroidered baseball caps and has an inventory

types of grain at prices of $200, $250, $300, $375, $450, $500 per ton. (a) If x = (x1 , . . . , x6 ) is the commodity bundle vector (meaning that xi is the number of tons of grain i to be purchased), express the total cost of the commodity bundle as a dot product of two vectors in R6 . (b) A customer has a budget of $100,000 to be used to purchase your grain. Express the set of possible commodity bundle vectors that the customer can afford. Also describe the relevant budget hyperplane in R6 . In Exercises 16–19, calculate the indicated matrix quantities where     1 2 3 −4 9 5 A= , B= , −2 0 1 0 3 0 ⎡

⎤ 1 −1 0 0 7 ⎦, C =⎣ 2 0 3 −2

 D=

16. 3A − 2B

17. AC

18. D B

19. B T D

1 0 2 −3

 .

20. The n × n identity matrix, denoted I or In , is the ma-

trix whose iith entry is 1 and whose other entries are all zero. That is, ⎡ ⎤ 1 0 ··· 0 ⎢ 0 1 ··· 0 ⎥ ⎢ ⎥ In = ⎢ . . .. ⎥ . . .. .. ⎣ .. . ⎦ 0 0 ··· 1 (a) Explicitly write out I2 , I3 , and I4 . (b) The reason I is called the identity matrix is that it behaves as follows: Let A be any m × n matrix. Then i. AIn = A. ii. Im A = A. Prove these results. (Hint: What are the ijth entries of the products in (i) and (ii)?)

60

Chapter 1

Vectors

Evaluate the determinants given in Exercises 21–23.   0 −1 0   7   0 1 3   2 21.   0 2   1 −3  0 5 1 −2    8   15 22.   −7  8      23.    

5 −1 0 2 0 0 0 0 0 0

0 0 1 0 6 −1 1 9

0 0 0 7

      

0 8 11 1 9 7 4 −3 5 0 2 1 0 0 −3

        

24. Prove that a matrix that has a row or a column con-

sisting entirely of zeros has determinant equal to zero. 25. An upper triangular matrix is an n × n matrix whose

entries below the main diagonal are all zero. (Note: The main diagonal is the diagonal going from upper left to lower right.) For example, the matrix ⎡ ⎤ 1 2 −1 2 3 4 3 ⎥ ⎢ 0 ⎣ 0 0 5 6 ⎦ 0 0 0 7 is upper triangular. (a) Give an analogous definition for a lower triangular matrix and also an example of one. (b) Use cofactor expansion to show that the determinant of any n × n upper or lower triangular matrix A is the product of the entries on the main diagonal. That is, det A = a11 a22 · · · ann .

Step 1. Exchange rows 1 and 2 (this entry in the upper left corner): ⎡ ⎤ ⎡ 0 2 3 1 ⎣ 1 7 −2 ⎦ −→ ⎣ 0 1 1 5 9

into one in upper triangular form in three steps:

⎤ 7 −2 2 3 ⎦. 5 9

Step 2. Add −1 times row 1 to row 3 (this eliminates the nonzero entries below the entry in the upper left corner): ⎡ ⎤ ⎡ ⎤ 1 7 −2 1 7 −2 ⎣ 0 2 3 ⎦ −→ ⎣ 0 2 3 ⎦. 1 5 9 0 −2 11 Step 3. Add row 2 to row 3: ⎡ ⎤ 1 7 −2 ⎣ 0 2 3 ⎦ −→ 0 −2 11

⎡

1 ⎣ 0 0

⎤ 7 −2 2 3 ⎦. 0 14

The question is, how do these operations affect the determinant? (a) By means of examples, make a conjecture as to the effect of a row operation of type I on the determinant. (That is, if matrix B results from matrix A by performing a single row operation of type I, how are det A and det B related?) You need not prove your results are correct. (b) Repeat part (a) in the case of a row operation of type III. (c) Prove that if B results from A by multiplying the entries in the ith row of A by the scalar c (a type II operation), then det B = c · det A. 27. Calculate the determinant of the matrix

⎡

⎢ ⎢ A=⎢ ⎣

26. Some properties of the determinant. Exercises 24

and 25 show that it is not difficult to compute determinants of even large matrices, provided that the matrices have a nice form. The following operations (called elementary row operations) can be used to transform an n × n matrix into one in upper triangular form: I. Exchange rows i and j. II. Multiply row i by a nonzero scalar. III. Add a multiple of row i to row j. (Row i remains unchanged.) For example, one can transform the matrix ⎡ ⎤ 0 2 3 ⎣ 1 7 −2 ⎦ 1 5 9

puts a nonzero

2 1 −1 0 −3

⎤ 1 −2 7 8 0 1 −2 4 ⎥ ⎥ 1 2 3 −5 ⎥ 2 3 1 7 ⎦ 2 −1 0 1

by using row operations to transform A into a matrix in upper triangular form and by using the results of Exercise 26 to keep track of how the determinant of A and the determinant of your final matrix are related. 28. (a) Is det(A + B) = det A + det B? Why or why not?

(b) Calculate

and

  1   3+2   0

  1 2   3 1   0 −2

2 1−1 −2 7 5 0

7 5+1 0

    1 2    +  2 −1     0 −2

and compare your results.

      7 1 0

   ,  

1.6

(c) Calculate

and

  1   0   −1

  1   0   −1

35. (a) Show that if A is invertible, then det A = 0. (In

 3 2 + 3  4 −1 + 5  0 0−2 

  3 2   1 4 −1  +  0 0 0   −1

fact, the converse is also true.) (b) Show that if A is invertible, 1 det(A−1 ) = . det A

3 3 4 5 0 −2

   ,  

and compare your results. (d) Conjecture and prove a result about sums of determinants. (You may wish to construct further examples such as those in parts (b) and (c).) 29. It is a fact that, if A and B are any n × n matrices, then

det(AB) = (det A)(det B). Use this fact to show that det(AB) = det(B A). (Recall that AB = B A, in general.) An n × n matrix A is said to be invertible (or nonsingular) if there is another n × n matrix B with the property that AB = B A = In , where In denotes the n × n identity matrix. (See Exercise 20.) The matrix B is called an inverse to the matrix A. Exercises 30–38 concern various aspects of matrices and their inverses.     1 0 1 0 30. (a) Verify that is an inverse of . 1 1 −1 1 ⎡ (b) Verify that ⎡

−40 16 ⎣ 13 −5 5 −2

⎤ 1 2 3 ⎣ 2 5 3 ⎦ is an inverse of 1 0 8 ⎤ 9 −3 ⎦. −1

2 inverse to ⎣ 0 0

⎤ 2 1 1 0 ⎦. 0 −1

⎡

0 32. Try to find an inverse matrix to ⎣ 0 0 What happens?

⎤ 2 1 1 0 ⎦. 0 −1

33. Show that if an n × n matrix A is invertible, then A can

have only one inverse matrix. Thus, we may write A−1 to denote the unique inverse of a nonsingular matrix A. (Hint: Suppose A were to have two inverses B and C. Consider B(AC).)

34. Suppose that A and B are n × n invertible matrices.

Show that the product matrix AB is invertible by verifying that its inverse (AB)−1 = B −1 A−1 .

then

36. (a) Show that,  if ad − bc = 0, then a general 2 × 2

matrix 1 ad − bc

a b c d 

has the matrix

d −b −c a



 =

d ad−bc c − ad−bc

as inverse.

b − ad−bc

(b) Use this formula to find an inverse of



a ad−bc



 2 4 . −1 2

37. If A is a 3 × 3 matrix and det A = 0, then there is

a (somewhat complicated) formula for A−1 . In particular,

A−1

⎤ ⎡ |A31 | 1 ⎣ |A11 | −|A21 | |A22 | −|A32 | ⎦ , −|A12 | = det A |A13 | −|A23 | |A33 |

where Ai j denotes the submatrix of A obtained by deleting the ith row and jth column (see Definition 6.5). Use this formula to find the inverse of ⎤ 2 1 1 A = ⎣ 0 2 4 ⎦. 1 0 3 ⎡

More generally, if A is any n × n matrix and det A = 0, then A−1 =

31. Using the definition of an inverse matrix, find an

⎡

61

Exercises

1 adj A, det A

where adj A is the adjoint matrix of A, that is, the matrix whose i jth entry is (−1)i+ j |A ji |. (Note: The formula for the inverse matrix using the adjoint is typically more of theoretical than practical interest, as there are more efficient computational methods to determine the inverse, when it exists.) 38. Repeat Exercise 37 with the matrix

⎡

⎤ 2 −1 3 2 −2 ⎦ . A=⎣ 1 3 0 1 Cross products in Rn . Although it is not possible to define a cross product of two vectors in Rn as we did for two vectors in R3 , we can construct a “cross product” of n − 1 vectors in Rn that behaves analogously to the three-dimensional cross

62

Chapter 1

Vectors

product. To be specific, if a1 = (a11 , a12 , . . . , a1n ),

a2 = (a21 , a22 , . . . , a2n ), . . . ,

an−1 = (an−11 , an−12 , . . . , an−1n ) are n − 1 vectors in Rn , we define a1 × a2 × · · · × an−1 to be the vector in Rn given by the symbolic determinant    e1 e2 ··· en    a11 a12 ··· a1n    a21 a · · · a2n . 22 a1 × a2 × · · · × an−1 =   .. .. ..  . ..  . . .    an−1 1 an−1 2 ··· an−1 n  (Here e1 , . . . , en are the standard basis vectors for R .) Exercises 39–42 concern this generalized notion of cross product. n

39. Calculate the following cross product in R4 :

(1, 2, −1, 3) × (0, 2, −3, 1) × (−5, 1, 6, 0). 40. Use the results of Exercises 26 and 28 to show that

(a) a1 × · · · × ai × · · · × a j × · · · × an−1 = − (a1 × · · · × a j × · · · × ai × · · · × an−1 ), 1 ≤ i ≤ n − 1, 1 ≤ j ≤ n − 1 (b) a1 × · · · × kai × · · · × an−1 = k(a1 × · · · × ai × · · · × an−1 ), 1 ≤ i ≤ n − 1.

y

1.7 P

x

(c) a1 × · · · × (ai + b) × · · · × an−1 = a1 × · · · × ai × · · · × an−1 + a1 × · · · × b × · · · × an−1 , 1 ≤ i ≤ n − 1, all b ∈ Rn . (d) Show that if b = (b1 , . . . , bn ) is any vector in Rn , then b · (a1 × a2 × · · · × an−1 ) is given by the determinant   b1   a11   ..  .   an−11

··· ··· ···

bn a1n .. . an−1 n

     .   

41. Show that the vector b = a1 × a2 × · · · × an−1 is

orthogonal to a1 , . . . , an−1 .

42. Use the generalized notion of cross products to

find an equation of the (four-dimensional) hyperplane in R5 through the five points P0 (1, 0, 3, 0, 4), P1 (2, −1, 0, 0, 5), P2 (7, 0, 0, 2, 0), P3 (2, 0, 3, 0, 4), and P4 (1, −1, 3, 0, 4).

New Coordinate Systems

We hope that you are comfortable with Cartesian (rectangular) coordinates for R2 or R3 . The Cartesian coordinate system will continue to be of prime importance to us, but from time to time, we will find it advantageous to use different coordinate systems. In R2 , polar coordinates are useful for describing figures with circular symmetry. In R3 , there are two particularly valuable coordinate systems besides Cartesian coordinates: cylindrical and spherical coordinates. As we shall see, cylindrical and spherical coordinates are each a way of adapting polar coordinates in the plane for use in three dimensions.

Cartesian and Polar Coordinates on R2 Figure 1.82 The Cartesian You can understand the Cartesian (or rectangular) coordinates (x, y) of a point coordinate system. P in R2 in the following way: Imagine the entire plane filled with horizontal and vertical lines, as in Figure 1.82. Then the point P lies on exactly one vertical line y and one horizontal line. The x-coordinate of P is where this vertical line intersects the x-axis, and the y-coordinate is where the horizontal line intersects the y-axis. Location y (See Figure 1.83.) (Of course, we’ve already assigned coordinates along the axes P (x, y) so that the zero point of each axis is at the point of intersection of the axes. We also normally mark off the same unit distance on each axis.) Note that, because x of this geometry, every point in R2 has a uniquely determined set of Cartesian coordinates. Location x Polar coordinates are defined by considering different geometric information. Now imagine the plane filled with concentric circles centered at the origin Figure 1.83 Locating a point P, using Cartesian coordinates. and rays emanating from the origin. Then every point except the origin lies on

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63

P P r

θ

(|r|, θ) θ

(−|r|, θ) Figure 1.86 Locating the point with polar coordinates (r, θ ), where r < 0.

Figure 1.84 The polar coordinate

Figure 1.85 Locating a point P,

system.

using polar coordinates.

exactly one such circle and one such ray. The origin itself is special: No circle passes through it, and all the rays begin at it. (See Figure 1.84.) For points P other than the origin, we assign to P the polar coordinates (r, θ ), where r is the radius of the circle on which P lies and θ is the angle between the positive x-axis and the ray on which P lies. (θ is measured as opening counterclockwise.) The origin is an exception: It is assigned the polar coordinates (0, θ), where θ can be any angle. (See Figure 1.85.) As we have described polar coordinates, r ≥ 0 since r is the radius of a circle. It also makes good sense to require 0 ≤ θ < 2π , for then every point in the plane, except the origin, has a uniquely determined pair of polar coordinates. Occasionally, however, it is useful not to restrict r to be nonnegative and θ to be between 0 and 2π . In such a case, no point of R2 will be described by a unique pair of polar coordinates: If P has polar coordinates (r, θ), then it also has (r, θ + 2nπ) and (−r, θ + (2n + 1)π) as coordinates, where n can be any integer. (To locate the point having coordinates (r, θ ), where r < 0, construct the ray making angle θ with respect to the positive x-axis, and instead of marching |r | units away from the origin along this ray, go |r | units in the opposite direction, as shown in Figure 1.86.) EXAMPLE 1 Polar coordinates may already be familiar to you. Nonetheless, make sure you understand that the points pictured in Figure 1.87 have the coor◆ dinates indicated.

θ

r = 6 cos θ

0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4

6 √ 3√3 3 2 3 0 −3 √ −3√2 −3 3 −6 √ −3√3 −3 2 −3 0 3 √ 3 2

(3√2, ⎯ π /4) (2, 5π /6)

(3√3, ⎯ π /6)

(2, π/6) (6, 0)

(5, 0)

(3, 3π /2)

(−1, 5π /6) or (1, 11π/6) or (1, −π/6)

Figure 1.87 Figure for

Example 1.

Figure 1.88 The graph of r = 6 cos θ in Example 2.

EXAMPLE 2 Let’s graph the curve given by the polar equation r = 6 cos θ (Figure 1.88). We can begin to get a feeling for the graph by compiling values, as in the adjacent tabulation.

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Thus, r decreases from 6 to 0 as θ increases from 0 to π/2; r decreases from 0 to −6 (or is not defined, if you take r to be nonnegative) as θ varies from π/2 to π ; r increases from −6 to 0 as θ varies from π to 3π/2; and r increases from 0 to 6 as θ varies from 3π/2 to 2π . To graph the resulting curve, imagine a radar screen: As θ moves counterclockwise from 0 to 2π, the point (r, θ) of the graph is traced as the appropriate “blip” on the radar screen. Note that the curve is actually traced twice: once as θ varies from 0 to π and then again as θ varies from π to 2π . Alternatively, the curve is traced just once if we allow only θ values that yield nonnegative r values. The resulting graph appears to be a circle of radius 3 (not centered at the origin), and, in fact, one can see (as in Example 3) that the graph is indeed such a circle. ◆ The basic conversions between polar and Cartesian coordinates are provided by the following relations:  x = r cos θ ; (1) Polar to Cartesian: y = r sin θ  2 r = x 2 + y2 Cartesian to polar: . (2) tan θ = y/x Note that the equations in (2) do not uniquely determine r and θ in terms of x and y. This is quite acceptable, really, since we do not always want to insist that r be nonnegative and θ be between 0 and 2π. If we do restrict r and θ, however, then they are given in terms of x and y by the following formulas: r=



x 2 + y2,

⎧ −1 tan y/x ⎪ ⎪ ⎪ ⎪ ⎪ tan−1 y/x + 2π ⎪ ⎪ ⎪ ⎨tan−1 y/x + π θ= ⎪π/2 ⎪ ⎪ ⎪ ⎪ ⎪3π/2 ⎪ ⎪ ⎩ indeterminate

if x if x if x if x if x if x

> 0, y ≥ 0 > 0, y < 0 < 0, y ≥ 0 . = 0, y > 0 = 0, y < 0 =y=0

The complicated formula for θ arises because we require 0 ≤ θ < 2π , while the inverse tangent function returns values between −π/2 and π/2 only. Now you see why the equations given in (2) are a better bet! EXAMPLE 3 We can use the formulas in (1) and (2) to prove that the curve in Example 2 really is a circle. The polar equation r = 6 cos θ that defines the curve requires a little ingenuity to convert to the corresponding Cartesian equation. The trick is to multiply both sides of the equation by r . Doing so, we obtain r 2 = 6r cos θ. Now (1) and (2) immediately give x 2 + y 2 = 6x.

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We complete the square in x to find that this equation can be rewritten as (x − 3)2 + y 2 = 9, ◆

which is indeed a circle of radius 3 with center at (3, 0). y

x Figure 1.89 The cylindrical coordinate system.

P(r, θ , z)

Cylindrical Coordinates Cylindrical coordinates on R3 are a “naive” way of generalizing polar coordinates to three dimensions, in the sense that they are nothing more than polar coordinates used in place of the x- and y-coordinates. (The z-coordinate is left unchanged.) The geometry is as follows: Except for the z-axis, fill all of space with infinitely extended circular cylinders with axes along the z-axis as in Figure 1.89. Then any point P in R3 not lying on the z-axis lies on exactly one such cylinder. Hence, to locate such a point, it’s enough to give the radius of the cylinder, the circumferential angle θ around the cylinder, and the vertical position z along the cylinder. The cylindrical coordinates of P are (r, θ, z), as shown in Figure 1.90. Algebraically, the equations in (1) and (2) can be extended to produce the basic conversions between Cartesian and cylindrical coordinates.

z θ

r

Figure 1.90 Locating a point P, using cylindrical coordinates.

The basic conversions between cylindrical and Cartesian coordinates are provided by the following relations: ⎧ ⎪ ⎨ x = r cos θ y = r sin θ ; Cylindrical to Cartesian: (3) ⎪ ⎩z = z

Cartesian to cylindrical:

r0

⎧ 2 2 2 ⎪ ⎨r = x + y tan θ = y/x . ⎪ ⎩z = z

(4)

As with polar coordinates, if we make the restrictions r ≥ 0, 0 ≤ θ < 2π , then all points of R3 except the z-axis have a unique set of cylindrical coordinates. A point on the z-axis with Cartesian coordinates (0, 0, z 0 ) has cylindrical coordinates (0, θ, z 0 ), where θ can be any angle. Cylindrical coordinates are useful for studying objects possessing an axis of symmetry. Before exploring a few examples, let’s understand the three “constant coordinate” surfaces.

Figure 1.91 The graph of the cylindrical equation r = r0 .

• The r = r0 surface is, of course, just a cylinder of radius r0 with axis the z-axis. (See Figure 1.91.) • The θ = θ0 surface is a vertical plane containing the z-axis (or a half-plane with edge the z-axis if we take r ≥ 0 only). (See Figure 1.92.) • The z = z 0 surface is a horizontal plane. (See Figure 1.93.)

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Half-plane only if r ≥ 0

z

θ0

z=c

x

z = z0

Figure 1.92 The graph of θ = θ0 .

Figure 1.93 The graph of

z = z0 .

EXAMPLE 4 Graph the surface having cylindrical equation r = 6 cos θ. (This equation is identical to the one in Example 2.) In particular, z does not appear in this equation. What this means is that if the surface is sliced by the horizontal plane z = c where c is a constant, we will see the circle shown in Example 2, no matter what c is. If we stack these circular sections, then the entire surface is a circular cylinder of radius 3 with axis parallel to the z-axis (and through the point (3, 0, 0) in cylindrical coordinates). This surface is shown in Figure 1.94. ◆

y

Figure 1.94 The graph of r = 6 cos θ in cylindrical coordinates.

z θ =c

y x

EXAMPLE 5 Graph the surface having equation z = 2r in cylindrical coordinates. Here the variable θ does not appear in the equation, which means that the surface in question will be circularly symmetric about the z-axis. In other words, if we slice the surface by any plane of the form θ = constant (or half-plane, if we take r ≥ 0), we see the same curve, namely, a line (respectively, a half-line) of slope 2. As we let the constant-θ plane vary, this line generates a cone, as shown in Figure 1.95. The cone consists only of the top half (nappe) when we restrict r to be nonnegative. The Cartesian equation of this cone is readily determined. Using the formulas in (4), we have z = 2r

=⇒

z 2 = 4r 2

⇐⇒

z 2 = 4(x 2 + y 2 ).

Since z can be positive as well as negative, this last Cartesian equation describes the cone  with both nappes. If we want the topnappe only, then the equation z = 2 x 2 + y 2 describes it. Similarly, z = −2 x 2 + y 2 describes the bottom nappe. ◆ Figure 1.95 The graph of z = 2r in cylindrical coordinates.

Spherical Coordinates Fill all of space with spheres centered at the origin as in Figure 1.96. Then every point P ∈ R3 , except the origin, lies on a single such sphere. Roughly speaking, the spherical coordinates of P are given by specifying the radius ρ of the sphere containing P and the “latitude and longitude” readings of P along this sphere. More precisely, the spherical coordinates (ρ, ϕ, θ ) of P are defined as follows: ρ is the distance from P to the origin; ϕ is the angle between the positive z-axis and the ray through the origin and P; and θ is the angle between the positive x-axis and the ray made by dropping a perpendicular from P to the x y-plane. (See Figure 1.97.) The θ-coordinate is exactly the same as the θ -coordinate used in cylindrical coordinates. (Warning: Physicists usually prefer to reverse the roles of ϕ and θ, as do some graphical software packages.)

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New Coordinate Systems

z

ϕ

P

ρ

y

θ

x Figure 1.96 The spherical

Figure 1.97 Locating the point

coordinate system.

P, using spherical coordinates.

It is standard practice to impose the following restrictions on the range of values for the individual coordinates: ρ ≥ 0,

0 ≤ ϕ ≤ π,

0 ≤ θ < 2π.

(5)

3

With such restrictions, all points of R , except those on the z-axis, have a uniquely determined set of spherical coordinates. Points along the z-axis, except for the origin, have coordinates of the form (ρ0 , 0, θ ) or (ρ0 , π, θ), where ρ0 is a positive constant and θ is arbitrary. The origin has spherical coordinates (0, ϕ, θ), where both ϕ and θ are arbitrary. EXAMPLE 6 Several points and their corresponding spherical coordinates are ◆ shown in Figure 1.98.

π /4

(2, π /4, π /2)

ϕ 0 < π/2

(1, π/4, 0) (2, π/2, π /2)

ϕ 0 = π/2

(2, π, π /4) or (2, π, π /3) or (−2, 0, 0) Figure 1.98 Figure for Example 6.

ϕ 0 > π/2

Figure 1.99 The graph of

ρ = ρ0 (> 0).

Figure 1.100 The spherical surface ϕ = ϕ0 , shown for different values of ϕ0 .

Spherical coordinates are especially useful for describing objects that have a center of symmetry. With the restrictions given by the inequalities in (5), the constant coordinate surface ρ = ρ0 (ρ0 > 0) is, of course, a sphere of radius ρ0 , as shown in Figure 1.99. The surface given by θ = θ0 is a half-plane just as in the cylindrical case. The ϕ = ϕ0 surface is a single-nappe cone if ϕ0 = π/2 and is the x y-plane if ϕ0 = π/2 (and is the positive or negative z-axis if ϕ0 = 0 or π). (See Figure 1.100.) If we do not insist that ρ be nonnegative, then the cones would include both nappes.

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The basic equations relating spherical coordinates to both cylindrical and Cartesian coordinates are as follows. Spherical/cylindrical: ⎧ ⎨ r = ρ sin ϕ θ =θ ⎩ z = ρ cos ϕ Spherical/Cartesian: ⎧ ⎨ x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ ⎩ z = ρ cos ϕ

(6)

⎧ 2 2 + y2 + z2 ⎨ρ = x  tan ϕ = x 2 + y 2 /z . ⎩ tan θ = y/x

(7)

ϕ

ρ

ϕ

r π /2 − ϕ

θ

⎧ ⎨ ρ2 = r 2 + z2 tan ϕ = r/z . ⎩θ = θ

z

r

Figure 1.101 Converting spherical to cylindrical coordinates when 0 < ϕ < π2 .

ϕ − π /2

ρ

z(< 0)

Figure 1.102 Converting spherical to cylindrical coordinates when π/2 < ϕ < π.

Using basic trigonometry, it is not difficult to establish the conversions in (6). From the right triangle shown in Figure 1.101, we have π

r −ϕ = . cos 2 ρ Hence,

π − ϕ = ρ sin ϕ. r = ρ cos 2 Similarly, π

z sin −ϕ = , 2 ρ so that π

− ϕ = ρ cos ϕ. z = ρ sin 2 Thus, the formulas in (6) follow when 0 ≤ ϕ ≤ π/2. If π/2 < ϕ ≤ π, then we may employ Figure 1.102. So π r = ρ cos ϕ − = ρ sin ϕ, 2 and π

π z = −ρ sin ϕ − = ρ sin − ϕ = ρ cos ϕ. 2 2

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69

Hence, the relations in (6) hold in general. The equations in (7) follow by substitution of those in (6) into those of (3) and (4). EXAMPLE 7 The cylindrical equation z = 2r in Example 5 converts via (6) to the spherical equation ρ cos ϕ = 2ρ sin ϕ. Therefore, 1 1 ⇐⇒ ϕ = tan−1 ≈ 26◦ . 2 2 Thus, the equation defines a cone (as we just saw). The spherical equation is especially simple in that it involves just a single coordinate. ◆ tan ϕ =

EXAMPLE 8 Not all spherical equations are improvements over their cylindrical or Cartesian counterparts. For example, the Cartesian equation 6x = x 2 + y 2 (whose polar–cylindrical equivalent is r = 6 cos θ) becomes 6ρ sin ϕ cos θ = ρ 2 sin2 ϕ cos2 θ + ρ 2 sin2 ϕ sin2 θ from (7). Simplifying, ⇐⇒ ⇐⇒

6ρ sin ϕ cos θ = ρ 2 sin2 ϕ (cos2 θ + sin2 θ ) 6ρ sin ϕ cos θ = ρ 2 sin2 ϕ 6 cos θ = ρ sin ϕ.

ϕ

ρ = 2a cos ϕ

This spherical equation is more complicated than the original Cartesian equation in that all three spherical coordinates are involved. Therefore, it is not at all obvious that the spherical equation describes a cylinder. ◆

0 π/6 π/4 π/3 π/2 2π/3 3π/4 π

√2a √3a 2a a 0 −a √ − 2a −2a

EXAMPLE 9 Let’s graph the surface with spherical equation ρ = 2a cos ϕ, where a > 0. As with the graph of the cone with cylindrical equation z = 2r , note that the equation is independent of θ. Thus, all sections of this surface made by slicing with the half-plane θ = c must be the same. If we compile values as in the adjacent table, then the section of the surface in the half-plane θ = 0 is as shown in Figure 1.103. Since this section must be identical in all other constant-θ half-planes, we see that this surface appears to be a sphere of radius a tangent to the x y-plane, which is shown in Figure 1.104.

θ =0

(a, π /3, 0)

(2a, 0, 0) Section of ρ = 2a cos ϕ

θ = π /2 θ =0

Figure 1.103 The cross section of ρ = 2a cos ϕ in the half-plane θ = 0.

θ = π /3 θ = π /4

Figure 1.104 The graph of ρ =

2a cos ϕ.

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The Cartesian equation of the surface is determined by multiplying both sides of the spherical equation by ρ and using the conversion equations in (7): ρ = 2a cos ϕ =⇒ ρ 2 = 2aρ cos ϕ ⇐⇒ x 2 + y 2 + z 2 = 2az ⇐⇒ x 2 + y 2 + (z − a)2 = a 2 by completing the square in z. This last equation can be recognized as that of a sphere of radius a with center at (0, 0, a) in Cartesian coordinates. ◆ EXAMPLE 10 NASA launches a 10-ft-diameter space probe. Unfortunately, a meteor storm pushes the probe off course, and it is partially embedded in the surface of Venus, to a depth of one quarter of its diameter. To attempt to reprogram the probe’s on-board computer to remove it from Venus, it is necessary to describe the embedded portion of the probe in spherical coordinates. Let us find the description desired, assuming that the surface of Venus is essentially flat in relation to the probe and that the origin of our coordinate system is at the center of the probe. z

Probe

10' y α

Surface of Venus

10/4'

5 5/2

Figure 1.105 The space probe of Example 10.

Figure 1.106 A slice of the probe

of Example 10.

The situation is illustrated in Figure 1.105. The buried part of the probe clearly has symmetry about the z-axis. That is, any slice by the half-plane θ = constant looks the same as any other. Thus, θ can vary between 0 and 2π. A typical slice of the probe is shown in Figure 1.106. Elementary trigonometry indicates that for the angle α in Figure 1.106, cos α =

z

cos−1 12

y z = −5/2

Figure 1.107 Coordinate view of

the cross section of the probe of Example 10.

5 2

5

=

1 . 2

Hence, α = = π/3. Thus, the spherical angle ϕ (which opens from the positive z-axis) varies from π − π/3 = 2π/3 to π as it generates the buried part of the probe. Finally, note that for a given value of ϕ between 2π/3 and π , ρ is bounded by the surface of Venus (the plane z = − 52 in Cartesian coordinates) and the spherical surface of the probe (whose equation in spherical coordinates is ρ = 5). See Figure 1.107. From the formulas in (7) the equation z = − 52 corresponds to the spherical equation ρ cos ϕ = − 52 or, equivalently, to ρ = − 52 sec ϕ. Therefore, the embedded part of the probe may be defined by the set    5 2π ≤ ϕ ≤ π, 0 ≤ θ < 2π . (ρ, ϕ, θ)  − sec ϕ ≤ ρ ≤ 5, ◆ 2 3

1.7

ez

eθ er

xi + yj Figure 1.108 The standard basis

vectors for the cylindrical coordinate system.

New Coordinate Systems

71

Standard Bases for Cylindrical and Spherical Coordinates In Cartesian coordinates, there are three special unit vectors i, j, and k that point in the directions of increasing x-, y-, and z-coordinate, respectively. We find corresponding sets of vectors for cylindrical and spherical coordinates. That is, in each set of coordinates, we seek mutually orthogonal unit vectors that point in the directions of increasing coordinate values. In cylindrical coordinates, the situation is as shown in Figure 1.108. The vectors er , eθ , and ez , which form the standard basis for cylindrical coordinates, are unit vectors that each point in the direction in which only the coordinate indicated by the subscript increases. There is an important difference between the standard basis vectors in Cartesian and cylindrical coordinates. In the former case, i, j, and k do not vary from point to point. However, the vectors er and eθ do change as we move from point to point. Now we give expressions for er , eθ , and ez . Since the cylindrical z-coordinate is the same as the Cartesian z-coordinate, we must have ez = k. The vector er must point radially outward from the z-axis with no k-component. At a point (x, y, z) ∈ R3 (Cartesian coordinates), the vector xi + yj has this property. Normalizing it to obtain a unit vector (see Proposition 3.4 of §1.3), we obtain xi + yj er =  . x 2 + y2 With er and ez in hand, it’s now a simple matter to define eθ , since it must be perpendicular to both er and ez . We take −yi + xj . eθ = ez × er =  x 2 + y2 (The reason for this choice of cross product, as opposed to er × ez , is so that eθ points in the direction of increasing θ.) To summarize, and using the cylindrical to Cartesian conversions given in (3),

xi + yj er =  = cos θ i + sin θ j; x 2 + y2

eρ

−yi + xj = − sin θ i + cos θ j; eθ =  x 2 + y2

eθ eϕ

(8)

ez = k.

xi + yj + zk

Figure 1.109 The standard basis

vectors for the spherical coordinate system.

In spherical coordinates, the situation is shown in Figure 1.109. In particular, there are three unit vectors eρ , eϕ , and eθ that form the standard basis for spherical coordinates. These vectors all change direction as we move from point to point. We give expressions for eρ , eϕ , and eθ . Since the θ -coordinates in both spherical and cylindrical coordinates mean the same thing, eθ in spherical coordinates is given by the value of eθ in (8). At a point (x, y, z), the vector eρ should point

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from the origin directly to (x, y, z). Thus, eρ may be obtained by normalizing xi + yj + zk. Finally, eϕ is nothing more than eθ × eρ . If we explicitly perform the calculations just described and make use of the conversion formulas in (7), the following are obtained: xi + yj + zk eρ =  = sin ϕ cos θ i + sin ϕ sin θ j + cos ϕ k; x 2 + y2 + z2 x zi + yzj − (x 2 + y 2 )k  eϕ =  x 2 + y2 x 2 + y2 + z2

(9)

= cos ϕ cos θ i + cos ϕ sin θ j − sin ϕ k; −yi + xj eθ =  = − sin θ i + cos θ j. x 2 + y2

Although the results of (8) and (9) will not be used frequently, they will prove helpful on occasion.

Hyperspherical Coordinates (optional) There is a way to provide a set of coordinates for Rn that generalizes spherical coordinates on R3 . For n ≥ 3, the hyperspherical coordinates of a point P ∈ Rn are (ρ, ϕ1 , ϕ2 , . . . , ϕn−1 ) and are defined by their relations with the Cartesian coordinates (x1 , x2 , . . . , xn ) of P as ⎧ x1 = ρ sin ϕ1 sin ϕ2 · · · sin ϕn−2 cos ϕn−1 ⎪ ⎪ ⎪ ⎪ ⎪ x2 = ρ sin ϕ1 sin ϕ2 · · · sin ϕn−2 sin ϕn−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ x3 = ρ sin ϕ1 sin ϕ2 · · · sin ϕn−3 cos ϕn−2 . (10) ⎪ x4 = ρ sin ϕ1 sin ϕ2 · · · sin ϕn−4 cos ϕn−3 ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ .. ⎪ ⎩ xn = ρ cos ϕ1 To be more explicit, in equation (10) above we take xk = ρ sin ϕ1 sin ϕ2 · · · sin ϕn−k cos ϕn−k+1

for k = 3, . . . , n.

Note that when n = 3, the relations in (10) become ⎧ x = ρ sin ϕ1 cos ϕ2 ⎪ ⎨ 1 x2 = ρ sin ϕ1 sin ϕ2 . ⎪ ⎩ x3 = ρ cos ϕ1 These relations are the same as those given in (7), so hyperspherical coordinates are indeed the same as spherical coordinates when n = 3. In analogy with (5), it is standard practice to impose the following restrictions on the range of values for the coordinates: ρ ≥ 0,

0 ≤ ϕk ≤ π for k = 1, . . . , n − 2,

0 ≤ ϕn−1 < 2π.

(11)

1.7

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73

Then, with these restrictions, we can convert from hyperspherical coordinates to Cartesian coordinates by means of the following formulas: ⎧ 2 ρ = x12 + x22 + · · · + xn2 ⎪ ⎪ ⎪ ! ⎪ ⎪ ⎪ 2 ⎪ tan ϕ = x12 + · · · + xn−1 /xn ⎪ 1 ⎪ ! ⎪ ⎪ ⎨ 2 tan ϕ2 = x12 + · · · + xn−2 /xn−1 . (12) ⎪ . ⎪ .. ⎪ ⎪ ⎪ ! ⎪ ⎪ ⎪ ⎪ tan ϕ = x12 + x22 /x3 ⎪ n−2 ⎪ ⎩ tan ϕn−1 = x2 /x1 Hyperspherical coordinates get their name from the fact that the (n − 1)dimensional hypersurface in Rn defined by the equation ρ = ρ0 , where ρ0 is a positive constant, consists of points on the hypersphere of radius ρ0 defined in Cartesian coordinates by the equation x12 + x22 + · · · + xn2 = ρ02 .

1.7 Exercises In Exercises 1–3, find the Cartesian coordinates of the points whose polar coordinates are given. √ 1. ( 2, π/4) √ 2. ( 3, 5π/6) 3. (3, 0)

In Exercises 4–6, give a set of polar coordinates for the point whose Cartesian coordinates are given. √ 4. (2 3, 2) 5. (−2, 2) 6. (−1, −2)

In Exercises 7–9, find the Cartesian coordinates of the points whose cylindrical coordinates are given. 7. (2, 2, 2) 8. (π, π/2, 1) 9. (1, 2π/3, −2)

In Exercises 10–13, find the rectangular coordinates of the points whose spherical coordinates are given. 10. (4, π/2, π/3) 11. (3, π/3, π/2) 12. (1, 3π/4, 2π/3) 13. (2, π, π/4)

In Exercises 14–16, find a set of cylindrical coordinates of the point whose Cartesian coordinates are given. 14. (−1, 0, 2) 15. (−1,

√

3, 13)

16. (5, 6, 3)

In Exercises 17 and 18, find a set of spherical coordinates of the point whose Cartesian coordinates are given. √ 17. (1, −1, 6) √ 18. (0, 3, 1) 19. This problem concerns the surface described by the

equation (r − 2)2 + z 2 = 1 in cylindrical coordinates. (Assume r ≥ 0.) (a) Sketch the intersection of this surface with the halfplane θ = π/2. (b) Sketch the entire surface.

20. (a) Graph the curve in R2 having polar equation r =

2a sin θ , where a is a positive constant. (b) Graph the surface in R3 having spherical equation ρ = 2a sin ϕ.

21. Graph the surface whose spherical equation is ρ =

1 − cos ϕ.

22. Graph the surface whose spherical equation is ρ =

1 − sin ϕ.

In Exercises 23–25, translate the following equations from the given coordinate system (i.e., Cartesian, cylindrical, or

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spherical) into equations in each of the other two systems. In addition, identify the surfaces so described by providing appropriate sketches. 23. ρ sin ϕ sin θ = 2 24. z = 2x + 2y 2

2

40. Use the formulas in (8) to express i, j, k in terms of er ,

eθ , and ez . 41. Use the formulas in (9) to express i, j, k in terms of eρ ,

eϕ , and eθ .

2

42. Consider the solid in R3 shown in Figure 1.110.

25. r = 0

In Exercises 26–29, sketch the solid whose cylindrical coordinates (r, θ, z) satisfy the given inequalities. 26. 0 ≤ r ≤ 3,

0 ≤ θ ≤ π/2,

27. r ≤ z ≤ 5,

0≤θ ≤π

(a) Describe the solid, using spherical coordinates. (b) Describe the solid, using cylindrical coordinates. z

−1 ≤ z ≤ 2

28. 2r ≤ z ≤ 5 − 3r

A portion of the sphere of radius 3 (centered at origin)

1

29. r 2 − 1 ≤ z ≤ 5 − r 2

In Exercises 30–35, sketch the solid whose spherical coordinates (ρ, ϕ, θ) satisfy the given inequalities. 30. 1 ≤ ρ ≤ 2 31. 0 ≤ ρ ≤ 1,

0 ≤ ϕ ≤ π/2

32. 0 ≤ ρ ≤ 1,

0 ≤ θ ≤ π/2

33. 0 ≤ ϕ ≤ π/4,

√8 ⎯

y

0≤ρ≤2

34. 0 ≤ ρ ≤ 2/ cos ϕ,

0 ≤ ϕ ≤ π/4

35. 2 cos ϕ ≤ ρ ≤ 3 36. (a) Which points P in R2 have the same rectangular

and polar coordinates? (b) Which points P in R3 have the same rectangular and cylindrical coordinates? (c) Which points P in R3 have the same rectangular and spherical coordinates? 37. (a) How are the graphs of the polar equations r = f (θ)

and r = − f (θ) related? (b) How are the graphs of the spherical equations ρ = f (ϕ, θ) and ρ = − f (ϕ, θ) related? (c) Repeat part (a) for the graphs of r = f (θ) and r = 3 f (θ). (d) Repeat part (b) for the graphs of ρ = f (ϕ, θ) and ρ = 3 f (ϕ, θ).

38. Suppose that a surface has an equation in cylindrical

coordinates of the form z = f (r ). Explain why it must be a surface of revolution.

39. (a) Verify that the basis vectors er , eθ , and ez for

cylindrical coordinates are mutually perpendicular unit vectors. (b) Verify that the basis vectors eρ , eϕ , and eθ for spherical coordinates are mutually perpendicular unit vectors.

x Figure 1.110 The ice-cream-

cone–like solid in R3 in Exercise 42.

In Exercises 43–47, you will use the equations in (10) to establish those in (12). 43. Show that tan ϕn−1 = x 2 /x 1 . 44. (a) Calculate x 12 + x 22 in terms of the hyperspherical

coordinates ρ, ϕ1 , . . . , ϕn−2 . (b) Assuming the inequalities ! in (11), use part (a) to show that tan ϕn−2 =

x12 + x22 /x3 .

45. (a) Calculate x 12 + x 22 + x 32 in terms of the hyperspher-

ical coordinates ρ, ϕ1 , . . . , ϕn−3 . (b) Assuming the inequalities ! in (11), use part (a) to show that tan ϕn−3 =

x12 + x22 + x32 /x4 .

46. (a) For k = 2, . . . , n − 1, show that x 12 + x 22 + · · · +

xk2 = ρ 2 sin2 ϕ1 · · · sin2 ϕn−k . (Note: This is best accomplished by means of mathematical induction.) (b) Assuming the inequalities in (11), use part (a) to ! show that, for k = 2, . . . , n − 1, tan ϕn−k = x12 + · · · + xk2 /xk+1 .

47. Show that x 12 + x 22 + · · · + x n2 = ρ 2 .

75

Miscellaneous Exercises for Chapter 1

True/False Exercises for Chapter 1 1. If a = (1, 7, −9) and b = (1, −9, 7), then a = b.

18. The volume of a parallelepiped determined by the vec-

tors a, b, c ∈ R3 is |(a × c) · b|.

3

2. If a and b are two vectors in R and k and l are real

numbers, then (k − l)(a + b) = ka − la + kb − lb.

3. The displacement vector from

P2 (5, 3, 2) is (−4, −3, −3).

P1 (1, 0, −1) to

19. a b − b a is a vector. 20. (a × b) · c − (a × c) · b is a scalar. 21. The plane containing the points (1, 2, 1), (3, −1, 0),

and (1, 0, 2) has equation 5x + 2y + 4z = 13.

4. Force and acceleration are vector quantities. 5. Velocity and speed are vector quantities.

22. The plane containing the points (1, 2, 1), (3, −1, 0),

and (1, 0, 2) is given by the parametric equations x = 2s, y = −3s − 2t, z = t − s.

6. Displacement and distance are scalar quantities. 7. If a particle is at the point (2, −1) in the plane and

moves from that point with velocity vector v = (1, 3), then after 2 units of time have passed, the particle will be at the point (5, 1).

8. The vector (2, 3, −2) is the same as 2i + 3j − 2k. 9. A set of parametric equations for the line through

(1, −2, 0) that is parallel to (−2, 4, 7) is x = 1 − 2t, y = 4t − 2, z = 7.

10. A set of parametric equations for the line through

(1, 2, 3) and (4, 3, 2) is x = 4 − 3t, y = 3 − t, z = t + 2.

line with parametric equations x = 2 − 3t, y = t + 1, z = 2t − 3 has symmetric form x +2 z−3 = y−1= . −3 2

11. The

12. The two sets of parametric equations x = 3t − 1,

y = 2 − t, z = 2t + 5 and x = 2 − 6t, y = 2t + 1, z = 7 − 4t both represent the same line.

13. The parametric equations x = 2 sin t, y = 2 cos t,

where 0 ≤ t ≤ π , describe a circle of radius 2.

14. The dot product of two unit vectors is 1. 15. For any vector a in Rn and scalar k, we have ka =

23. If A is a 5 × 7 matrix and B is a 7 × 7 matrix, then B A

is a 7 × 5 matrix. ⎡ ⎤ 1 2 0 3 1 ⎥ ⎢ −1 0 2 24. If A = ⎣ , then 5 9 2 0 ⎦ 0 8 0 −6     1 0  −1 2 3 1     −1 2   1 5 2 0 + 9 det A = 2     0 0 −6  0 0 −6     1 0 3    − 8  −1 2 1  .  5 2 0 

25. If A is an n × n matrix, then det (2A) = 2 det A. 26. The surface having equation r = 4 sin θ in cylindrical

coordinates is a cylinder of radius 2. 27. The surface having equation ρ = 4 cos θ in spherical

coordinates is a sphere of radius 2. 28. The surface having equation ρ cos θ sin ϕ = 3 in spher-

ical coordinates is a plane. 29. The surface having equation ρ = 3 in spherical coor-

dinates is the same as the surface whose equation in cylindrical coordinates is r 2 + z 2 = 9.

k a . 16. If a, u ∈ Rn and u = 1, then proju a = (a · u)u.

     

30. The surface whose equation in cylindrical coordinates

17. For any vectors a, b, c in R3 , we have a × (b × c) =

(a × b) × c.

is z = 2r is the same as the surface whose equation in spherical coordinates is ϕ = π/6.

Miscellaneous Exercises for Chapter 1 1. If P1 , P2 , . . . , Pn are the vertices of a regular polygon

having n sides and if O is the center of the polygon, "n −−→ show that i=1 O Pi = 0. The case n = 5 is shown in

Figure 1.111. (Hint: Don’t try using coordinates. Use instead sketches, geometry, and perhaps translations or rotations.)

76

Vectors

Chapter 1

P1

P5

(a) Give a set of parametric equations for the perpendicular bisector of the segment joining the points P1 (−1, 3) and P2 (5, −7). (b) Given general points P1 (a1 , a2 ) and P2 (b1 , b2 ), provide a set of parametric equations for the perpendicular bisector of the segment joining them.

P2

2π /5 O

6. If we want to consider a perpendicular bisector of a

(1, 0, −2) that is parallel to the line x = 3t + 1, y = 5 − 7t, z = t + 12.

line segment in R3 , we will find that the bisector must be a plane. (a) Give an (implicit) equation for the plane that serves as the perpendicular bisector of the segment joining the points P1 (6, 3, −2) and P2 (−4, 1, 0). (b) Given general points P1 (a1 , a2 , a3 ) and P2 (b1 , b2 , b3 ), provide an equation for the plane that serves as the perpendicular bisector of the segment joining them.

3. Find parametric equations for the line through the

7. Generalizing Exercises 5 and 6, we may define the per-

P4

P3

Figure 1.111 The case n = 5.

2. Find parametric equations for the line through the point

point (1, 0, −2) that intersects the line x = 3t + 1, y = 5 − 7t, z = t + 12 orthogonally. (Hint: Let x0 = 3t0 + 1, y0 = 5 − 7t0 , z 0 = t0 + 12 be the point where the desired line intersects the given line.)

4. Given two points P0 (a1 , a2 , a3 ) and P1 (b1 , b2 , b3 ), we

have seen in equations (3) and (4) of §1.2 how to parametrize the line through P0 and P1 as r(t) = −−→ −−→ O P0 + t P0 P1 , where t can be any real number. (Recall −→ that r = O P, the position vector of an arbitrary point P on the line.) −−→ (a) For what value of t does r(t) = O P0 ? For what −−→ value of t does r(t) = O P1 ? (b) Explain how to parametrize the line segment joining P0 and P1 . (See Figure 1.112.)

pendicular bisector of a line segment in Rn to be the hyperplane through the midpoint of the segment that is orthogonal to the segment. (a) Give an equation for the hyperplane in R5 that serves as the perpendicular bisector of the segment joining the points P1 (1, 6, 0, 3, −2) and P2 (−3, −2, 4, 1, 0). (b) Given arbitrary points P1 (a1 , . . . , an ) and P2 (b1 , . . . , bn ) in Rn , provide an equation for the hyperplane that serves as the perpendicular bisector of the segment joining them. 8. If a and b are unit vectors in R3 , show that

a × b 2 + (a · b)2 = 1. 9. (a) If a · b = a · c, does it follow that b = c? Explain

your answer. (b) If a × b = a × c, does it follow that b = c? Explain.

z

P1

10. Show that the two lines

P0

y x Figure 1.112 The segment joining P0

l1 : l2 :

x = t − 3, x = 4 − 2t,

y = 1 − 2t, y = 4t + 3,

z = 2t + 5 z = 6 − 4t

are parallel, and find an equation for the plane that contains them. 11. Consider the two planes x + y = 1 and y + z = 1.

(c) Give a set of parametric equations for the line segment joining the points (0, 1, 3) and (2, 5, −7).

These planes intersect in a straight line. (a) Find the (acute) angle of intersection between these planes. (b) Give a set of parametric equations for the line of intersection.

5. Recall that the perpendicular bisector of a line seg-

12. Which of the following lines whose parametric equa-

ment in R2 is the line through the midpoint of the segment that is orthogonal to the segment.

tions are given below are parallel? Are any the same? (a) x = 4t + 6, y = 2 − 2t, z = 8t + 1

and P1 is a portion of the line containing P0 and P1 . (See Exercise 4.)

Miscellaneous Exercises for Chapter 1

(b) x = 3 − 6t, y = 3t, z = 4 − 9t (c) x = 2 − 2t, y = t + 4, z = −4t − 7 (d) x = 2t + 4, y = 1 − t, z = 3t − 2 13. Determine which of the planes whose equations are

given below are parallel and which are perpendicular. Are any of the planes the same? (a) 2x + 3y − z = 3 (b) −6x + 4y − 2z + 2 = 0 (c) x + y − z = 2 (d) 10x + 15y − 5z = 1 (e) 3x − 2y + z = 1 14. (a) What is the angle between the diagonal of a cube

and one of the edges it meets? (Hint: Locate the cube in space in a convenient way.) (b) Find the angle between the diagonal of a cube and the diagonal of one of its faces.

77

(b) Let P be the point of intersection of B M1 and C M2 . −→ −→ −→ −→ Write B P and C P in terms of AB and AC. −→ −→ −→ −→ −→ (c) Use the fact that C B = C P + P B = C A + AB to show that P must lie two-thirds of the way from B to M1 and two-thirds of the way from C to M2 . (d) Now use part (c) to show why all three medians must meet at P. 17. Suppose that the four vectors a, b, c, and d in R3 are

coplanar (i.e., that they all lie in the same plane). Show that then (a × b) × (c × d) = 0. 18. Show that the area of the triangle, two of whose

sides are determined by the vectors a and b (see Figure 1.114), is given by the formula ! 1

a 2 b 2 − (a · b)2 . Area = 2

15. Mark each of the following statements with a 1 if you

agree, −1 if you disagree: (1) Red is my favorite color. (2) I consider myself to be a good athlete. (3) I like cats more than dogs. (4) I enjoy spicy foods. (5) Mathematics is my favorite subject. Your responses to the preceding “questionnaire” may be considered to form a vector in R5 . Suppose that you and a friend calculate your respective “response vectors” for the questionnaire. Explain the significance of the dot product of your two vectors.

16. The median of a triangle is the line segment that joins

a vertex of a triangle to the midpoint of the opposite side. The purpose of this problem is to use vectors to show that the medians of a triangle all meet at a point. −−→ −−→ (a) Using Figure 1.113, write the vectors B M1 and C M2 −→ −→ in terms of AB and AC. A M1

C

M2

b

a Figure 1.114 The triangle in Exercise 18.

A(1, 3, −1), B(4, −1, 3), C(2, 5, 2), and D(5, 1, 6) be the vertices of a parallelogram. (a) Find the area of the parallelogram. (b) Find the area of the projection of the parallelogram in the x y-plane.

19. Let

20. (a) For the line l in R2 given by the equation ax +

by = d, find a vector v that is parallel to l. (b) Find a vector n that is normal to l and has first component equal to a. (c) If P0 (x0 , y0 ) is any point in R2 , use vectors to derive the following formula for the distance from P0 to l: |ax0 + by0 − d| . Distance from P0 to l = √ a 2 + b2 To do this, you’ll find it helpful to use Figure 1.115, where P1 (x1 , y1 ) is any point on l. (d) Find the distance between the point (3, 5) and the line 8x − 5y = 2.

21. (a) If P0 (x 0 , y0 , z 0 ) is any point in R3 , use vectors

B Figure 1.113 Two of the three medians

of a triangle in Exercise 16.

to derive the following formula for the distance from P0 to the plane having equation Ax + By + C z = D: |Ax0 + By0 + C z 0 − D| . Distance from P0 to = √ A2 + B 2 + C 2

78

Chapter 1

Vectors

y

Describe the configuration of points P that satisfy the equation. P0

25. Let a and b be two fixed, nonzero vectors in R3 , and let

c be a fixed constant. Explain how the pair of equations,  a·x = c a×x = b,

n l: ax + by = d P1

x

Figure 1.115 Geometric construction for

Exercise 20.

Figure 1.116 should help. (P1 (x1 , y1 , z 1 ) is any point in .) P0

z

completely determines the vector x ∈ R3 . 26. (a) Give examples of vectors a, b, c in R3 that show

that, in general, it is not true that a × (b × c) = (a × b) × c. (That is, the cross product is not associative.) (b) Use the Jacobi identity (see Exercise 30 of §1.4) to show that, for any vectors a, b, c in R3 , a × (b × c) = (a × b) × c if and only if (c × a) × b = 0.

Distance

27. (a) Given an arbitrary (i.e., not necessarily regular)

n P1 Π : Ax + By + Cz = D y

x Figure 1.116 Geometric construction for

tetrahedron, associate to each of its four triangular faces a vector outwardly normal to that face with length equal to the area of that face. (See Figure 1.117.) Show that the sum of these four vectors is zero. (Hint: Describe v1 , . . . , v4 in terms of some of the vectors that run along the edges of the tetrahedron.)

Exercise 21.

(b) Find the distance between the point (1, 5, −3) and the plane x − 2y + 2z + 12 = 0.

v1

v4

v2

22. (a) Let P be a point in space that is not contained in the

plane that passes through the three noncollinear points A, B, and C. Show that the distance between P and is given by the expression

|p · (b × c)| ,

b × c

−→ −→ −→ where p = A P, b = AB, and c = AC. (b) Use the result of part (a) to find the distance between (1, 0, −1) and the plane containing the points (1, 2, 3), (2, −3, 1), and (2, −1, 0). 23. Let A, B, C, and D denote four distinct points in R3 .

(a) Show that A, B, and C are collinear if and only if −→ −→ AB × AC = 0. (b) Show that A, B, C, and D are coplanar if and only −→ −→ −→ if ( AB × AC) · C D = 0. −→ 24. Let x = O P, the position vector of a point P in R3 . Consider the equation x·k 1 = √ .

x

2

v3 Figure 1.117 The tetrahedron of part

(a) of Exercise 27.

(b) Recall that a polyhedron is a closed surface in R3 consisting of a finite number of planar faces. Suppose you are given the two tetrahedra shown in Figure 1.118 and that face ABC of one is congruent to face A B  C  of the other. If you glue the tetrahedra together along these congruent faces, then the outer faces give you a six-faced polyhedron. Associate to each face of this polyhedron an outward-pointing normal vector with length equal to the area of that face. Show that the sum of these six vectors is zero.

Miscellaneous Exercises for Chapter 1

79

C

C

A

A

B

B

Figure 1.118 In Exercise 27(b), glue the two tetrahedra shown along congruent faces.

(c) Outline a proof of the following: Given an n-faced polyhedron, associate to each face an outwardpointing normal vector with length equal to the area of that face. Show that the sum of these n vectors is zero. 28. Consider a right tetrahedron, that is, a tetrahedron

that has a vertex R whose three adjacent faces are pairwise perpendicular. (See Figure 1.119.) Use the result of Exercise 27 to show the following three-dimensional analogue of the Pythagorean theorem: If a, b, and c denote the areas of the three faces adjacent to R and d denotes the area of the face opposite R, then a 2 + b2 + c2 = d 2 .

(b) Conjecture the general form of An for the matrix A of part (a), where n is any positive integer. (c) Prove your conjecture in part (b) using mathematical induction. 32. A square matrix A is called nilpotent if An = 0 for

some positive power n. ⎡ ⎤ 0 1 1 (a) Show that A = ⎣ 0 0 0 ⎦ is nilpotent. 0 0 0

T (b) Use a calculator or computer to show that A = ◆ ⎤ ⎡

0 0 0 ⎢ 1 0 0 ⎢ ⎢ 0 1 0 ⎣ 0 0 1 0 0 0

0 0 0 0 1

0 0 0 0 0

⎥ ⎥ ⎥ is nilpotent. ⎦

T 33. The n × n matrix H whose i jth entry is 1/(i + j − 1) ◆ is called the Hilbert matrix of order n. n

R

Figure 1.119 The right tetrahedron

of Exercise 28. The three faces containing the vertex R are pairwise perpendicular.

29. (a) Use vectors to prove that the sum of the squares

of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the four sides. (b) Give an algebraic generalization of part (a) for Rn . 30. Show that for any real numbers a1 , . . . , an , b1 , . . . , bn

we have

#

n  i=1

$2 ai bi

≤

# n  i=1

$# ai2

n 

$ bi2

.

i=1

31. To raise a square (n × n) matrix A to a positive integer

power n, one calculates An as A · A · · · A (n times). (a) Calculate successive powers A, A2 , A3 , A4 of the   1 1 . matrix A = 0 1

(a) Write out H2 , H3 , H4 , H5 , and H6 . Use a computer to calculate their determinants exactly. What seems to happen to det Hn as n gets larger? (b) Now calculate H10 and det H10 . If you use exact arithmetic, you should find that det H10 = 0 and hence that H10 is invertible. (See Exercises 30–38 of §1.6 for more about invertible matrices.) (c) Now give a numerical approximation A for H10 . Calculate the inverse matrix B of this approximation, if your computer allows. Then calculate AB and B A. Do you obtain the 10 × 10 identity matrix I10 in both cases? (d) Explain what parts (b) and (c) suggest about the difficulties in using numerical approximations in matrix arithmetic. As a child, you may have played with a popular toy called a Spirograph® . With it one could draw some appealing geometric figures. The Spirograph consists of a small toothed disk with several holes in it and a larger ring with teeth on both inside and outside as shown in Figure 1.120. You can draw pictures by meshing the small disk with either the inside or outside circles of the ring and then poking a pen through one of the holes of the disk while turning the disk. (The large ring is held fixed.)

80

Chapter 1

Vectors

y

y

2

4

1

2 x

−1

1

2

−1

4

6

Figure 1.122 Hypocycloids with a = 3, b = 2 and a = 6, b = 5.

Figure 1.120 The Spirograph.

y

An idealized version of the Spirograph can be obtained by taking a large circle (of radius a) and letting a small circle (of radius b) roll either inside or outside it without slipping. A “Spirograph” pattern is produced by tracking a particular point lying anywhere on (or inside) the small circle. Exercises 34–37 concern this set-up. 34. Suppose that the small circle rolls inside the larger

circle and that the point P we follow lies on the circumference of the small circle. If the initial configuration is such that P is at (a, 0), find parametric equations for the curve traced by P, using angle t from the positive x-axis to the center B of the moving circle. (This configuration is shown in Figure 1.121.) The resulting curve is called a hypocycloid. Two examples are shown in Figure 1.122. y

B t

2

−4

−2

a

x

−6 −4 −2 −2

3

P b A

x

4 2 −4

x

−2

2

4

−2 −4

Figure 1.123 An epicycloid with

a = 4, b = 1.

this happens whenever the smaller circle rolls through 2π . Assuming that a/b is rational, how many cusps does a hypocycloid or epicycloid have? (Your answer should involve a and b in some way.) (b) Describe in words and pictures what happens when a/b is not rational. 37. Consider the original Spirograph set-up again. If we

Figure 1.121 The coordinate

configuration for finding parametric equations for a hypocycloid.

35. Now suppose that the small circle rolls on the outside

of the larger circle. Derive a set of parametric equations for the resulting curve in this case. Such a curve is called an epicycloid, shown in Figure 1.123. 36. (a) A cusp (or corner) occurs on either the hypocy-

cloid or epicycloid every time the point P on the small circle touches the large circle. Equivalently,

now mark a point P at a distance c from the center of the smaller circle, then the curve traced by P is called a hypotrochoid (if the smaller circle rolls on the inside of the larger c