Wexler - Analytic Geometry a vector approach

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OUTSTANDING TEXTBOOKS AVAILABLE IN WEXLER WORLD STUDENT SERIES WORLD STUDENT SERIES EDITIONS MATHEMATICS APOSTOL: Mathematical A Modern Approach to Advanced Calculus (1957) AND SHANKS: Fundamental Concepts of Elementary Mathe- matics (1962) KAPLAN: Advanced Calculus (1952) MORREY: University Calculus with Analytic Geometry (1962) THOMAS: Calculus 2nd ed. (1961) Calculus and Analytic Geometry Complete 3rd ed. (1960) VANCE: Modern Algebra and Trigonometry (1962) VANCE: Modern College Algebra (1962) WEXLER: Analytic Geometry: A Vector Approach (1962) PHYSICS The Mainstream of Physics (1962) Study Guide for the Mainstream of Physics (1962), paperbound DICKE & WITTKE: Introduction to Quantum Mechanics (1960) GOLDSTEIN: Classical Mechanics (1950) KAPLAN: Nuclear Physics 2nd ed. (1963) PANOFSKY AND PHILLIPS: Classical Electricity and Magnetism 2nd ed. (1962) REITZ AND MILFORD: Foundations of Electromagnetic Theory (1960) RICHARDS, SEARS, AND ZEMANSKY: Modern University Physics (1960), one volume complete, hardbound Part 1: Mechanics and Thermodynamics, paperbound Part Fields, Waves, and Particles, paperbound Optics (1957) SEARS: Electricity and (1951) SEARS: Mechanics, Heat, and Sound 2nd ed. (1950) SEARS: Optics 3rd ed. (1949) A VECTOR APPROACH ANALYTIC GEOMETRY: ANALYTIC GEOMETRY A VECTOR APPROAC CHARLES WEXL SEARS: Thermodynamics 2nd ed. (1953) SEARS AND ZEMANSKY: College Physics 3rd ed. (1960), one volume complete, hardbound Part 1: Mechanics, Heat, Sound (1960), paperbound Part Electricity and Magnetism, Light, and Physics (1960), paper- bound SYMON: Mechanics 2nd ed. (1960) WORLD ADDISON WESLEY. WEHR AND RICHARDS: Physics of the Atom (1960) ENGINEERING STUDENT CHENG: Analysis of Linear Systems (1959) Guy: Physical Metallurgy for Engineers (1962) JOYCE AND CLARKE: Transistor Circuit Analysis (1961) LEE AND SEARS: Thermodynamics 2nd ed. (1963) WESLEY PUBLISHING COMPANY, ADDISON WESLEY(U.S. Edition $ ANALYTIC GEOMETRY: A VECTOR APPROACH By CHARLES WEXLER Arizona State University This textbook for a one-semester course introduces the student to analytic geometry by the very natural and in- teresting vector approach. Thus, solid analytic geometry studied simultaneously with plane analytic geom- etry, and polar coordinates are brought in at an early stage and used continually. Furthermore, introducing vectors early and helping students learn to think in terms of them is a very valuable preparation for later work in science and engineering. Only those portions of vector analysis which are natural THE AUTHOR in the development of analytic geometry are used, up to and including the dot and cross products of vectors and Charles Wexler received his S.B., A.M., and Ph.D. their geometric properties. However, some material on degrees from Harvard University. He has been an In- linear algebra is included, giving a development of ab- structor at Harvard, an Associate in Mathematics at the vector spaces more or less parallel with that of University of Georgia as well as at George Washington ordinary three-dimensional vector space. Thus, mathe- University, and is currently Professor of Mathematics at maties majors are provided with a preview of this very Arizona State University. important modern trend if the instructor desires it, and yet it can be omitted for the non-mathematics majors During World War II, Dr. Wexler served in the Army without affecting the continuity. and is now in the Army Reserve with the rank of Lieu- tenant Colonel. He has been a Research Associate at Los The book includes a brief review of determinants, occa- Alamos Scientific Laboratories, at UCLA in conjunction reviews of trigonometry, and ample problems, many with the National Bureau of Standards, and at Woods with answers. The study of tangents to curves and nor- Hole Oceanographic Dr. Wexler is a past mals to surfaces, by vector methods, provides a smooth governor of the Mathematical Association of America transition to the type of thinking used in the calculus. and is currently a Regional Representative for its pro- None of the vocabulary of calculus is used, but the dif- gram of Visiting Lecturers for Secondary Schools, spon- ference quotient and its limit stand out as a very power- sored by the National Science Foundation. ful tool for the study of curves and surfaces. A surprising amount of differential calculus is thus developed rather naturally.ANALYTIC GEOMETRY This book is in the ADDISON-WESLEY SERIES IN A Vector Approach INTRODUCTORY MATHEMATICS RICHARD S. PIETERS AND GEORGE B. THOMAS, JR. Consulting editors by CHARLES WEXLER Department of Mathematics Arizona State University WORLD STUDENT SERIES ADDISON-WESLEY PUBLISHING COMPANY, INC. READING, MASSACHUSETTS PALO ALTO LONDONTHIS BOOK IS AN ADDISON-WESLEY WORLD STUDENT SERIES EDITION PREFACE First printed in 1964 In my own teaching I have found that to introduce the student to A complete and unabridged reprint of the original American analytic geometry by way of vectors not only furnishes elegant proofs textbook, this World Student Series edition may be sold only in but also gives him early familiarity and training in vector concepts that those countries to which it is consigned by Addison-Wesley or are invaluable to future scientific work. By means of vectors, solid analytic by its authorized Trade Distributor, The Book Centre, Ltd., geometry can be studied practically simultaneously with plane analytic London. It may not be re-exported from the country to which it geometry, and polar coordinates can be brought in early. Not the least has been consigned, and it may not be sold in the United States of of many advantages is that early introduction to vector methods in America or its possessions. analytic geometry makes later work in physics or engineering much easier, and in fact would save much time in those courses that must otherwise be devoted to developing vectors. A recent trend in mathematics has been to combine, or "integrate," analytic geometry and calculus, usually in a three-semester sequence. This Edition is distributed in Japan, Korea, Hong Kong, Okinawa, Taiwan, The Philippines, Indonesia, Vietnam, Laos, Most texts now available which combine these two subjects underempha- Thailand, Malaya, Singapore, Pakistan, India, size the analytic geometry. After a very meager introduction to analytic geometry, they move into calculus at full speed, not returning to analytic Burma, and Ceylon by geometry until 150 or 200 pages farther on. It seems to me that this JAPAN PUBLICATIONS TRADING COMPANY, LTD. scanty preparation in analytic geometry is not an adequate background CENTRAL P.O. BOX 722, TOKYO, JAPAN for the calculus. Perhaps I may be considered too conservative, but I have felt that the emphasis during the first semester, or at least the first quarter, should be on the analytic geometry, with calculus first being developed for finding tangents to curves and normals to surfaces; and that there should be too much analytic geometry to choose from rather than too little. Copyright © 1961 This book was originally begun with this aim in mind: to combine analytic ADDISON-WESLEY PUBLISHING COMPANY, INC. geometry and calculus in this manner for a three-semester course, but when the analytic geometry subject matter was completed together with ALL RIGHTS RESERVED. THIS BOOK, OR PARTS the necessary calculus for tangents and normals, it seemed appropriate to THEREOF, MAY NOT BE REPRODUCED IN ANY FORM publish it as a straight analytic geometry text for a course of one semester. WITHOUT WRITTEN PERMISSION OF THE PUBLISHER. Incidentally, only the difference quotient limit is used; calculus language such as derivative, etc. is avoided. The constant repetition of its use Library of Congress Catalog Card No. 61-10974 emphasizes that the difference quotient is a basic concept of the calculus. These three features, the thorough development of vector methods from the very beginning, the generous amount of analytic geometry to choose from, and the not inconsiderable amount of calculus motivated by the need to find tangents and normals, may enhance the value of the TOSHO INSATSU regular analytic geometry course as to save it from fading into oblivion. PRINTING LTD. This book is not, and is not intended to be, an exhaustive treatise on TOKYO, JAPAN vectors. It uses only that much of vector analysis that goes naturally with the development of analytic geometry and calculus at the elementary Vvi PREFACE PREFACE vii level. A remarkably small amount of vector analysis is needed for this those on the fine points concerning conic sections. But the vector material purpose, only up to and including the dot and cross products of vectors at the beginning and the calculus material at the end should not be cur- tailed. and their geometric properties. It was thought worthwhile to include a little linear algebra, some I wish to take this opportunity to express my deep appreciation and material on abstract vector spaces is woven in, paralleling the three- gratitude to Mr. M. L. Clabaugh, who spent many hours typing and re- dimensional development to a certain extent, up to the point of defining typing the manuscript and who suggested many improvements. My inner product spaces. This material may be too difficult for most be- thanks also to several anonymous reviewers for their constructive criticism ginning students, but some instructors may wish to experiment to see how and suggestions, most of which I tried to follow. much of it can be absorbed. It is purely background material, and no C. W. exercises are based on it. I shall appreciate receiving comments as to how successful this feature is. Incidentally, some knowledge of determinants is needed, and a short summary of their properties and main theorems is given, mostly without proof. In several places in the text, the conditions for nontrivial solutions of linear homogeneous equations are needed, and these are developed more fully (with proofs), together with Cramer's Rule. Every instructor knows how "unfair" of him it is to expect students to remember any trigonometry after they have passed that course, the student will have to be prodded into recalling certain exotic items such as the functions of 0°, 30°, 45°, 60°, 90°, 180°, the addition formulas, the double-angle formulas, right-triangle relations, etc., as they are needed. Some of these topics are reviewed here and there in the text, some from the point of view of vectors. It is very important that students be given regular out-of-class assign- ments and that some method be found to insure that these are seriously attempted, because then the class explanations take on much more mean- ing for the student. It is especially important in the case of the vector assignments that all or nearly all of the problems not only be assigned, but also be done in class by the instructor. Many of the problems extend the theory. Vectors are a strange and new subject for the student; it is not a forbidding subject, in fact most students are quite intrigued and often fascinated by the power and elegance of vectors; but since vectors are new and unfamiliar, the student should be led by the hand a bit more than usual. After all, the text states a fact only once or twice, while the repetition of the basic given by the instructor in doing the problems in class helps enormously in clarifiying vectors for the student and makes them appear natural to use. There is probably more material in the book than can be handled comfortably in one semester. The general equation of the second degree and abstract vector spaces could be postponed to a later course (or assigned as a reading project for the good students), and the material on radical axes could be omitted. Discussions of families of curves could be omitted. Also, if time is limited, fewer exercises could be done in class, especiallyCONTENTS CHAPTER 1. INTRODUCTION 1 CHAPTER 2. FUNDAMENTAL CONCEPTS 5 2-1 Vectors. Definitions and operations 5 2-2 Abstract vector spaces 19 2-3 Basic concepts of analytic geometry 24 CHAPTER 3. THE STRAIGHT LINE 47 3-1 Locus problems in general 47 3-2 The straight line in three dimensions 48 3-3 The straight line in two dimensions 50 3-4 The equation of the first degree 56 3-5 Analytic properties of straight lines 58 3-6 The straight line in polar coordinates. Rotation of axes 68 3-7 The plane 74 3-8 Families of lines. Families of planes 80 3-9 Summary and review 84 CHAPTER 4. CIRCLES AND SPHERES 94 4-1 The standard equations 94 4-2 Circle through three points 97 4-3 Tangent lines and tangent planes 101 4-4 Families of circles 106 4-5 The circle in polar coordinates 112 4-6 Regions bounded by circles and straight lines 119 4-7 Miscellaneous topics 126 A. Parametric Equations of a Circle 126 B. Translation of Axes. Mappings 128 C. Surfaces and Curves Related to the Circle 133 CHAPTER 5. THE CONIC SECTIONS AND OTHER CURVES 139 5-1 Introduction 139 5-2 The ellipse 141 5-3 The hyperbola 151 5-4 The parabola 166 5-5 A unifying principle for the conic sections 171 Some special curves in polar coordinates 177 5-6 The general equation of the second degree 187 ixCONTENTS X 199 5-7 Other curves and 220 5-8 Tangents to curves. The difference-quotient 244 Normals to surfaces CHAPTER 1 249 CHAPTER 6. SUMMARY INTRODUCTION 259 ANSWERS TO PROBLEMS Analytic geometry is a wedding of algebra and geometry. The ancient 289 Greek mathematicians, with their marvelous minds, concentrated on INDEX geometry and perfected it to a high degree. But, to our great loss, they more or less ignored arithmetic (and the beginnings of algebra) because it was developed to serve trade and commerce. Trade and accounts, and hence arithmetic, were the province of slave supervisors, while mathemati- cians devoted their thoughts to the higher arts. It took about 1500 years for arithmetic and algebra to spread across Europe, and it remained for René Descartes to recognize what a powerful tool algebra would be in simplifying and extending geometry. It was only after a sufficient interval for developing this idea that the time was ripe for Isaac Newton to create an even more powerful new tool, the calculus. Descartes' fundamental idea, like most great inventions, was simple. Why not set up two basic directions, such as a "horizontal" and a "verti- cal" direction (from a central starting point 0, called the origin), which would serve as measuring sticks to locate any desired point P in the plane? The distance along the horizontal direction could be denoted by a negative, positive, or zero number, depending on whether one had to go to the left, or to the right, or neither to reach a given point. Similarly, the number denoting the vertical distance could be positive, negative, or zero, according to whether one had to go up, down, or neither to reach the point. Following Descartes' construction, illustrated in Fig. 1-1, we find that to every point in the plane there corresponds a pair of real numbers, x and y; and, conversely, to every pair of real numbers there corresponds one Vertical (-2,1) P:(x, 1 -2 0 Horizontal (0, -1) FIGURE 1-1 12 INTRODUCTION [CHAP. 1 INTRODUCTION 3 y-axis P:(x, a 2 y 0 I 0 y FIGURE 1-3 FIGURE 1-2 the z-axis, perpendicular to the plane of the and the y-axis at the origin (Fig. 1-3). and only one point. The horizontal line is called the x-axis, its positive The basic idea of directed line segments is very important, not only in direction being to the right; the vertical line is called the y-axis, its positive analytic geometry and mathematics in general, but also in physics and direction being up. The two numbers written "(x, are called the engineering. Directed line segments are called vectors. We shall devote coordinates of the point P, respectively the and y-coordinate the next chapter to building up an "algebra" of vectors and using it to (sometimes called the abscissa and the ordinate). The origin 0 has co- develop the fundamental concepts of analytic geometry. The reader will ordinates (0, 0); any point on the x-axis has coordinates (a, 0); just as any find vector analysis a somewhat strange but very stimulating subject. If point on the y-axis has coordinates (0, b). Thus the x-axis is characterized he applies himself and learns to incorporate it into his thinking, he will by the equation y = 0. In words, every point on the x-axis has as its find it a wonderful help in finding his way about in space. But the thrill y-coordinate the value of zero; conversely, any point whose y-coordinate of knowing that he is beginning to master the subject will come only is zero lies on the x-axis. when he finds he can do many of the exercises, some of which are not easy. When there is such a relation between a curve and an equation, namely, that the coordinates of any point on the curve satisfy the equation and EXERCISE GROUP 1-1 any point whose coordinates satisfy the equation must lie on the curve, we say that the equation is the equation of the curve. Thus y = 0 is the 1. Draw a two-dimensional frame of reference; marking the and equation of the Similarly, x = 0 is the equation of the y-axis. y-axes. Plot the following points and write the coordinates next to each It can be seen that the idea of a "coordinate system" or "frame of refer- point in your drawing: (3, 0), (0, 2), (-3, -2), (-2, 3). ence," as this scheme of having an origin and two basic directions is called, 2. Draw a three-dimensional frame of reference, marking the three presents great possibilities. For example, to study a circle of radius a, axes. Plot the following points: (2, 0, 0), (0, 1, 0), (1, 2, 3). we can choose the origin to be at the center of the circle (Fig. 1-2). We see by the Pythagorean theorem that the coordinates (x, y) of any point on the 3. What is the equation of the straight line through (2, 1) and (2, -3)? circle satisfy the equation x² + = a², and that any point whose coordi- through (2, 1) and (-1, 1)? Draw each line first. nates satisfy the equation + = a² is the vertex of a right triangle 4. What is the equation of the xy-plane in three dimensions? the of hypotenuse a, at distance a from the origin, and hence lies on the the yz-plane? the plane through the points (3, 1, 2), circle. All properties of a circle are presumably embodied in its equation. (-1, 2, 2), (0, 0, 2)? We shall study the circle in detail from this point of view later. Going back to Descartes' fundamental idea, we see that we can reach 5. Find two equations that together characterize completely the x-axis in three dimensions; likewise for the y-axis; the z-axis. any point in the plane by means of two directed line segments, right or left and up or down. Similarly, we can reach any point in three-dimen- 6. Find the equation of the line containing the origin and the point sional space with three directed line segments by providing a third axis, (1, 1).4 INTRODUCTION [CHAP. 1 7. 9 = 0 is the equation of what curve(s)? 8. + y² 9 = 0 is the equation of what curve? CHAPTER 2 A remark is in order about the use of our phrase "the equation of a FUNDAMENTAL CONCEPTS curve." It was pointed out above that = 0 is the equation of the x-axis. The y-coordinate of every point on the x-axis is certainly zero, 2-1 Vectors. Definitions and operations. A vector is a directed line and any point whose y-coordinate equals zero certainly lies on the segment. Note that there are two aspects to a vector: it has direction, But the same is true of 5y = 0 or of + 1)y = 0. Every point on the and it has length or magnitude. Any entity that has these two qualities x-axis satisfies each of these equations; also, any point that satisfies either can be represented by a vector. Thus if we use a scale of inch = 10 of these equations must lie on the x-axis. The reason is that 5y or + 1)y miles, the speed of a car traveling northeast at 40 mi/hr can be expressed can equal zero only when y = 0. So, strictly speaking, y = 0 is only one as a one-inch arrow pointing in a direction 45° above the horizontal to the possible equation of the x-axis. There are many. Hence when we speak right (east) (Fig. 2-1). The weight of a 165-lb man (the gravitational pull of "the equation" of a certain curve, we shall mean that the particular of the earth) can be represented by an arrow of appropriate length pointed equation being discussed satisfies the two-part definition of "equation in a downward direction. of a curve," namely, that every point on the curve satisfies the equation In print, vectors are usually denoted by boldface type, such as A or r and, conversely, any point that satisfies the equation must lie on the and their length by |A| or |r|, sometimes called the absolute value of A curve. or of In handwriting we usually indicate the vector by a letter with an arrow above it: A, Sometimes, even in print, we shall use a beginning point and an endpoint with an arrow, for example, OP, to designate a vector; its magnitude is then designated by See Fig. 2-2. 40 mi/hr A A NE A B OP OP P 45° E 0 FIGURE 2-1 FIGURE 2-2 FIGURE 2-3 (a) Definition of equality. Two vectors are equal if and only if they have the same length and the same direction. Note (Fig. 2-3) that equal vectors do not necessarily start from the same point, nor must they be along the same line. The vectors we will deal with are free vectors, unless otherwise specified. They can be transported from place to place; and so long as they remain of the same length and keep the same direction, we consider them to be the same vector or equal vectors. This may seem to be a rather loose way of thinking of "same" or "equal." Some writers prefer the term "equivalent," reserving "equal" to mean identical or coincidental vectors. But we prefer to use the words "equal" or "same" in the loose sense; in the few cases where we mean to indicate strict equality, we shall be careful to point it out. 5FUNDAMENTAL CONCEPTS [CHAP. 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 7 6 B B N C A A P A+B+C B Intersecting Non-intersecting N A (b) (a) FIGURE 2-6 FIGURE 2-7 FIGURE 2-4 A A -A 2A A P = A A FIGURE 2-5 FIGURE 2-8 We shall use the word "parallel" with the same freedom. The term a direction vector by any convenient scalar not zero; and just as we spoke parallel vectors will include the special case where the two vectors are along of the equation of a line even though it was one of many, we shall speak the same line or even when they coincide. Thus a vector is parallel to of the direction vector of a line and the normal to a plane. In the next itself. Also, two vectors in opposite directions, as well as vectors in the paragraph we discuss further the concept of a scalar times a vector. same direction, can be called parallel. (b) Definition of addition. Two or more vectors are added by the Consistent with the idea of free vectors is that of angle between two vec- polygon rule: from the arrowhead of one vector, the second vector (of tors, even if the vectors do not meet. We can transport them parallel to correct length and direction) is drawn; from its tip, the third vector is themselves to emanate from the same point. Then the angle between drawn; and so on, as illustrated in Fig. 2-7. The vector sum is the vector them that is between 0° and 180° inclusive is called the angle between that extends from the beginning of the first vector to the tip of the last the two vectors (Fig. 2-4b). Thus if one's right arm extends upward and vector, i.e., the vector (A+B+C in the figure) which completes or the left arm forward, the angle between them is 90° in spite of the fact "closes" the polygon. Note that this could be a twisted polygon in three- that they do not meet in a vertex. Since we cannot distinguish between dimensional space; i.e., not in a plane. clockwise and counterclockwise in three dimensions, we do not need Referring to Fig. 2-8, we see that A + A = 2A is a vector in the same negative angles. direction as A and of twice the length. Likewise KA is k times as long as One other term that we shall use occasionally is direction vector. A and in the same direction if k is positive. However, if k is negative, Very often we are concerned only with the direction of a vector and not is in the opposite direction from A and is |k| times as long. As noted with its size. For example, a straight line is completely determined by a above, k is called a scalar. If two vectors are parallel, one is a scalar point P which lies on it and any vector A along it (Fig. 2-5). The mag- times the other; and conversely, if nA = mB (m and n being scalars not nitude of A does not matter; a vector twice as long, or in general equal to zero), then A and B are parallel vectors. Further, A/|A| = i' would do just as well, where k is a pure number not zero and may even is the unit vector (length = 1) in the same direction as A, and A = |A|i'. be negative. The pure number k is called a scalar. Likewise, a plane (Fig. 2-6) is determined by a given point P which lies in it and any vector (c) Definition of subtraction. is the vector which added to B N perpendicular to it. Such a vector is called a normal of the plane. N gives vector A. It is therefore the vector from the tip of to the tip of A. may be of any length; AN would do just as well. We shall freely multiply (See Fig. 2-9.) One could, of course, add to A to give A B, as[CHAP. 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 8 FUNDAMENTAL CONCEPTS 9 B A C A B B B+A A+B+C B B A+B B A B A A FIGURE 2-9 FIGURE 2-10 FIGURE 2-13 FIGURE 2-14 C P to the other vector, these lines intersecting at point P. Thus a parallelo- D gram is formed. The diagonal vector OP is called the resultant or sum B B of the vectors A and B. This method agrees with the polygon method, A since OP divides the parallelogram into two congruent triangles, either -A of which considered by itself is a special case of our polygon method of A 0 A addition. The polygon rule is more general in that any number of vectors FIGURE 2-11 FIGURE 2-12 can be added at once, while by the parallelogram law a new parallelo- gram, formed by the preceding resultant with one additional vector, shown in Fig. 2-10. The second method is better for problems like must be painstakingly drawn after each addition. 2B 4C - 2D. It is fairly evident from the definition of addition of vectors that: Exercise: Prove that the two vectors given by these two methods, both (1) A + B = A (called the Commutative Law for Addition). alleged to be B, are actually the same (i.e., have the same length and same direction). See Figs. 2-9 and 2-10. = A+B+C (the Associative Law for Addition). (3) + + A3 + An can be added in any order; for example, The operations of addition and subtraction lead to an important special it is equal to A3 + + + vector. In the case where the addition of the vectors leads to a closed polygon, the endpoint of the last vector coinciding with the beginning The Commutative Law follows from the fact that A+B and B+ A, point of the first vector, as in Fig. 2-11, we define the sum to be a vector emanating from a common point, form a parallelogram; and the diagonal thereof, as shown in Fig. 2-13, is simultaneously the vector and of zero length, called the null vector, and we write the vector B+ A (the Parallelogram Law again). = The Associative Law follows directly from the fact that when vectors are added, the same vector closes all three polygons, as shown in Fig. 2-14. Thus + (-A) = 0, the null vector; also, = = A; and As to property (3), which is a generalized commutative law, our method if mA + nB = 0, we can say that of addition enables us to say that = mA = -mA (the vector which added to mA gives 0) by the definition of subtraction. is the same no matter what set or sets of A's are enclosed in parentheses. For example, we can write This justifies "transposing" a vector from one side of an equation to the other, as in ordinary algebra. + A2) + A4 + + (Aₖ + One other remark should be made. Many readers may have already met addition of vectors in earlier work, using the "parallelogram law" a generalized associative law, since the vector necessary to close the of addition to find the "resultant" of two forces. The method is as follows: polygon is always the same. Hence by placing parentheses around pairs Two vectors A and emanate from a common point as shown in Fig. and applying the Commutative Law repeatedly, we can put the A's in 2-12. Then from the endpoint of each vector, a line is drawn parallel any order we please.10 FUNDAMENTAL CONCEPTS [CHAP. 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 11 The following properties also follow readily from the definition of addi- More precisely, we should say that in our number system multiplication tion and that of a scalar times a vector: is distributive over addition. (Many beginning students of algebra wrongly (4) m(nA) = (mn)A (another Associative Law, for multiplication of a try to make multiplication distributive over multiplication; in doing vector by scalars). 2(3 5), they multiply both the 3 and the 5 by the 2, getting 6. 10 or 60!) Whatever the signs of m and n are, the two vectors m(nA) and (mn) A If we think of the m as a scalar, we also have for vectors must have the same direction (same as A if m, n have the same sign, B) = mA + mB. opposite to A otherwise). The lengths are the same because The proof of this is like those of (4) and (5) above, in which both direc- = = |A| tion and lengths of vectors are considered, and will be a problem in holds, these being ordinary real numbers. Exercise Group 2-1. Thus, in our vector system, multiplication by a scalar is distributive (5) (m + n)A = mA + nA (a type of Distributive Law described over vector addition. Later, we will learn how to "multiply" vectors by below). vectors (in fact, there will be two kinds of multiplication), and we shall Again, the directions are the same. This is obvious if m and n are have to determine if that kind of multiplication is also distributive over of the same sign, or if m = Otherwise, the direction is deter- vector addition. mined by the sign of whichever scalar has the larger numerical value, Last, each system has a special element, called the zero element, such |m| or As for the lengths, that a = a = a; and corresponding to each element a of = |m| + |A| the system, there exists an element in the system, its inverse, such that a (-a) = 0 and Elements a and are if m and n are of the same sign; while inverses of each other, that is, -(-a) = a. In fact, if we think of the numbers as being represented by vectors in |A| = |(|m| |n|)| |A| one dimension, we can consider the number system a special case of the if m and n are of opposite sign, since these are all real numbers. system of vectors. For example, corresponding to the number 3, there is a vector 3, of length 3, whose direction is to the right. Likewise, -5 is We have thus far made a good start toward building an algebra of of length 5 pointing to the left. From the rule of addition of vectors, we see vectors. It is very similar in its laws to the algebra of our real number that 3+ (-5) = -2, which closes the polygon (see Fig. 2-15). All the system, as the following comparison will show. usual combinations in the addition of signed numbers follow easily from First, addition is always possible in each system. If a and b are any this vector point of view. For example, two elements in the system of numbers or the system of vectors, then = c, where is another element in the same system. This is called and (-3) = 1, the Closure Law, and we say the system is closed with respect to addition. Second, each system is commutative with respect to addition: as seen in Fig. 2-16. -5 Third, each system is associative with respect to addition: -2 3 FIGURE 2-15 Fourth, our number system is distributive, meaning that 5 4 For example, -2 -3 1 -3 FIGURE 2-1612 FUNDAMENTAL CONCEPTS 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 13 P +3 +3 P 2B D +6 C mA nC -6 nB A 0 -3 -3 Q B FIGURE 2-17 FIGURE 2-18 mB OK Even the usual rules of multiplication of signed numbers, such as FIGURE 2-20 "minus times minus" and the like, that are difficult to justify in early algebra follow plausibly and without difficulty from the vector point of B, namely nB. Thus C is a linear combination of A and B. As to unique- view. In multiplying two numbers, ab, we have only to think of the first ness, suppose as being the scalàr a and the second as the vector b. Thus (see Fig. 2-17) C = + = + 2. 3 means the vector twice as long as 3 and pointing in the same direc- tion, namely, the vector 6; and 2(-3) is twice as long and pointing Then either (m₁ = and A and are parallel, a in the same direction; is a vector three times as long as -4 contradiction, or else each side is the null vector, in which case = but pointing in the opposite direction, namely, +12. All this follows from and n₁ = n2 since A 0 and 0, which proves that C can be ex- the meaning of scalar times a vector, KA, discussed earlier. pressed in only one way as a linear combination of A and B. Note that we Let us now consider two given vectors A and B both lying in a plane transposed vectors, as justified earlier. (Fig. 2-18). Starting at a given point 2B reaches a certain point BASIS THEOREM, part (b). In three-dimensional space let A, B, and C be P in the plane, which means that OP is the sum. Likewise, mA + nB, any three non-null vectors not parallel to the same plane, with no two of where m and n are any two scalars, reaches some point Q. The question them parallel to each other. Then any vector D can be written as a arises: Can any point in the plane be reached by starting at 0 and taking unique linear combination of A, B, and C: D = IA + mB + nC. an appropriate number of A's and B's? An expression like mA + nB, where m and n are any scalars, is called a linear combination of A and B. Proof. The proof is similar to that of part (a). If the vectors are made The question just asked is answered by part (a) of the following theorem, to emanate from the same point 0 (Fig. 2-20), a line parallel to C through and a similar question in three dimensions is answered by part (b). the tip P of D = OP will cut the plane of A, B in point K. Then OK, in the same plane as A, B, is a unique linear combination of A, B [by BASIS THEOREM, part (a). Any vector C can be written as a unique linear part (a)]; KP, being parallel to C, is a scalar times C; and, of course, combination of any two given nonparallel non-null vectors A and B in the D = OK + KP. Hence D is a linear combination of A, B, and C. As same plane: = mA + nB, m and n being unique scalar coefficients. to uniqueness, again suppose Proof. Through the tip P of = OP, draw a line parallel to A, and through the other end 0 of C, draw a line parallel to B. (See Fig. 2-19.) Hence These lines meet in a point K. Then = OK + KP. Since KP is parallel to A, it is a scalar times A, namely mA; and likewise OK is a scalar times (l₁ + = and if n₁ n2, we have C as a linear combination of A and B, and thus C parallel to the plane of A and B, a contradiction. Hence n₁ = n2. And mA we now see that l₁ = l₂ and m₁ = by the same reasoning as in the A proof of part (a). B nB The vectors A and in part (a) and the vectors A, B, and C in part (b) form what is known as a base (also basis, hence the name of the theorem) FIGURE 2-19 of the corresponding vector system or vector space, as it is sometimes14 FUNDAMENTAL CONCEPTS [CHAP. 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 15 called. The number of vectors in a base is the same as the number of mB dimensions of the space involved. The scalars l, m, and n are called the components of the vector IA + mB + nC. No base vector may be the mA + mB null vector. A+B B Although two- and three-dimensional vector spaces are sufficient for our purposes in this book, and are closely allied to the geometry we wish to study, vector spaces can be generalized to any number of dimensions A mA and, in fact, to an infinite number of dimensions (called Hilbert spaces). FIGURE 2-25 While we began with our geometry and have started to build a vector al- gebra, in the general case a complete algebra of scalars and sets of scalars 4. Given Fig. 2-25, (the components of the vectors) is organized as a set of axioms to form the (a) prove that m(A + B) = mA + mB. This is the Distributive starting point for further development of the theory, divorced from geo- metric considerations. These axioms define the space. Later we will show Law mentioned earlier. Considering m to be positive, one would need to show two things: (i) that mA + mB is in the same direc- in detail how this is done. tion as A+ B, and (ii) that |mA + mB| is m times as long as + EXERCISE GROUP 2-1 (b) discuss when m is negative. 1. In Fig. 2-21, = 3, |B| = 2, and = 6, with the angles as in- [Hint: Prove that the two triangles are similar, in each case.] dicated. To what linear combination of A and B is C equal? 5. Show how subtraction and division with signed numbers can be done with one-dimensional vectors. Illustrate with 2. In Fig. 2-22, |A| = 2, |B| = 3, and = 4, with the angles as in- dicated. C is equal to what linear combination of A and B? (-2), -2(3) 3(-2), -6 2 = -2 6 -3 -6 3. Given Figs. 2-23 and 2-24, (a) find the formulas for the lengths + B| and B| in terms of and 0, where is the angle between the two vectors A It will be advantageous if in two dimensions we choose as the base two and B. [Hint: Use the Law of Cosines.] perpendicular vectors, each of length 1: one horizontal, called i, the other (b) prove that |A| ≤ ± B ≤ |A| + |B|, in general. vertical, called j, as in Fig. 2-26. Then any vector r from a fixed point 0 When does the equals sign hold? to point P in a plane can be expressed uniquely as a linear combination of i and j: C = xi + yj. C 4 We see that 6 B x = 0 and 30° 120 2 3 30° y-axis A 3 B FIGURE 2-21 FIGURE 2-22 P:(x, y) r Aj B A B yj B A+B A 0 A i FIGURE 2-23 FIGURE 2-24 FIGURE 2-2616 FUNDAMENTAL CONCEPTS [CHAP. 2 2-1] VECTORS. DEFINITIONS AND OPERATIONS 17 where stands for and 0 is the angle from i to Γ. The indefinite lines through 0 along i and j are called the coordinate axes, the x-axis and y, 2 y-axis respectively, or a frame of reference. The scalar coefficients (x, y) are called the rectangular coordinates (or the cartesian coordinates, after I Descartes) of point P referred to 0 as the origin. Note that x and y may k zk k j be positive, negative, or zero. 0 y The vector = OP emanating from the origin is called the position i 0 i vector of P because its components are the rectangular coordinates (x, y) ri j of its terminal point P. Also, (r, where |r|, are called the polar coordinates of the point P. The relationship between the rectangular co- y = 1 ordinates and the polar coordinates, in the same frame of reference, FIGURE 2-27 FIGURE 2-28 should be kept in mind: = r 0, = r = |r| = = tan = sin y = , provided, of course, that x and y are not both zero. Of course, 0 may have many values, each differing from any other by a multiple of 2π nk radians (or 360°), but its quadrant is unique if we remember that x and y y are signed numbers. Thus for the point (-1, 2) is tan⁻¹ which li is in the second quadrant, and not which is ambiguous as to the quadrant. Note that r, being a distance, is always positive for any mj point (except the origin). However, see page 176 for an extended defini- FIGURE 2-29 tion allowing r to be negative. Similarly, in three dimensions, there are many advantages if three mutually perpendicular vectors of length 1, called i, j, and k, are taken in two dimensions even when x or y or both are negative. Similarly, in as the base. Then the position vector three dimensions, as is evident in Fig. 2-27, = = = has for its components the rectangular coordinates (x, y, 2) of the terminal point P. The unit vectors i, j, and k are also called the fundamental triad. and in general, the vector mj + nk anywhere in space has length Note that when we look down upon them from the positive z-axis, i and j + + We call m, and n the direction components of the vec- along the x-axis and the y-axis in three dimensions are oriented to each tor mj + nk. Sometimes they are referred to as direction numbers. other as the x- and y-axes are in two dimensions (see Fig. 2-27). The The angles B, and (Fig. 2-29) that A = mj + nk makes with three-dimensional frame of reference in this case is referred to as a right- the three coordinate axes are called the direction angles of the vector. hand system because the thumb, index finger, and middle finger of the Also, right hand correspond to the 2-, x-, and y-axes, respectively, when the m fingers are held mutually perpendicular. A left-hand system is also possible, = , = as shown in Fig. 2-28. It is evident from Fig. 2-26 that by the Pythagorean theorem, and COS Y = =18 FUNDAMENTAL CONCEPTS [CHAP. 2 2-2] ABSTRACT VECTOR SPACES 19 are known as the direction cosines of A. By squaring each and adding, we 2-2 Abstract vector spaces. We now take a glimpse at abstract vector see that spaces, to see one direction in which modern mathematics is moving. We + + = 1. can of course do no more than scratch the surface and present a few fundamental definitions and concepts. We can also speak of direction angles and direction cosines of a line, but Any set of elements (from now on often called vectors) on which the a line has two directions, hence two sets of direction angles and of direc- following two operations can be performed, and which satisfy the follow- tion cosines: Y and 180° - and their cosines. ing axioms, or postulates, is called a vector space. Usually we try to avoid this by considering a vector in one of the direc- tions. The operations: EXERCISE GROUP 2-2 (1) There is a process called addition by which two vectors of the sét 1. What is the unit vector in the direction 4i - 12j + 3k? can be combined, the result of which, called the sum, is also a (unique) vector in the set: x+y= (the Closure Law for Addition). 2. What is the unit vector in the direction = + yj + zk? (2) There is another process called multiplication whereby a vector X 3. Find the two vectors in the perpendicular to 3j and can be multiplied by any real number a, called a scalar, to give another of length 10. vector: = ax. 4. (a) A certain line has two of its direction angles given: 30° and B = 60°. Find its other direction angle and draw the line. How The postulates: many answers are there? (1) x+ y = X (Commutative Law for Addition). (b) Work the same problem for 45° and = 120°. (2) X + (y + z) = y) + Z (Associative Law for Addition). 5. What is the locus of all lines through the origin for which = 30°? (3) There is a unique vector, 0, in the set, called the zero element or null 6. What are the direction cosines of: vector, such that X + = X for each X in the set. (4) To every element X there corresponds a unique inverse element (a) each axis, in the set such that X + (-x) = 0. (b) the 45° line in the xy-plane through the origin, (c) the line through the origin in the first octant making equal (5) a(x + y) = ax + ay (a Distributive Law). (6) (a + b)x = ax + bx (another Distributive Law). angles with the three axes, and (7) a(bx) = (ab)x (another Associative Law). (d) the vector from (2, -3, 5) to (-1, 1, -7)? (8) X = X. 7. Find the angle between the two lines through the origin, in the first (9) = 0. octant, (a) for one of which a₁ = 45°, = 45° and for the other of which Such spaces are called linear spaces and their elements points, but their = = 60°; algebra is so much like our ordinary vector algebra that they are often (b) for one of which = = 60° and for the other of which called vector spaces. Strictly speaking, since the scalars are real numbers, COS = cos = the spaces should be called real vector spaces. We could allow the scalars (c) for one of which = = and for the other to be complex numbers (of the form a + bi, with a and b real), and the of which cos = cos γ₂ = resulting space would be a complex linear, or vector, space. (d) for one of which = = and for the other Note that vector spaces have a purely algebraic definition; geometry of which is not involved. = 1 Studying a subject from an abstract, or postulational, point of view is = helpful because it reveals the underlying mathematical structure. Very often topics in mathematics which appear to be quite different turn out [Hint: Each pair of lines above forms a plane in which one of the to have the same basic postulational structure, and study of one subject coordinate axes lies. Why?] helps us extend the boundary of our knowledge in the other.20 FUNDAMENTAL CONCEPTS [CHAP. 2 2-2] ABSTRACT VECTOR SPACES 21 Some immediate conclusions from the postulates, and their proofs, pendent. More generally, if a₁x₁ + + + = 0, where the follow: are not all zero, the being elements of a vector space S, the set of vectors X2, Xₖ is said to be linearly dependent. THEOREM (the inverse of x). 2. The outstanding property of A, B, C is that they are not linearly de- For + (-1 = + (-1.x) = (1 1) = 0; but the in- pendent: the only way that + mB + nC can equal 0 is for m, n all verse element is unique. (We used postulates 8, 6, 9.) to be zero; for otherwise, if for example, THEOREM 2. a0 = 0. = m n C For ax + = a(x + 0) = ax; but the zero element is unique. (We used postulates 5, 3, 3.) and hence A would be parallel to the plane of B and C. We say that THEOREM 3. If = x+ z, then = A, B, C are linearly independent. More generally, if + Add the inverse of X to each side (operation 1, postulates 1, 4, 3). = 0 only when a₁ = a₂ an = 0, the x/s being vectors in a space S, the set of vectors X2, is said to be linearly inde- THEOREM 4. If ax = ay, a 0, then = y. pendent. Multiply both sides by the scalar 1/a (operation 2, postulates 7, 8). The concepts of linear dependence and independence allow us to de- THEOREM 5. If ax = bx, X 0, then a = b. fine the number of dimensions of a vector space S, or more simply, dimen- Add -bx to both sides, hence b)x = 0. If a b 0, then sion of vector space S: b)x = 0 = b)0; hence = 0, a contradiction. (We used DEFINITION. If S contains a set of n vectors which are linearly inde- operation 1, postulates 6, 9, Theorems 2 and 4.) pendent, whereas every set of n + 1 vectors in S is linearly dependent, There often exists a subset M of a vector space such that + y and ax we say that vector space S is finite dimensional and that n is its dimension. are in M if y are. Such a subset M is called a linear manifold in S. If S is not finite dimensional, it is said to be infinite dimensional. We have This is why the Basis Theorems (a) and (b) gave rise to vector spaces of THEOREM 6. If M is a linear manifold in S, M is also a linear space dimension 2 and 3 respectively. (called a subspace of S). Note that in infinite dimensional spaces there exist linearly independent For, if X is in M, then - = is in M, and hence + (-x) = 0 sets of vectors, no matter how many in the set. For example, many sets is also in M; thus the nine postulates defining a vector space hold. of 100 or 1000 vectors can be found each of which is linearly independent (but this does not mean that all sets of 100 vectors in such a space are EXAMPLE. In our three-dimensional space, the subset of all vectors linearly independent). parallel to a given direction form such a subspace. In a finite dimensional space, the n vectors that are linearly independent Now we come in abstract linear spaces to what corresponds to linear (there may be many such sets) are called a basis of the vector space S, combinations of base vectors, considered in the Basis Theorems (a) and giving rise to this theorem: that any vector in S can be expressed as a (b) earlier. linear combination of a particular set of basis vectors. For if the set Two aspects of abstract algebra are inherent in the Basis Theorem, is a basis, then any vector X in S added to this set makes that any vector D could be represented as a unique linear combination a set of n + 1 vectors which is therefore linearly dependent; hence the coefficient of X must be different from zero when we write of the base vectors A, B, C, where A, B, C were any three nonzero vectors, no two parallel, and no one parallel to the plane of the other two. + 1. = IA + mB + nC becomes IA + mB + = 0. In words, we have a linear combination of the four vectors A, B, C, D result- (otherwise would not be linearly independent), and ing in the null vector, where the coefficients m, are not all zero. When this happens, we say that the vectors in question are linearly de- X = a₁ a a₂ a an aFUNDAMENTAL CONCEPTS [CEAP. 2 2-2] ABSTRACT VECTOR SPACES 23 22 This brings the subject of abstract vector spaces up to the point of de- It is seen that the nine postulates are now easily verified: velopment corresponding to the Basis Theorems in our geometrically based space. In this abstract case, everything was done purely alge- X braically with no reference whatever to geometry. Of course, in building since up the algebraic structure of abstract spaces we kept glancing back at our (x₁ In + = + Yn familiar geometric space, extracting what we thought might be the essen- tial features of identical structure in different sets of elements, even though X + z) = (x + the elements of one set may be of totally different character from the ele- similarly; ments of another set and even though the process of combination, called (-x) = (0, 0, 0) = 0, addition, may be totally different in the two sets. We will conclude by setting up the simplest, nevertheless important, and on for all nine. Hence we have a vector space. example of a real vector space, of which our ordinary vector space is a This is an n-dimensional vector space, because the n vectors special case. But it must be realized that there are many kinds of vector spaces, and very sophisticated ones, which we cannot even define here, = (1,0,0, 0), e₂ = (0,1,0,0, 0) since they are based on mathematics much beyond the scope of this en = (0, 0, 0, 1) course. The elements of this vector space we wish to set up are the so-called are linearly independent, since the only way in which n-tuples of real numbers, such as the set of all real number pairs (a, b), or all real number triples, (a,b,c), or the set of all number + = (a₁, an) quadruples, (a₁, etc., hence in general the set of all n-tuples can equal 0 is for a₁, an, each to equal 0, and because real. Now before we can be sure that this alleged vector space is one, we have to define the two operations, the zero ele- = + + xₙeₙ ment, the inverse of any element, and show that the nine postulates hold. means that all sets of n + 1 vectors are linearly dependent. Hence The operations: e₁, e₂, , en form a basis for this n-space. If n = 3, then e₁, e₂, e₃ become what we called i, j, k in our ordinary If X = space, and hence any vector = + But we also have spaces of any dimension: n = 4, or 5, or 100, etc., and all similar in structure to our three-space; and we even have an infinite y = dimensional space merely by having our elements consist of an infinite then number of scalars: (1) X + y shall mean (x₁ + + + X = ) and instead of a finite n-tuple. We see that X will also equal (2) ax shall mean (ax₁, + where The special elements: e₁ = e₂ = (1) 0 shall be (0,0, 0). (2) shall be en = (0, 0,1,0, The definition of equality: and any finite set of e/s is linearly independent. A basis therefore con- sists of an infinite number of elements. if and only if x₁ = = = Yn.24 FUNDAMENTAL CONCEPTS [CHAP. 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 25 We have thus been introduced to the concept of abstract linear spaces. P₁:(x₁, Going much further into the subject would take us far beyond the in- tended scope of this book. But as we continue to develop the vector P₁P₂ algebra of our three-space, we will apply, or annex, special features of this 22) algebra to that of the linear spaces. In fact, we can point out now that the concept of length of a vector in M 0 y our ordinary three-space has its counterpart in abstract linear spaces. The abstract properties of "length" are carried over into vector spaces to define what is called a norm for such a space. A norm on a linear space S is any real-valued expression, dependent on the vector X in vector space FIGURE 2-30 FIGURE 2-31 S, and we use as its value at point X the notation ||x|| (read: norm of x), having the following four properties: Subtracting these, we get (by the commutative, associative, distributive PROPERTY (1). + y ≤ X + The norm of a sum is less laws) than or equal to the sum of the norms. (Recalls PROPERTY (2). ||ax|| = X The norm of a scalar times a vector is the numerical value of the scalar times the norm of the vector. (A good way of remembering how to get a vector from one point to an- PROPERTY (3). 0. The norm of a vector is never negative. other point is to think "ending coordinate minus beginning coordinate" PROPERTY (4). ||x|| if 0. The only vector whose norm is 0 times i, j, and k, respectively, and adding). Hence the distance between is the null vector 0. the two points is As a matter of fact, property (3) follows from the other three properties, P₁P₂ = since 0 = = + (-x)| ≤ X + = In two dimensions, of course, this distance formula reduces to What we have called the length of namely, |A|, is thus a special case of a norm. Any real-valued function of A having the above four properties can be called a norm of A. In other words, then, the distance between two points is: For example, ||A|| could be + since this expression satisfies the four conditions above. (diff of the + (diff of the + (diff of the (if any) 2-3 Basic concepts of analytic geometry. We are now ready to de- EXAMPLE. The distance from (-1,7) to (3, 5) is = velop some of the basic ideas and formulas of analytic geometry. We do PROBLEM II. To find the midpoint of a line segment. this by setting up certain fundamental problems and solving them. There will be no need, for a while at least, to separate the discussion of Solution. Given P₁P₂, the vector from to Y2, two dimensions from that of three dimensions. and the midpoint of this vector (see Fig. 2-31). We see that PROBLEM I. To find the vector P₁P₂, and hence the length of P₁P₂ If since they are in the same direction, and since is this vector from a given point to another given point twice as large as |P₁M|. Hence, (see Fig. 2-30). Solution. It is evident that P₁P₂ = (see definition of subtraction). Since they are position vectors, and = = 2(y + 2(z26 FUNDAMENTAL CONCEPTS 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 27 -2,3) SUMMARY 1. The vector between two points is the difference of the position 3 equal parts to 5 equal parts vectors to the two points ("position vector to endpoint minus position vector to beginning point" or "ending coordinates minus beginning co- D:(x, ordinates" or simply "endpoint minus beginning point"). 2. The length of the vector between two points is P₁:(2,3,1) (differences of corresponding FIGURE 2-32 where Σ stands for "sum" (in words, the square root of the sum of the squares of the differences of corresponding coordinates). by the distributive laws. Since a vector is a unique linear combination of 3. The coordinates of the midpoint are each one-half the sum of the i,j, and k, we see that corresponding coordinates of the endpoints. = Y2 = 2(y 21 = 2(z 3 to 2 P₂ which, when solved for x, y, and give x = x₁ 2 y = z = + 2 D FIGURE 2-33 Hence, in words, the midpoint M of P₁P₂ is the point whose coordinates are 4. To find a point dividing a line segment in any given ratio, the sum of x's sum of y's , sum of z's method, rather than a formula, should be remembered, namely that the 2 2 2 vector from an endpoint to the point of division is a certain scalar (frac- of the endpoints. Naturally there are no to be considered in two tion) times the whole vector. Note that a line segment may be divided not only internally in a certain ratio, but also externally. In Fig. 2-33 dimensions. |P₁D| is to as 3 is to 5. D divides P₁P₂ externally in the ratio EXAMPLE. The point halfway from (-1, 7) to (3, 5) is (1, 6). of 3:5. The method used to find the midpoint of a line segment can also be used EXERCISE GROUP 2-3 for more general problems, as in the following example. To find the point D dividing the line segment P₁P₂ in the ratio of 3:5 where the end- 1. Given A:(-3, 2), B:(5, 4), draw AB and find: points are P₁:(2, 3, 1) and (see Fig. 2-32). Here P₁D = (a) The distance AB. (b) The midpoint. If D's coordinates are (x, z), we have (c) The vector BA. (x 2)i + 3)j + 1)k 5j + 2k) (d) The point on AB three times as far from A as from B. That is, the point which divides AB in the ratio of 3:1. (Two answers: (ending coordinates minus beginning coordinates). Thus because of the one internal, the other external, to AB.) uniqueness of representing the vector in terms of i,j, and k, we can solve (e) The tangent of the angle between i and AB (this is called the for D: (x, y, 2). It is Since D is between P₁ and we say that "slope" of line AB). D divides P₁P₂ internally in the ratio of 3:5. There is another external (f) The point on AB whose is 2. point D₁ to the left of P₁ (on the extended line) such that P₁D₁ and (g) The point on AB (extended) whose y-coordinate is 5. P₂D₁ are vectors in the same direction with lengths in the ratio of 3:5 (h) The point on the x-axis equidistant from A and B; also the point (see below). on the y-axis equidistant from A and28 FUNDAMENTAL CONCEPTS 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 29 (i) If (x, y) moves that PA = PB, simplify the equation that (d) d(x, y) ≥ 0 [this follows from (a), (b), (c): (x, y) satisfies. From high-school geometry, tell what curve P:(x, y) traces. 0 = d(x, x) d(x, + d(y, x) = 2d(x, y) (j) Change the equation in (i) above to one involving the polar coordinates (r, of point P. by setting = X in (b)]. A set S with such a distance function is called a metric space. 2. Given (4, 8, 11), B:(-3, 1, 4), C:(2, 3, -3), draw a schematic Now let S be a linear space that has a norm defined for it (page 24). figure and find: Then the linear space S becomes a metric space if we define d(x, y) to be (a) The lengths of the three sides of the triangle. yll. We must of course show that satisfies all four con- (b) The midpoints of the three sides. ditions for a distance function: (c) Vectors AB, BC, CA; why should their sum be the null vector? (a) d(x, = d(y, x) follows from property (2) of a norm: ||ax|| = (d) The point on AB whose y-coordinate is 5. ||x||, thus giving (e) The three points in the coordinate planes on AB (extended). (f) The angle between AB and BC [Hint: Use the Law of Cosines.] = = = (g) The two points of trisection of BC (internally). (h) The length of the altitude from B to side AC. (b) d(x, z) ≤ d(x, y) + d(y, z) follows from property (1) of a norm: (i) The area of triangle ABC. (j) The length of the line that bisects angle C. [Hint: 0/2 = X + X + y + 0)/2; use right-triangle trigonometry.] thus making (k) The radius and the center of the circle that circumscribes the triangle. [Hint: The hypotenuse is the diameter.] X = y +y ≤ + (1) Point D such that ABCD is a parallelogram (three answers). (c) d(x, y) = 0 if and only if = y follows from property (4) of a norm: 3. (a) (4,5) is the midpoint of a line segment, one endpoint of which is (-1, 2). Find the other endpoint. 0, if 0. (b) If (4, 5) divides the line segment internally in the ratio of 3:2 and one endpoint is (-1,2), what is the other endpoint (two answers)? Thus y 0 if X y 0; while, if X = we have 4. Given that the vector 2)i + (y 3)j + 1)k is a scalar, = = = X = 0. t, times the vector 3i 2k as point (x, y, moves. What geometrical figure does point (x, y, z) trace? Find x, y, each in terms of t. (d) d(x, y) ≥ 0 follows immediately from property (3) of a norm: ||x|| ≥ 0; hence 5. A point (x, z) moves that its distance to (3, 2, 4) is always 5. yll ≥ 0. What figure does (x, y, z) trace? Draw part of it (one octant). Write in simplified form the equation that (x, y, must satisfy. A linear space which is thus made into a metric space (by defining d(x, y) = is called a normed linear space or normed vector space. We can now extend to abstract spaces the idea of distance between two The most important normed linear spaces are those that have one points. additional property, that of completeness, a property which is rather Given any set S, not necessarily a space. A real-valued expression, difficult to understand, but which we now try to describe. Suppose we d(x, y), dependent only on pairs of elements, X, y of S, is called a distance have given a sequence of points, X2, Xₙ, from a metric space function on S, if S, such that the distance function d(xₙ, or in a normed linear space, approaches 0 as m, n increase indefinitely. Such a sequence (a) d(x, = d(y, x); of points is called a Cauchy sequence. It can be shown (in the calculus) (b) d(x, z) ≤ d(x, y) + d(y, z); that in our ordinary space this implies the existence of a limit point X (c) d(x, = 0 if and only if = y; of this sequence, namely, that there is a unique fixed point X for which30 FUNDAMENTAL CONCEPTS [CHAP. 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 31 L₂ 2 Simplifying, we get b3) 0 = a₁b₁ + a₂b₂ + B AB B A A In two dimensions b₃ are both zero. We call a₁b₁ + + a₃b₃ the 180° 0 y dot product of A by B and it is written B ("A dot B"); since it is a pure number, it is also called the scalar product of A by B. Sometimes it is called the inner product. Note that A. = B. A; also A. (B + C) = A B + A x Thus the formula for the angle between two vectors FIGURE 2-34 FIGURE 2-35 = + + and + d(x, approaches 0 as n becomes infinite. We say that = X in this case (the notation is the abbreviation for "limit of can be written as n becomes infinite"); also we say that the sequence of points a₁b₁ + a₂b₂ + converges to or is convergent. But in an abstract space, a Cauchy = AB = + + + , sequence need not converge: there is no guarantee that there exists a point X in the space to which the sequence of points converges. When every Cauchy sequence in S converges to a point in the space S, where we realize that in two dimensions the z-components are zero. An- we say the metric space is complete (otherwise incomplete). Normed linear other form is spaces that are complete are called Banach spaces, and these have become a₁b₁ increasingly important in modern mathematics. PROBLEM III. To find the angle between two lines L₁ and L₂ in EXAMPLE 1. A = 3i + 12j + 4k and = 5i + 3j 2k. Then Fig. 2-34. (Note: In space the two lines do not have to meet; they may be "skew" to each other.) = AB B = 169 38 = 43 Solution. We find a vector along each line, perhaps between two given or points on each line, and then transport the vectors to emanate from the origin. Then the angle between these two vectors, or its supplement, is = 43 the one desired. Suppose the two vectors thus obtained are A = a₁i and EXAMPLE 2. = k k = 1. Since we transported these to the origin, making position vectors of EXAMPLE 3. A. A = usually written A². them, the endpoints have coordinates (a₁, and (b₁, b₂, Joining these two points by a straight line, we get a triangle whose three vertices Since two vectors are perpendicular when and only when the angle are known and hence the lengths of the three sides (Fig. 2-35). The between them is 90°, we obtain the following important corollary, or "test angle 0 can be found by the Law of Cosines: for perpendicularity": Two vectors are perpendicular if and only if their dot product is zero. or EXAMPLE 1. and 2i - 3j + 6k are perpendicular because their dot product vanishes: 3 (-3) + 1 6 = 0. = + + + + + 2|A| |B| 8. EXAMPLE 2. i.j = j k = k = 0.32 FUNDAMENTAL CONCEPTS 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 33 EXERCISE GROUP 2-4 It immediately follows from (2) and (1) that 1. (a) By the formula for find the three angles of the triangle (x, + z) = (x,y) + (x, z) whose vertices are (4, 7, 11), (-3, 1, 4), and (2,3,-3). (b) Prove that 21j + 4k is perpendicular to the plane of the and from (2) and (3) that (x, = a(x, y). triangle in (a). 2. Prove that the vectors 3j + 6k, B = 3i + 3j 2k, These have their easily proved counterparts in our three-space, such as and C = 6i - 16j - 15k are mutually perpendicular. A. (B + C) = (Distributive Law), 3. Prove that the angle between any two vectors is also given by A. B = B. A (Commutative Law), 0 = cos + + cos Y₂ (mA) B = m(A. B) (a type of Associative Law); etc. (the direction cosines of the vectors-see pages 17-18). If we first prove a preliminary theorem, or lemma, we can then show 4. If vector li + mj + nk is perpendicular to 4i + k and is also that behaves like a norm. First the lemma: perpendicular to 2i - 3j + 4k, set up two equations for m, n according to the perpendicularity test and solve for and m in terms of n. Then (x, ≤ (x, find the vector proportional (parallel) to li + mj + nk whose com- ponents are all whole numbers by choosing an appropriate whole number To prove this, we consider value for n. (ax + by, ax + by) [by property (4)] 5. (a) Find the three angles of the triangle whose vertices are (3, 1), (5, -2), (6, 3). which, by properties (1), (2), (3), reduces to (b) Also find the area of the triangle. To find an altitude, the sine of the appropriate angle is needed. Recall that sin = (x, x) + 2ab(x, y) + y) ≥ 0. 6. If is the angle between If ab 0, we can rewrite this as A = + + and = (x, + 2(x, + (y, y) ≥ 0. prove that sin (= can be manipulated into the form + + This quadratic expression on (a/b), if >0, has no real roots, and the condi- tion for this is that the discriminant be negative: A |B|[CHAP. 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 35 34 FUNDAMENTAL CONCEPTS We can now show that can be defined to be ||x||, that is, it satisfies We remember that the set of e's was also a basis for the corresponding the four properties that define a norm: space (either the n-space or the infinite dimensional space). These corre- sponded to i, j, and k in our three-space: = k. = 0, (1) Is + Yes; for = = |k| = 1. If we were to pursue this study of abstract spaces further, one of the tasks would be to study the linear manifolds, or subspaces, generated by = (x, + 2(x, + orthonormal sets taken from a given space, and their relation to the whole space. Also it should be noted that inner product could be defined a bit ≤ + + more generally, to apply to complex linear spaces. and hence We recall that the most interesting normed linear spaces were the Banach ≤ + spaces, those which were complete (every Cauchy sequence having a limit point which is a point of the given space). The most interesting and im- (2) Does ||ax|| Yes, for portant spaces are those that are not only complete, but also infinite dimensional. Complete, infinite dimensional, inner-product ||ax|| = = spaces are called Hilbert spaces, in honor of the late David Hilbert, a giant (3) Is ||x|| ≥ 0? Yes, because ≥ 0 by property (4) above; figure in modern mathematics, who was among the first to use these spaces in attacking unsolved problems. and finally Of course, it is realized that in his first exposure to this subject, the (4) 0, if again follows from property (4) above. reader cannot help but be somewhat bewildered by the many types of In our ordinary space, if = mj + nk, we recall that the dot prod- spaces presented to him: metric spaces, vector spaces, complete spaces, uct (or inner product) incomplete spaces, normed spaces, Banach spaces, Hilbert spaces! But in spite of this, it is hoped that this necessarily short introduction to the A = + = or VA.A = concepts of abstract vector spaces has been interesting. They are a major field of study in present-day mathematics. corresponding to the norm of A (see page 24). In our space made up of n-tuples, where = (x₁, and PROBLEM IV. To find a vector perpendicular to two given nonparallel = we can define the inner product (x,y) to be vectors. + a generalized dot product for this n-dimen- Solution. Let L = + mj nk be the desired vector, perpendicular sional space. The norm of would then be to = + + and also perpendicular to B = + + From the test for perpendicularity, we have = and = a generalized vector length for n-space. If we try to do this in the corre- or sponding infinite dimensional space where = we = 0 and + = 0. would have to limit this space to include only those elements for which + (meaning that the sequence Since A is not parallel to B, A is not a scalar times B, and hence a₂, + + must be a Cauchy are not proportional to b₂, b₃, respectively. Therefore at least one of sequence). the two row determinants is not zero, say If (x,y) = 0, we say that X and y are orthogonal (corresponding to A and B being perpendicular when = and we use the notation In the above spaces, e₁ = e₂ = = (0,0, are mutually orthogonal; and indeed the Another way of saying this is to say that the matrix norm of each is 1: = 1. The set of e's is an example of what is called a₂ an orthonormal set (ortho indicating 1, normal indicating each norm = 1).36 FUNDAMENTAL CONCEPTS 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 37 is of rank 2. Hence we can-solve for m in terms of n; and using determi- If C = 2j + 4k and D = 3j - k, to find nants (Cramer's rule), we get we write 4 -3 -1 a₂ b₂ b₃ 4 = = , m Then a₁ a₂ a₁ a₂ b₁ b₂ b₂ b₁ b₂ b₁ b₂ = -2 -3 -1 4 = 14, |31| = = 19, = 1, Note that we factored out n, and interchanged the two columns of each numerator, thus changing the sign (see determinant review below). Since we are interested only in the direction of L, not its length, we and would be may take which is in the opposite direction, but of course still perpendicular n = b₁ to C and D (see Exercise 1, page 43). Further examples are: = i, k = j, as indicated by to avoid fractions, and hence we obtain 100 010 001 = a₁ 010 001 100' m = b₂ b₃ Thus respectively. Also, in Exercise 2, page 32, each vector L = is easily shown to be a scalar times the cross product of the other two. is a vector perpendicular to both A and B. SUMMARY This vector L is known as the cross product of A by B and is written 1. = a₁b₁ + + = cos = A. ("A cross B"). Since it is a vector, it is also called the vector product 2. The test for two vectors to be perpendicular is that their dot product of A by B. Sometimes it is called the outer product of A by B. be zero; or the sum of the products of the corresponding components The quick way to obtain the cross product of two vectors, is must vanish. to line up the components thus: 3. To construct a vector to two given vectors, find their a₁ a₂ cross product: b₁ b₂ with those of A on top. The components of the cross product are then the three determinants made up of columns respectively 4.i.j = j k = i.k = 0; = j.j = k k = = k, (the three determinants of the matrix in cyclic order 2, 3, 1 as shown in = i, k X i = j. Fig. 2-36). Also see Exercise 4, page 43. 5. A. + C) = (b₁ (b₂ (b₃ = + A - C. i's It will be worthwhile for us at this point to briefly review determinants. We give the main properties mostly without proof; the proofs can be 3 2 found in any good algebra book. First, a determinant is a square array of numbers that can be evaluated, i.e., a value can be assigned to it, by the following inductive process. (1) The value of the 2-row determinant k's a₁ a₂ FIGURE 2-36 b₂FUNDAMENTAL CONCEPTS [CHAP. 2 2-3] 38 BASIC CONCEPTS OF ANALYTIC GEOMETRY 39 is easily remembered by drawing the two diagonals THEOREM 2. The value of a determinant is unchanged if the rows and the columns are interchanged: (1) a₁ (2) a₁ a₂ e.g., a₂ = b₁ b₃ and thinking "diagonal (1) minus diagonal (2)." Note that the value of b₃ a 1-row determinant, is a by definition. (2) To evaluate a determinant of more than two rows, we alternately THEOREM 3. The value of a determinant changes sign if two rows add and subtract each element of a single row (or column) times its corre- (or columns) are interchanged: sponding minor (the minor of an element is the determinant left after deleting from the original determinant both the row and the column in = la b which the element lies). Thus if we let be the abbreviation for the THEOREM 4. If two rows (or columns) of a determinant are identical, following 3-row determinant, we have the value of the determinant is 0. b₂ THEOREM 5. The value of a determinant is unchanged if a multiple la b = b₂ = b₁ + b₃ of any row (or column) is added term by term to any other row (or b₃ column): b₂ b₁ b₁ also a₂ + b = la + + Whether one starts this alternate process by adding, then subtracting, It is this last theorem that helps us cut down on the computation in etc., or by subtracting, then adding, etc., depends upon which row (or evaluating a determinant, because by its use we can make all but one of column) we use to evaluate by minors: an element times its minor is the elements of a row (or column) zero. added if the row number of the element plus the column number of the Illustration: To evaluate element is even, and subtracted if that sum is odd. For example, the 3 2 4 -5 reason we have b₁ 2 4 -3 2 minus 4 3 -2 6 above is that is in the second row, first column, 2+ 1 being odd. Thus 5 -4 3 2 a₁ b₁ d₁ It would be convenient to have at least one element equal to 1; one a₂ b₂ d₂ way would be to add column (2) to column (3), giving a3 b₃ d₃ 3 2 6 -5 b₄ d₄ 2 4 1 2 its its its 4 3 d₂ 1 6 = minor 5 -4 -1 2 We see that the work involved in evaluating a determinant by minors Now "anchor" the second row and add (or subtract) sufficient multiples of a row (or column) would be decreased if some of the elements of the of it to the other rows to make all the other elements of column (3) zero: row (or column) were zero. -9 -22 0 17 THEOREM 1. If all the elements of a row (or column) have a common row (1) - 6(2) 2 4 1 2 factor h, this factor can be "taken out": 2 -1 0 4 (3) (2) a b ch = h/a 7 00 4 (4) + (2)40 FUNDAMENTAL CONCEPTS 2 BASIC CONCEPTS OF ANALYTIC GEOMETRY 41 Next evaluate by minors of column (3): the only element 0 is 1 in row These answers do satisfy the equations; for substitution in the ith (2), column (3), the sum of row and column being odd. Hence we have equation gives, after transposing the constant to the left and making one determinant out of elements times minors, -9 -22 -17 9 22 17 -1 2 -1 4 = 4 7 0 4 7 0 4 by Theorem 1. Now anchor column (2), add its double to column (1) and its quadruple to column (3), giving bi 53 22 105 53 105 = 0, 1 = 1(53 105 7) = 523. 7 0 4 since two rows are identical. Consider now the n equations on unknowns: THEOREM 6. Cramer's theorem. Given n first-degree equations and n unknowns: = + = k₂, + + = These are called linear homogeneous equations: linear, because they are of the first degree; homogeneous, because the constant in each is zero. where the determinant of the coefficients of the unknowns is not zero: We know that if la the only solution is = y 2 = = 0, la Then by Cramer's Theorem (each numerator determinant has a column of 0's). This is called the trivial solution. What will happen if la = 0? x = y = , 2 = abk... abc... etc., Will there be solutions? If so, will there be nontrivial solutions, that is, solutions other than 0, = 0, = 0, etc.? The following theorem answers these questions. where the numerator of each fraction is the determinant of the coefficients of the unknowns but with the column of coefficients of the corresponding THEOREM 7. The linear homogeneous equations on n unknowns: unknown being replaced by the column of constants on the right (the k's). Proof. To solve for (assuming there are solutions), multiply the first equation by the minor of a₁ in la b the second equation by the negative of the minor of the third by the minor of etc., and add. By putting together these elements times the minors as a single determi- have nontrivial solutions if and only if the determinant of the co- nant, we find that the coefficient of in the sum is la but the efficients vanishes: la = coefficient of y is 16 b = 0 (by Theorem 4); the coefficient of 2 To prove this theorem, suppose that (n 0 (referring is = 0, etc., while the constant on the right is to the first columns and 1 rows); we can then solve for Hence if there is a set I, y, 2, etc., that satisfies the equations, then I in the first 1 equations in terms of the nth column con- must be sidered as the column of constants, by Cramer's rule. If is the nth variable, we get Likewise, for y, 2, etc. x = (nu) y = (nu) , = b (nu) , etc.,42 FUNDAMENTAL CONCEPTS 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 43 where n stands for the column of the coefficients of (minus because the This checks with the solutions obtained when we found the cross product nu column is negative when transposed). L of two vectors A and B, where we had Let us interchange two consecutive columns at a time in each numera- tor, until the (nu) column the last column; this of course changes the sign each time. But in the numerator for y there will be one inter- + + = 0 change fewer than in that for I, and in the numerator for z one less than the number for y, and so on; so that the final signs alternate for the suc- (from 0 and = 0). Thus cessive numerators (whose columns are now in normal order). Hence = = 13 = when these fractions for x, 2, etc. are substituted into the nth equation, we will get for the combined numerator (over a common denominator) the various coefficients of the nth equation times their respective minors, with EXERCISE GROUP 2-5 signs alternating, giving 1. Prove that but that B = (Cross product is noncommutative; dot product is commutative.) on the left (with appropriate sign). Hence the solutions of the first 1) 2. Prove: equations, as given by Cramer's rule, will satisfy the nth equation if and only if = 0, which is what we set out to prove. That the and solutions are nontrivial is seen from the fact that the nth variable can be taken as any value say (n The k is an arbi- trary constant, meaning that all the sets of nontrivial solutions are pro- (The dot product and the cross product are both distributive over portional to each other; the ± is chosen to make = 1)|. addition.) (n = 0, but some other determinant of 1 rows and columns is not zero, we could solve for the unknowns in these columns 3. (a) Prove: A. = 0 and = 0 in two ways: in terms of the remaining one, in the same way. If all the 1) row (i) by evaluating, and (ii) by recalling the property of A X B. determinants = 0, meaning that rank of is less than 1), (b) Prove: in general can be written as the three-row we look for a nonzero determinant of the greatest number of rows pos- determinant of their components (in order): sible, and solve for the corresponding unknowns in these equations in terms of all the others. That these are also solutions of the remaining equations can be checked by substituting them; in each case a determinant of more rows is obtained, whose value is therefore zero. We omit the details, since they are similar to the case treated above where the rank [Hint: Write the determinant in terms of the minors of the first was 1. row, and compare with (B This proves (a) again. Note that the nontrivial solutions when the rank 1, turn out to be Why? It also proves that A. = B. = C. (A B). Why? = y = u = (n 4. Prove: B can be written as the symbolic determinant with signs alternating; or, in terms of numbered columns, ijk y = 2 = = 23 (n by expanding in terms of the minors of the elements of the first row. where we have omitted the proportionality factor k. One column of coef- 5. Prove: = sin 0 (length of the vector A X B). [Hint: ficients out of the columns is missing in each answer, the column of Equivalent to Exercise 6, page 32.] This is a very useful property; we coefficients of that unknown which is being found. shall find it helpful later.44 FUNDAMENTAL CONCEPTS [CHAP. 2 2-3] BASIC CONCEPTS OF ANALYTIC GEOMETRY 45 y 6. Prove: C = = C)A (a linear combination of A and B; remember that A. C and are scalars). This is the so- called triple cross product. 7. Prove: (A (nonassociative). [Hint: Use Exercises 1, 6 above.] What is the similar formula for A X (B C) 8. If A = 3i + 4j + k, B = 2i + 3j + 2k, = 4i + 2j + 3k; find I 2A + 4C. Also find C; B; A X B; B A; (A X B) X 0 0 and then check the theorems in Exercises 1 through 7 with these vectors. (a) (b) 9. (a) Find a vector perpendicular to the plane of the triangle whose FIGURE 2-37 vertices are (3, 1, 2), (4, 5, 3), (-2, 4, 6). (b) Find the three altitudes of this triangle, using the theorem in PROBLEM V. Choosing a coordinate system when a choice is permissible. Exercise 5. Solution. Note that a frame of reference can be chosen with any point 0 10. (a) Prove that and (-22, 35, -52) lie on as the origin, with the x-axis (or i) pointing in any direction. If we are a straight line by each of the following two methods: directions asked to prove properties of general geometric figures, say in two dimen- of vectors, AB = scalar times AC; lengths of vectors sions, we can choose the x- and y-axes 80 that certain vertices in the figure have rather simple coordinates. The resultant algebra necessary to AB + = prove the proposition would then be greatly simplified. For example: (b) Why do these two methods each prove that the three points In a triangle, one side could be made to lie along the x-axis and the op- lie on a line? posite vertex on the y-axis. Three zeros would thus be distributed among (c) Find the three points where this line intersects the three co- the coordinates, leaving three arbitrary quantities a, b, and in ordinate planes: (0, ?, ?), (?, 0, ?), (?, ?, 0). Fig. 2-37(a), as compared with six in Fig. 2-37(b). The midpoints of two sides in Fig. 2-37(a), 11. Do the same as in Exercise 10(a) and for the points (-2, 1), (3, -1), (-7, 3), reformulating (c) appropriately. 2 12. Find in two dimensions where li + mj. y y 13. In each of the following sets of linear homogeneous equations, check c) that the determinant of the coefficients vanishes, and solve by the method above (the 1 row determinants with alternate signs). (a) 3x 2y + 4z = 0 0 2x + y = 0 (b,0) 0 B:(b,0) + 4y 10z = 0 Trapezoid Parallelogram Here x, y, are 3/4 and respectively (numbers re- y (d,e) ferring to columns). Check these solutions in all three equations. (0,c) (b) 2x - 3y + 4u = 0 y 2z + = 0 9x - 4y + 2z 5u = 0 I 0 2y + 2u = 0 Here x, 2, are 3 3 4/4 2 and - Quadrilateral respectively. Check these solutions in all four equations. FIGURE 2-3846 FUNDAMENTAL CONCEPTS 2 make it immediately evident that a line joining the midpoints of two sides of a triangle is parallel to the third side (same y-coordinates); whereas to prove the same property in Fig. 2-37(b) would require a bit more work. This would be a good exercise for the student. Other geometrical figures and possible choices of axes (when a choice CHAPTER 3 is permissible) are shown in Fig. 2-38. [The coordinates of D in the paral- lelogram are as shown because CD = AB = a)i.] THE STRAIGHT LINE EXERCISE GROUP 2-6 3-1 Locus problems in general. A point moves according to a rule, thus tracing a curve. Finding the equation of the curve and its properties is 1. Choose axes skillfully, draw the figures, and find the coordinates of a locus problem. We look for a mathematical statement, usually in the the vertices and pertinent points for the following figures. Parts (a) form of an equation, which is true for every point on the curve and for no through (e) are in two dimensions; parts (f) through (j) are in three di- other point, and from which we can deduce properties of the curve. mensions. This is basic problem of analytic geometry. (a) An isosceles triangle, (b) a rectangle, (c) an isosceles trapezoid, For example: the rule according to which the point moves might be (d) parallelogram (vertex at 0), (e) an equilateral triangle. that its distance from a fixed point is always the same, thus tracing what (f) A tetrahedron (pyramid with four triangular faces), (g) a cube of we call a circle (we assume two-dimensional motion). To arrive at the side a, (h) rectangular parallelepiped, (i) a cone [two ways: mathematical statement in this case, let (a, b) be the fixed point and (x, y) (i) vertex at (ii) base in xy-plane; draw first octant only], the point moving always at the same distance (the radius) from (a, b) (j) a cylinder (draw first octant only). in Fig. 3-1. The distance formula from (a, b) to (x, or the magnitude 2. Prove that if a triangle is isosceles, the medians to the two equal sides of the vector r from (a, b) to y), is + and the are equal. (By median we mean the line from a vertex to the midpoint mathematical statement is that this distance always equals r. Thus we of the opposite side.) have + which simplifies to 3. Prove the converse: if two medians of a triangle are equal, the triangle is isosceles. 4. Prove that the diagonals of a parallelogram bisect each other. (In In three dimensions, of course, if (x, y, z) is always at distance from other words, próve that they each have the same midpoint.) (a, b, c), the surface of a sphere is traced with center at (a, b, c), and the equation satisfied by all points of the sphere (and by no other point) 5. Prove that the lines joining the midpoints of adjacent sides of any is quadrilateral form a parallelogram (even in three dimensions). 6. Prove that the sum of the squares of the three medians of any tri- angle is 1 the sum of the squares of the three sides. 7. Taking the vector yj + zk to any point (x, on the conical surface in Exercise 1(i), find the relation between x, y, 2 and h, a (the height and radius of the cone). [Hint: Use similar right triangles.] 8. Three vertices of a parallelogram are (-1, 2), (3, 1), (1, 5). Find (a,b) the fourth vertex. (Three answers.) 9. Do Exercise 8 for points (2,-1, 3), (5, 2, 6), (-1, 4, 2). 0 FIGURE 3-1 4748 THE STRAIGHT LINE [CHAP. 3 3-2] THE STRAIGHT LINE IN THREE DIMENSIONS 49 3-2 The straight line in three dimensions. It takes some thought to formulate a rule by which a moving point traces straight line. The usual P:(z, phrase which comes to mind is that two points determine a straight line. But what does the moving point, the point that generates the line, do in A:(a, b, c) relation to these two fixed points? In terms of vectors the answer is simple. In place of the two fixed points we think of fixed vector joining them, in L mj + nk the fixed vector either direction. Then the vector from one of the fixed points to the moving point must be in the same or opposite direction as this fixed vector, and hence a scalar times the fixed vector. B 0 If, as in Fig. 3-2, one of the fixed points is A:(a, b, c), the moving point is and the fixed vector is + mj + nk, we have = (x a)i + (y b)j + (z c)k = t(li + mj + nk) = FIGURE 3-2 t being the scalar. Hence, by uniqueness, In two dimensions, of course, z-coordinates are not considered and we x a = y b = tm, = get or = y-b l m a = y b = m n as the "equation of the line." (See Fig. 3-3.) Every point on the line, We call and no other, satisfies this equation. This is usually written b = a), where m/l = tan and (see page 51). Strictly x a y b = = speaking, even in two dimensions we should say "equations of the line" m n (plural), since the second equation is always = 0; but we take this for granted. the "standard equations of the line" (note the plural); while EXAMPLE 1. To find the equation of the line through A:(5, 4, 1) and = a + It, B:(3, 1, 6). (We use Fig. 3-2 as a schematic drawing.) y = b + mt, Solution. The fixed direction vector is = BA = 2i + 3j 5k = + nt, ("ending point minus beginning point"). The vector AP with moving y are called the "parametric equations of the line," t being the parameter. Fixed vector We can explicitly find the coordinates of points on the line by letting take on different values. In this form m, or n may equal zero, but not Moving point:(x, all three, of course, since the fixed vector cannot be the null vector. Note that the first column above contains the coordinates of the moving point, li + mj (a,b) the second column the fixed point, and the third column the coefficients mj of the fixed vector times the parameter. Its condensed vector form, as above, is 0 where r, are the position vectors to the moving point and fixed point respectively, and L the fixed vector. FIGURE 3-350 THE STRAIGHT LINE [CHAP. 3 3-3] THE STRAIGHT LINE IN TWO DIMENSIONS 51 y endpoint is 5)i + 4)j + 1)k. Hence the standard equa- tions of the line are 5 = (a,b) = 2 3 -5 mj We could, if we desire, leave them in the equivalent parametric form mj = 0 I = 4 + 3t 0 i 2 = 1 5t If we set t = 1, we get the point (7, 7, -4); t = 2 gives (9, 10, -9); FIGURE 3-4 FIGURE 3-5 and so on, various values of giving us various points on the line. EXAMPLE 2. To find the line through (0, 0, 1) and (2, 3, 1). the direction cosines. We note that if meaning 0, then tan 0 = m/l, where mj is a fixed vector along the line (Fig. 3-5). Solution. This line is parallel to the zy-plane, since the two points have This is true not only when m are both positive but also when either is the same The moving vector, using (0,0,1) as the anchor negative. Angle may be taken between 0° and 180° by taking the fixed point, is + yj + 1)k; this is a scalar, t, times 2i + 3j + Ok, vector so that m is positive (or zero), but there is no objection to taking the fixed vector. Hence the parametric equations are 2t, y = 3t, the opposite vector and having 0 clockwise. We call tan 0 the slope = 1; the other form is x/2 = y/3, = 1 [it would not do to write of the line. It is denoted by the Greek letter (lambda): = tan = m/l. = 1)/0!]. Since mj = + we can take as the direction vector of the line. As (x, y) traces the line through (a, b) of slope the vector it EXAMPLE 3. Identify the locus whose equations are makes with (a,b), a)i + b)j, is a scalar times hence 4x the components are proportional: = = 5 4 2 Solution. We may put these in standard form by dividing the numer- 1 = y-b or y b = ator and denominator of each fraction by the coefficient of the letter, getting Every point on the line through (a, b) and of slope satisfies this equa- tion. Conversely, any point (x, y), not (a, b), that satisfies this equation H must lie on this line; for y-b y b a) or = Hence the locus is a straight line through the point (1, having a the direction vector - + k, or - 15i + 16j + 12k if we lengthen says that the vector from (a, b) to (x, y), it appropriately so as to have whole number coefficients (multiplying it by 12). (x + (y b)j = (x a) + 3-3 The straight line in two dimensions. If we restrict ourselves to two is a scalar 0 times the fixed vector + and hence always is in the dimensions, there are some distinctive and simplifying features of straight same (or opposite) direction as that the point (x, y) must lie lines. First, the line always intersects the x-axis, except when parallel on the line. The point (a, b) obviously satisfies the equation and lies on to it, the direction angle from i to the line being 0 (Fig. 3-4). Even if the the line. Thus any point satisfying the equation b = a), line is parallel to the x-axis we can say It is easier and more ele- including (a,b) itself, lies on the line. We call b = a) the gant to work with the trigonometric tangent of this angle rather than with point slope equation of the line.52 THE STRAIGHT LINE 3 3-3] THE STRAIGHT LINE IN TWO DIMENSIONS 53 y y y y y 3 (-1,3) (-1,3) 4 (-1,2) (2,1) -3 150° 0 I 0 (1,0) 0 FIGURE 3-6 FIGURE 3-7 FIGURE 3-8 FIGURE 3-9 Of course, in the above discussion it was taken for granted that always well have gone 4 to the left (-4i) and 3 up (3j), starting at (-1,2). z = 0, which is in reality the second equation of "the equations of the The equation, of course, is line." y 2 = or EXAMPLE 1. To find the equation of the line through (1, 0) at an angle of 150° with the (Fig. 3-6). Note that since the fixed vector joining two points has components obtained by respectively subtracting beginning coordinates from ending Solution. coordinates, and since slope is the ratio m/l(=tan a formula for the = tan 150° = 1 slope of a line joining two points is: difference of the y-coordinates Hence the equation of the line is = difference of the x-coordinates y 0 = 1 (x 1) or each difference being taken in the same order. Thus the slope of the line in Example 2 above is when simplified. 2 EXAMPLE 2. To find the equation of the line through the two points 2 = 3 (-1,3) and (2, 1) (Fig. 3-7). It should be mentioned that lines parallel to coordinate axes are rather Solution. The direction vector is + 2j and the slope The special. A line parallel to the y-axis has no slope, since tan 90° does not equation of the line is exist. In the slope formula given above there would be a zero denominator, since the z-coordinates of any two points on the line are the same. This y = 2) or last fact, that the x-coordinates of all points on the line are the same, gives The same equation would ensue if the point (-1, 3) had been used instead us the equation of the line, namely, x = constant. Thus (Fig. 3-9) the line through (-1, 3) parallel to the y-axis has for of (2, 1). its equation, = -1. EXAMPLE 3. To draw a line through (-1,2) of slope - (Fig. 3-8). Likewise, the equation of a line parallel to the x-axis is y = constant. The line through (-1,3) parallel to the x-axis has the equation y = 3. Solution. Starting at the point (-1,2) we go parallel to the x-axis a Note that horizontal lines do have a slope. Since = 0°, tan 0 = 0. distance of 4 to the right (note that 4 is the denominator), and from Thus any line always has an equation of the first degree. (We will (the numerator) parallel to the y-axis (down three units), and prove the converse in the next section.) This is why a first-degree equa- join. This is equivalent to taking the vector 4i - 3j. We could equally tion is called a "linear" equation.54 THE STRAIGHT LINE 3 3-3] THE STRAIGHT LINE IN TWO DIMENSIONS 55 y SUMMARY The straight line is characterized by having a fixed point (a, b, c) and h a fixed direction vector mj + nk which are either given or can be found from other given information. The equations of the line in para- 0 metric form are: = a + FIGURE 3-10 y = b + mt, 2. Find the point on the line in Exercise 1 above whose y-coordinate = + nt. is 3; also find where the line cuts the two coordinate axes. (These last two points are called the and the "y-intercept" of the line.) If we solve for t, we get the standard equations: 3. Draw the line through (-1, 2) and (3,0). Find its equation by = y b = = t, the point slope formula; also its two intercepts. m n 4. (a) The x-intercept of a line is (a, 0) and its y-intercept is (0, b), neither of which is the origin. Starting with the point slope provided m, n 0. In two dimensions, equation, show that the equation can be written as = y-b m which is another standard equation worth remembering. can be written as the point slope equation: (b) Find the equation of the line at distance h from the origin, α being the angle between the x-axis and the segment of length y b = a), h (Fig. 3-10). [Hint: Find the two intercepts in terms of h, a and use the answer in Exercise 4(a).] where 5. Draw, and find the equation of, the line through: = m = tan = difference difference of of the the y's x's (a) (3, 1), (3, 4). (b) (1, 0), (5, 0). of two different points. A line parallel to the x-axis has for its equation (c) (0,0) at an angle of 45° with the x-axis. (d) (0, 2) and making an angle of 60° with the y-axis. (Two answers.) (e) 4, -3. = constant 6. Prove that the points (6, 4), (-3, 5), (24, 2) are collinear (lie on and its slope A line parallel to the y-axis has for its equation same line) by proving that one point satisfies the equation of the line de- termined by the other two. Describe two other methods of proving that these points are collinear. = constant 7. Find the equations of the three medians of the triangle with vertices (a, 0), (b, 0), (0, c). and has no slope (the slope does not exist). 8. Find the trisection point of one median (the one of the way from the vertex to the midpoint of the opposite side) and prove not only that EXERCISE GROUP 3-1 it satisfies the equation of each median, but that it is also the trisection 1. Draw, and find the equation of, the line through (2, -3), with point of each of the other two medians (use vectors). Thus the three slope of (Ans. 2x - 3y 13). medians of any triangle are concurrent (go through a common point).56 THE STRAIGHT LINE [CHAP. 3 3-4] THE EQUATION OF THE FIRST DEGREE 57 9. The point in which two nonparallel lines meet must satisfy both Conversely, it is also true that given any first-degree equation, a straight equations. Find the point of intersection of 4y = 1, 4x + 5y = 22. line can be found which has this equation. To prove this, let us find the 10. Do Exercise 8 above by finding the point of intersection of two of equation of a line through the point (0, -C/B) with the medians by solving their equations simultaneously and proving that y = (x 0), it satisfies the third equation. 11. Prove that the two nonparallel sides of a trapezoid and the line assuming B 0. Clearing of fractions, we transpose all terms to one joining the midpoints of the parallel sides are concurrent. (See page 45.) side, and we get the equation we started with, By + = 0, as 12. Prove also, that the two diagonals of the trapezoid and the line the equation of the line of slope - A/B and y-intercept (0, through the midpoints of the parallel sides are concurrent. If = 0, the equation becomes Ax + = 0. To match a line to this equation, we take the point (-C/A, 0) and find the line through this SOME THREE-DIMENSIONAL EXERCISES point parallel to the y-axis: it is -C/A, which simplifies to the given equation. Note that A, B cannot both be zero, or the equation would 13. Find the equations of the line through (3, 5, 1) and (-2, 3, 2) both not be of the first degree. in standard form and in parametric form. being the equation of a straight line, is for this 14. The same for (2, 4, 3), (4, 2, 3). reason called a linear equation. 15. The same for (4, 1, 2), (6, 1, 2). The above proof, although it appears to be somewhat sly, is really not unnatural. When we meet a linear equation, we like to know if there is a 16. Identify the line whose equations are 2x - 1 = 4y + 8 = 5. point that satisfies it. The and y-intercepts, if any, are the most obvious; Find its direction vector; also three distinct points on it. and the point slope form b = a) suggests dividing by B in 17. The same for the line 2x = 3, 4y = 5 is arbitrary). + By = 0 to get the coefficient of x (after transposing). Thus we 18. Find standard equations for the line 2y + 5z = 6, 2x + y start with a line through a given point and having a given slope, pulled, 3z = 0. [Hint: Find two points on it and the direction vector.] as it seems, out of thin air, and this legitimate line turns out to have the same equation we began with. 19. Find the equations of the line perpendicular to the plane of (3, 4, 2), Here is another proof, by vectors. Let (x₀, yo) be a point that satisfies (-1, 5, 3), (2, 1, 4) and passing through the origin. [Hint: The cross the equation. There is at least one such point: (-C/A, 0) or (0, -C/B) product of two vectors in the plane is perpendicular to the plane.] or both, since A, B are not both zero. We then have Ax + By + = 0, 20. Find the equations of the line through (2, 1, 5), perpendicular to = 0; and subtracting we get + B(y yo) = 0. and meeting the line This says that (x, y) moves that the vector + yo)j is 1 = 3 always perpendicular to Ai + Bj, since their dot product is zero. Hence = 3 4 2 the locus of (x,y) is the line through (x₀, yo) perpendicular to Ai + Bj; its direction vector is Aj, and therefore its slope (= m/l). [Hint: It is also perpendicular to the normal of the plane determined by the given point and given line.] y 3-4 The equation of the first degree. It was noted in the preceding 2j section that the point slope equation, as well as the equations of lines (0,2) parallel to the axes, were all of the first degree: -3i y b = a); = constant; y = constant I 0 can all be put into the form + By + = 0 where A, B, C are con- stants, A, B not both zero. FIGURE 3-1158 THE STRAIGHT LINE [CHAP. 3 3-5] ANALYTIC PROPERTIES OF STRAIGHT LINES 59 EXAMPLE. To identify and draw the curve whose equation is 2x (3) the equations of two parallel lines can be written 3y = 6. Solution. The equation, being of the first degree (linear), represents a straight line. Its slope A/B = Its y-intercept is (0,2). To draw merely having different constant terms. the line (Fig. 3-11) we start at (0, 2), go 3 to the left (-3i), then up For example: The line through parallel to 2x + 5y = 11 starts 2 (2j), and join. Note that the slope indicates that a fixed vector off the same way: 2x + 5y = constant; and the constant must be of such on it is + 2j. In general, the line Ax + By + = 0 has a direction a size that must satisfy the equation, giving 3 + 5(-1) = vector that can be written + Aj or Bi - Aj, since the vector constant = 1, all of which can be done mentally. The answer is mj means the slope is m/l = tan which in this case A/B. 2x 5y = 1. To find a line perpendicular to Ax + By 0 we seek a vector EXERCISE GROUP 3-2 + mj whose dot product with - Bi + Aj vanishes: -Bl + Am = 0. 1. Find the slope, the y-intercept, and draw. When slope and inter- Hence m/l, the slope of the desired line, is B/A; and the equation of a line cept do not exist, so state. perpendicular to + By = 0 can be written Ay = const. (a) 3x - 4y = 6 (b) + 3y = 6 (e) 4y + 5 = 0 In other words, the coefficients of I and y are interchanged and the sign (d) 3x + 2y 0 (e) (f) 4x + 3 = 0 of one of them is changed. = Another way of saying this is to say that the slopes of two perpendicular (g) + y b = 1 (another standard equation) lines are negative reciprocals of each other: a (h) y = mx + b (still another standard equation) (i) bx + ay = 0 = versus (j) = 9 or that + = 0. (k) y² = 0 (two lines; why?) For example, the line through (3, 4) perpendicular to 2x + 5y = 9 is (1) xy(2x 3y + 4) = 0 (three lines) 5x - 2y = 7 (since = (m) a + sin = h (indicate h and a in your figure) Moreover, the whole family of lines having slope is given by (n) x = 3 + 2t, y = 3t 2x + 5y = C, while every line perpendicular to each member of this (o) = a + y = b + family, of slope is given by 5x - 2y = C', where C and C' are arbitrary 2. If and are each a linear expression: constants. EXAMPLE. To find the line perpendicular to 3x + 4y = 1 that makes + with the coordinate axes a triangle of area 8. what does the graph of = 0 represent? What about mu + no = 0, Solution. The temporary answer is 4x - k and its intercepts (Fig. 3-12) are (k/4,0) and (0, -k/3). Of course k may be positive or m and n constants, not both 0 (see next exercise)? 3. Draw the two lines u 2x - 1 = 0, v 3x + 4y 10 = 0 IV and find their point of intersection. Now draw = 0, 3u + = 0, 30 = 0. From your observation of all five of these lines (in one figure), what can you say of mu + = 0? k negative 3-5 Analytic properties of straight lines. Since parallel vectors are a scalar times one another and conversely, it is obvious that 0 (1) parallel lines have the same slope; (2) lines with the same slope are parallel; FIGURE 3-1260 THE STRAIGHT LINE 3 3-5] ANALYTIC PROPERTIES OF STRAIGHT LINES 61 negative. The lengths of the two sides of the triangle are |k|/4, y B and the area is one-half the base times the altitude or A 1 k k = 2 4 3 24 and this must equal 8. Hence = 8, k = The final answers are 3y = one of these lines making a triangle in the second quadrant, and the other in the fourth quadrant. The absolute values were 0 necessary to insure positive numbers for lengths. FIGURE 3-13 EXERCISE GROUP 3-3 7. Through the point (4, 1) there are two lines that make with the co- In the following exercises, find the equations of the lines and draw a ordinate axes a triangle of area in the first quadrant. Find their equa- figure for each. Where possible, set up the temporary answer and find the tions. value of the constant mentally, as indicated in the examples above. 8. There are two other lines through (4, 1) that make triangles of area 1. The line through (-5,7) perpendicular to 5y = 10. in quadrants II and IV. Find their slopes. 2. The two lines through (-1, 2), one parallel and one perpendicular = PROBLEM I. To find the angle between two given lines. If the two slopes are we find two fixed vectors on the 3. The line through (a, 0) perpendicular = 1. lines (Fig. 3-13). These could be + B = The angle between these vectors is given by: 4. In the triangle with vertices (a, 0), (b, 0), (0, c) find the equations of (a) the three altitudes. = or (b) the three perpendicular bisectors of the three sides. AB 5. In the preceding problem, prove: When this is positive, would be the angle less than 90°; when negative, (a) that the three altitudes meet in a point H (called the orthocenter would be between 90° and 180° and in either the clockwise or the counter- of the triangle). clockwise direction, since = We see again that the con- (b) that the three perpendicular bisectors also meet in a point dition for perpendicular lines is = 0 (cos 90° = 0). (called the circumcenter). We can also find angle by remembering that (c) that these two points in (a) and (b), together with the point of intersection of the three medians, G: sin = (Exercise 5, page 43). Lining up the components of A and B we get (called the centroid) that were found in Exercise 10, page 56, are 0 collinear (all lie on a straight line); and that G divides the line 1 0 segment in the ratio 2:1. Note: Keep a record of the coordinates from which |23| = 0, |31| = 0, |12| = (see page 36). Hence of these three points and the equation of their straight line for or = - and Exercise 2, pages 100-101. sin = 6. Find the two lines of slope that make with the coordinate axes a ; triangle of area square units.62 THE STRAIGHT LINE [CHAP. 3 3-5] ANALYTIC PROPERTIES OF STRAIGHT LINES 63 y L₂ L₁ L₂ 0 0 FIGURE 3-14 FIGURE 3-15 FIGURE 3-16 EXAMPLE 1. The angle from the line whose equation is 3x - 2y = 5 and in this case must be taken as a counterclockwise angle less than 180°, to the line whose equation is 4x + 3y = 2 (Fig. 3-16) is given by but there are two values for it since sin (180° Φ) = sin Another approach that produces a neater formula is to recall that the tan = -1- = 17 6 slope stands for the tangent of the angle counterclockwise from the 1 + positive direction of the x-axis to the line: = tan = tan (Fig. 3-14). We see that = since the exterior angle is the sum EXAMPLE 2. There are two lines through (2, 1) that make 45° angles with 2x - 3y = 6. Find their equations. of the opposite interior angles. Hence the counterclockwise angle from the line L₁ to L₂ is = and Solution. Let the slope of the line we seek be (Fig. 3-17). Consider the answer to be L₁ and we get tan = tan = 1 tan + tan tan tan or tan = + tan 45° = 1 = 1-1 It is to be noted that is the counterclockwise angle from line 1 to line 2. The proof above might have to be slightly modified for a different figure. or 1 + = and = and = making one answer: For example in Fig. 3-15 is not However, does equal y 1 = 2) or I + 5y = 7 (line L₁ in the figure). Now con- + (180 = (180° + and since tan (180° + = sider the answer to be line L₂ and we get tan the same formula results. The student should experiment with other figures. It will be seen that always equals + a multiple tan 45° = 1 = of 180°, hence tan = tan or À = 5 and y = 9 (which is the line L₂ shown in the figure). This formula for tan checks with the preceding two formulas, since Slope sin = tan = 4 but the formula Slope (2,1) tan = 1 45% I 0 45° actually gives the counterclockwise angle from line 1 to line 2, while the other two are noncommittal as to which line is the initial side of the angle and which is the terminal side. FIGURE 3-17THE STRAIGHT LINE [CHAP. 3 3-5] ANALYTIC PROPERTIES OF STRAIGHT LINES 65 64 EXERCISE GROUP 3-4 for the length of the cross product is = KL sin (see Exer- cise 5, page 43) and we note that = K sin so that Draw all figures. 1. Find the acute angle between 3x - 4y + 1 = 0 and 2x + 3y = 5. d = K sin = K L L sin 0 = LXK L 2. What is the angle between the and 5x + 12y = 1? 3. Find the angle between 4x + 3y = 5 and the y-axis. We obtain LXK by the + |31|j + k process. Lining up the coefficients of i, j, k in matrix form 4. Find the two lines through making 45° angles with 3x - 4y = 7. B A 0 5. Find the two lines through (2, 2) making 60° angles with + 2y = 3. C + 0 6. Find the three angles of the triangle whose vertices are (2, 1), (-1, 2), A (3, -2). As a check, see if they add up to 180°. This requires use of a we get table of trigonometric functions. 7. Do the same for the triangle whose vertices are (1, 2, -3), (4, 6, 9), LXK = Bs + A k. (2, -3, 4). Hence = + + which is the coefficient of k (made positive), and which is seen to be the linear expression PROBLEM II. To find the distance from a given point (r,s) to a given in the equation of the line L, with r, 8 replacing x, y in it. Thus line + By = 0. Note that "distance" of course means per- pendicular distance. d = Solution. If the line is parallel to the the answer is obvious: merely the difference of the y-coordinates. If the line is not parallel to is the distance from the point (r, s) to the straight line whose equation is the x-axis, it has the x-intercept (-C/A, 0). (If the line is parallel to the = 0. An alternative form is y-axis, the answer is also obvious.) The formula for distance d can be nicely obtained by the use of vectors d = ± (Fig. 3-18). The vector from (-C/A, 0) to is + K = where the ± means that the correct sign must be chosen to make d positive. EXAMPLE 1. The distance from (2, 1) to 5x - 12y 3 = 0 is (ending point minus beginning point), and a vector along the given line is L = Aj since the slope is -A/B. We recall that the formula 12 3 = 5 13 L EXAMPLE 2. There are two points, whose r-coordinates at a distance of 6 from the same line, 5x - 12y 3 = 0 (Fig. 3-19). Find K their y-coordinates. Solution. We have 0 ± 5(-3) 13 12s 3 = 6, or = FIGURE 3-18 Hence 5 or -8, and the two points are (-3,5) and (-3, -8).[CHAP. 3 3-5] ANALYTIC PROPERTIES OF STRAIGHT LINES 67 66 THE STRAIGHT LINE y y (-3,5) 6 (2,1) 5 0 13 5x - 12y 3 0 6 FIGURE 3-20 in one of these regions, positive in the other region, and zero only on the line itself. FIGURE 3-19 9. Find the inequalities describing the region completely in quadrant I between the lines 3x - 5y + 2 = 0 and 2x - y = 1. Check by a point. EXERCISE GROUP 3-5 10. Indicate in a figure the region given by the inequalities In Exercises 1 through 4, find the distances from the given points to the given lines. Draw the figures. and 3x - 2y > 0. 1. (-3, 4) to 5x - 2y = 3 Check by a point. 2. (-2, 5) to 7x + 3 = 0 11. Use the method similar to that in the text, namely, the absolute value of the cross product of two vectors, to obtain in three dimensions 3. (1, -3) to 4y + 5 = 0 the perpendicular distance from the point (7, 3, 7) to the line (x - 3)/4 4. Origin to 3x - 2y + 6 = 0 2)/3 = 1)/2. [Hint: Take vector K from point (3, 2, 1) to (7, 3, 7); sin wanted.] 5. Find the altitudes of the triangle with vertices (a, 0), (b, 0), (0, c). 12. Solve Exercise 11 for the general case: the distance from (r, in 6. Find the distance between the two parallel lines: 3x + 2y = 6 and Fig. 3-21 to the line 4y = 9. I : = = 7. Prove that the distance between the parallel lines whose equations m n are + By + C = 0 and Ax + By + C" = 0 is - K 8. Prove that if + By + C is negative for point (r, then Ax + By is negative for every point on the same side of the line Ax + + mj + nk By 0 as (r, 8) is, and positive for every point on the other side (a,b,c) (Fig. 3-20). Hence the straight line Ax + By + C = 0 divides the plane into two parts, in which the linear expression Ax + By + is negative FIGURE 3-21THE STRAIGHT LINE 3 3-6] THE STRAIGHT LINE IN POLAR COORDINATES 69 68 -3 a 3-6 The straight line in polar coordinates. Rotation of axes. If we superimpose a polar coordinate frame of reference on a rectangular frame A Moving point 2 of reference so that the pole and origin coincide and the prime direction, OA, falls along the positive x-axis (Fig. 3-22), we see that the relationship between the two sets of coordinates for the same point is h -4 = 0 A = T sin 0. Thus by means of these substitutions the equation of any curve in rec- FIGURE 3-24 FIGURE 3-25 tangular coordinates can be transformed into a polar coordinate equation or (cos a + sin 0 sin a) = h where of the same curve. The straight equation Ax + By = C then becomes, in polar coordinates, h = r(A 0 + B sin = C, and the equation of the straight line finally reduces to where C can be thought of as being positive (by changing signs throughout the equation if necessary). Thus 3x + 4y + 10 = 0 becomes = h r(-3 4 sin = 10. as the standard equation in polar coordinates. The significance of this is that h, which we were careful to make posi- Manipulation on A B sin 0 will prove to be interesting. tive, is the distance from the origin to the line and a is the angle from If we construct in the appropriate quadrant a right triangle of sides A, B to this distance h. This can easily be seen by reversing the problem, (Fig. 3-23 illustrates the case where A is negative, B positive), the hy- namely, starting with a line at distance h from 0 (Fig. 3-24). As the point (r, traces the line we see that h, T and the line form a right triangle potenuse is Hence with hypotenuse T, adjacent side h, and angle Therefore B T cos a) = h. Note that when (r, is on the other side of (h, sin the angle is but since = α), the same equa- tion is obtained. If we divide both sides of the given equation by we get Thus in our example 3x + 4y + 10 = 0, which became r A + sin = B² r(-3 4 sin = 10, we divide by = 5, to get T cos = 2, where a = tan y is the third quadrant angle of a 3, 4, 5 triangle of reference, and hence our line is at a distance 2 from 0 turned through angle a as shown in B y a Fig. 3-25. There is no ambiguity of quadrant if we write A 0 0 tan FIGURE 3-22 FIGURE 3-23 not cancelling the minus signs.THE STRAIGHT LINE 3 3-6] THE STRAIGHT LINE IN POLAR COORDINATES 71 70 y y L 330° 60° 0 0 330° A h I 1 A 0 2 0 (0, -3) h FIGURE 3-26 FIGURE 3-27 FIGURE 3-28 FIGURE 3-29 FIGURE 3-30 Since T COS = h (or = h) is the equation of the line L perpendicular to h units from 0, as shown in Fig. 3-26, we see that a) = h EXAMPLE 1. To identify and draw the curve whose equation is merely results in rotating line L about 0 through angle a. In other words, sin = 3. replacing in a polar equation by results in a rotation about 0 Solution. Of course one method is to change to rectangular coordinates, through an angle a; or, to put it another way, the equations r' = T and where = T and y = sin giving y = 3, a straight line = a transform the coordinates (r, in one frame of reference to of slope the angle with the x-axis being tan⁻¹ = 60°, y-inter- (r', the coordinates in a rotated frame of reference. cept = 3 (Fig. 3-28). Going back to the superimposed rectangular frames of reference (Fig. For curves other than straight lines this method may not work well, 3-27) and remembering that as will be seen later. One should know the standard equations of certain important curves in polar coordinates as well as their standard equations = r' = in rectangular coordinates. The straight line is a curve for which both = r' sin = sin α), should be known. In polar coordinates the standard equation is we get r α) = h, and we try to make the given equation = cos + r sin sin α, = r sin T sin a, look like it. by the trigonometric addition formulas, and we see that The adjacent side of and opposite side of - 1 puts in the fourth x' = x + y sin and y' = sin + quadrant (Fig. 3-29) as either 330° or -30°. The hypotenuse is 2; and dividing by it we get since = 0, y = T sin Conversely, since = (0 we have = + a and r cos sin = or 330°) = = T 0 = r sin sin a, thus the curve is a straight line perpendicular to the hypotenuse at a = r sin = T sin a + sin a, distance of from the origin (Fig. 3=30). This distance of is informa- by the addition formulas; and since x' = T COS y' = T sin we see tion the first method did not give. (However, see Exercise 4(b), page 55.) that EXAMPLE 2. r sin (which is = in rectangular coordi- = x' cos a - y' sin nates) becomes r cos + 90°) = 2 in polar coordinates; therefore h = 2 and a = -90°, the line (Fig. 3-31) being parallel to = x' sin a + y' cos Note: r a) = h is the standard equation of a line not through are the equations of transformation from a rectangular frame of reference the origin, when h 0. If h = 0, the line goes through the origin; and (x, y-axes) to rotated set of I', (through angle a). we can find a simpler equation as follows: A half-line or "ray" emanating72 THE STRAIGHT LINE [CHAP. 3 3-6] THE STRAIGHT LINE IN POLAR COORDINATES 73 y 0 y 90° A Ray 2 Constant V19 0 sin a + 2 = 0 A 5 FIGURE 3-31 FIGURE 3-32 2 from 0 is characterized by the fact that is the same for all points on it (Fig. 3-32). Its equation is thus = constant. For example, = π/3 is the 60° ray from A full line through the origin can be characterized FIGURE 3-34 FIGURE 3-35 by the equations of two rays, = a and a + π, or combined, such as tan = constant (tan equation of a curve we shall study in Chapter 5, an ellipse (Fig. 3-35), EXAMPLE 3. What is the graph of the equation sin = if we write it in the form Solution. Solving for 0, we get = 30° or 150° (Fig. 3-33), or two rays, the v-shaped figure shown. 19 1. We shall also show in Chapter 5 how the angle of rotation a (in this 150° case tan⁻¹ can be determined in advance that rotation of axes will 30° eliminate the in the new equation. 0 A FIGURE 3-33 EXERCISE GROUP 3-6 EXAMPLE 4. A curve has the equation 12xy + = 76. To Find the new equation after the axes have been rotated through the determine its equation after the axes have been rotated (Fig. 3-34) indicated angle. through the angle a= 1. + 4xy + = 36, a = Solution. 2. 13x² + + = 99, = = cos y' sin α, 3. + 8xy = 72, = = x' sin + y' α, 4. 5xy + 13y² = 68, = [Hint: Use the formula become = 2x' y' y = taking 0/2 = a, to find tan and hence a and sin α.] Substituting in the equation we get 5. + Bry + Cy2 = F, a = 2 1 cot A B C 76 Identify and draw the following curves. Do by standard equations or in polar coordinates; check Exercises 6 through 9 with rectangular co- + ordinates. This equation is simpler than the one given in the original frame of = 7. T sin = 3 reference since there is no and indeed it is like the standard 8. r(5 cos sin = 9. r(5 12 sin = 2674 THE STRAIGHT LINE 3 3-7] THE PLANE 75 Ai + Bj + Ck 0 + 0 6 A (x, moves FIGURE 3-36 (a,b,c) 10. 2 sin = 11. 30 + = 0 12. 2 cos² = 0 (4 rays) FIGURE 3-37 FIGURE 3-38 In Exercises 13 and 14, the graphs are not straight lines. Some thought is necessary to analyze them. always perpendicular to a fixed vector Ai + Bj + Ck (Fig. 3-37). There- 13. = 3 fore lies in the plane through (a,b,c) perpendicular to Ai + Bj + Ck. 14. = 6 [Hint: (r, moves so that the length 6 subtends a Conversely, the vector from (a, b, c) to any point in this plane right angle there (Fig. 3-36).] (Fig. 3-38) is perpendicular to Ai + Bj + Ck and hence the dot product 15. By plotting 6 or 8 points trace each of the following curves. + (a) 3(1 or (b) T = 2 30. (Remember that = (e) = 4 20 16. Derive a formula for the distance between the two points Thus every first-degree equation, Ax + By + + D = 0, always and [Hint: Use the Law of Cosines.] represents a plane, and this plane is perpendicular to a fixed vector N = Ai + Bj + Ck. N is called the normal of the plane. Note that the components of the normal are the coefficients of x, and 3-7 The plane. We should not close this chapter without discussing 2 in the equation of the plane. + By + + D = 0, the first-degree equation in three dimensions, with A, B, C not all zero. The reader has probably already guessed that Note: All points that satisfy two linear equations simultaneously form a straight line L, the line of intersection, whose direction vector is the cross it is the equation of a plane. To prove this, we first find one point that satisfies the equation. Many product of the two normals, since it is perpendicular to each of the two such points exist. For example: if let y, 2 be any two numbers normals (Fig. 3-39). and solve for I; similarly if A = 0 but B or C are different from zero. Let such a point be (a, b, c), so that Aa + Bb + Cc + D = 0. Hence Cc, and substituting for D in the original equation, we get Ax + By + Cc = 0 or a) + B(y b) + C(z c) = 0. This says that the dot product of non-null vector Ai + Bj + Ck by a)i b)j + c)k is zero or that the two vectors are per- pendicular. L N1 X In other words (x, y, z), in order to satisfy the linear equation, can move only in such a way that the vector it makes with a fixed point (a, b, c) is FIGURE 3-3976 THE STRAIGHT LINE [CHAP. 3 3-7] THE PLANE 77 N process. Hence N = X L is the cross product obtained from (2,5,1) N L 23 18 (2,4,3) N1 or N = 34i + 40j - 31k, and the desired plane has the equation (3,2,1) =3i j + + 40y - 31z = 237 (the coefficients are the components, and (2, 5, 1) must satisfy the equation). (4,1,5) EXERCISE GROUP 3-7 FIGURE 3-40 FIGURE 3-41 In each of Exercises 1 through 6, first draw a schematic figure; then find EXAMPLE 1. To find the equation of the plane through (3, 2, 1), the equation of the plane: (4, 1, 5), (2, 4, 3). 1. through 1, 4), whose normal is N = Solution. We need the normal and a point, of which we have three 2. through (4, 2, -3), 2), (7, 4, -1). (Fig. 3-40). If we can find two vectors to which the normal is perpendic- ular, the cross product will give the direction of the normal. The normal 3. having intercepts (a, 0, 0), (0, b, 0), (0, 0, c). Show that its equation can be written is obviously perpendicular to 4k and to -i + 2j + 2k [vectors from (3, 2, 1) to (4, 1, 5) and (2, 4, 3), respectively]. The cross product is a y b = 1, 10i k. Hence the equation of the plane is if abc 0 (see Exercise 4(a), page 55). + const. = 41, 4. through (2, 1, 3) containing the line the constant, 41, being determined by the fact that (3, 2, 1) or (4, 1, 5) or (2, 4, 3) must satisfy the equation. = y 2 = 2 3 5 EXAMPLE 2. To find the equation of the plane through (2, 5, 1) per- pendicular to the plane 3x - = 1 and parallel to the line 5. through (3, 2, -5) containing the line + 5y = 7, 3y + 4z = 1. 3x 4y + = 2, Solution. We need merely the normal N, since a point is given (Fig. 3-41). N is perpendicular to the normal of the given plane, which [Hint: Find a point on the line.] is 3i - 2k; and N is also perpendicular to L, since the line L is parallel 6. through (1, -2,3) perpendicular to the plane 3x + 2y + 5z = 1, to the desired plane. Hence N is the cross product X L. We see that and parallel to the line L is the intersection of two planes and lies in each, and is therefore per- pendicular to the normal of each. Thus L is the cross product of the normals of the two given planes in whose intersection it lies. They are + 2y + 3z = 6. 7. A line segment from the origin has direction angles a, B, and is of + 3j + 4k (the coefficients of x, and length h (Fig. 3-42). A plane is perpendicular to this line segment at the other endpoint (not the origin). Find its three intercepts and its equa- and their cross product is L = 18j + 2k, by the tion. (See Exercise 4(b), page 55, and Exercise 3 above.)78 THE STRAIGHT LINE 3 3-7] THE PLANE 79 N (0,0,?) (r,s,t) 0 K d |d| (0,?,0) (?,0,0) FIGURE 3-43 FIGURE 3-42 with the sign so chosen that d is positive. Note that the numerator is the linear expression Ax + By + + D in the equation of the given plane, with r, t replacing x, y, PROBLEM III. To find the perpendicular distance d from a given point (r, to a given plane Ax + By + + D = 0. EXERCISE GROUP 3-8 Solution. N of course is Ai + Bj + Ck. If the point -D/A, 0, 0) lies in the plane (Fig. 3-43). Let K be the vector from this 1. Prove that the same formula for distance d is obtained if K is taken point to the point (r, 8, t): from point D/B, 0) or (0,0, D/C) to 2. Find the distance from: K = (a) (3, 2, 1) to 4x - 5y 3z 1. (b) (-1, 3, 4) to 3x - y - = 6. (ending point minus beginning point). If = 0, see Exercise 1 below. (c) (4, 1, 2) to 3y 4z = 7. If 0 is the acute angle between N and K, we see that |d| = |K| cos (d) origin to 5x - y - = 10. Vector d is a scalar times N. (e) (3, 2, 5) to 3x + 4 = 0 (by inspection). Since d is perpendicular to the plane, is also the angle (or its supple- (f) (4, -1,2) to the plane determined by (2, 3, 1), (4, 2, 5), ment) between K and the normal N, where N = = Ai + Bj + Ck, and (5, 2, -1). (g) origin to x COS + y COS + z h = 0 (see Exercise 7, = N K K or K cos = N = A[r + (D/A)] + + page 77). 3. (a) Find the distance, d, between the two parallel planes making + 4y = 3 and 6x 8y 24z + 9 = 0. d = Why are they parallel? + (b) Find the vector d, perpendicular to these two planes, of length d. 4. (a) Find the distance from (3, -2, 4) to the line We use the absolute value to insure that d, a distance, is positive, since K might be negative. Here again, there is an alternative form for d, namely, + 3y = 10. d = ± + + Ct + [Hint: d = K sin 0; use K X L| = |K| sin 0.] + (b) Find vector d also, of correct magnitude in terms of i, j, k.80 THE STRAIGHT LINE 3 3-8] FAMILIES OF LINES. FAMILIES OF PLANES 81 5. Find the equation of the plane determined by the point and the line m(3x + 2y 8) + 3) = 0, where m, n can be determined. in Exercise 4(a). For example, if we want to find the line through the point of intersection 6. Find the shortest distance between the two skew (nonmeeting and and the point (4, 2), we know that (4, 2) must make the above expression zero: nonparallel) lines: + 8) + n(2 4 - 3) = 0 5y + = 74, or and 2y 6z = 0, 5y 4z + 61 = 0. 8m + 3n = 0 and (or any two numbers proportional and 8). Putting these values for [Hint: The line segment of shortest distance joining these two lines is m and n in the equation above we get perpendicular to each; use its direction vector as the normal of two parallel planes, one through each of the two given lines; the distance be- + 2y 8) + 8(2x 3) = 0 or 7x - 14y = 0, tween these planes.] which reduces to 2y = 0 as the equation of the unique straight line through the intersection of 3-8 Families of lines. Families of planes. We have seen that to find 2y 8 = 0, the point of intersection of two lines in two dimensions, we solve the two 3 = 0, equations simultaneously. Thus (2, 1) is the intersection of the lines and the point (4, 2). It also happens to go through the origin. In case the reader has raised his eyebrows at this solution of the straight- y 3 = 0. forward problem of finding the equation of a line through two points, when use of the point slope formula is so easy, we hasten to point out that The point (2, 1) is intimately tied up with these lines; they intersect by this new method we avoid having to solve for the point of intersection in one and only one point, and (2, 1) is that point (Fig. 3-44). It satisfies in order to find other lines through it. Also we have in compact form the both equations. equations of all lines through the point of intersection, by the use of the It would seem that all the straight lines through (2, 1) should be related parameters m, n. to each other, and to these two given lines in particular. This is indeed In general, if u(x, y) = 0, v(x, = 0 are the equations of two curves, the case. then the parametric equation mu(x, y) + nv(x, = 0, m and n not both If we take two constants m and n and construct a new linear equation, zero, is the equation of a curve which passes through each intersection namely, n(3x + 2y 8) + n(2x - 3) = 0, which is the equation of the two given curves and has no other point in common with either. of a straight line, we see that the point (2, 1) satisfies it, since it makes It is obvious that any point of intersection satisfies the equation each expression in parentheses zero. We thus have found many lines mu + = 0, since the point separately makes and each vanish. through the point (2, 1) (Fig. 3-45). For example, when we collect the Moreover, any point making the expression mu + no vanish, and also I's and y's, m = 3, gives x + 10y = 12, which is obviously making vanish, must make and therefore v vanish also, n 0. satisfied by (2, 1). Moreover, any line through (2, 1) can be expressed by Hence any point that mu + = 0 has in common with either = 0 or = 0 (Fig. 3-46) is a point of intersection of = 0 and = 0. If y (4,2) = 0, = 0 have no point of intersection, then mu + = 0 does not y intersect with either one. X (2,1) 0 (2,1) mu + - 0 I I 0 0 0 FIGURE 3-44 FIGURE 3-45 FIGURE 3-4682 THE STRAIGHT LINE [CHAP. 3 3-8] FAMILIES OF LINES. FAMILIES OF PLANES 83 We call mu(x, y) + nv(x, = 0, with m, n taking on all possible (3, 2, y values not both zero, a family of curves. If and v are first-degree expres- sions of and y, mu + = 0 is the family of all lines through the inter- section of = 0, = 0 if they have one, or the family of all lines parallel to u(x, = 0, v(x, = 0 if they do not have a point of intersection. (3) Sometimes mu + = 0 is called a pencil of lines. One parameter would (1) do almost as well as two, namely, + kv = 0 as the family of lines (2) through the intersection. Here k would have a unique value, instead of any proportional sets of values for m, n. We prefer to use the two param- L eters m, n because whole number values can be chosen for m, n when k is a fraction, and also because mu + = 0 gives all lines through the FIGURE 3-47 FIGURE 3-48 intersection, whereas + kv = 0 never gives the line v = 0, no matter what k is. Using the point slope form, the three medians have the equations: EXAMPLE 1. To find the line through the intersection of y = a (x b), or (1): + (2b a)y be = 0, y = 1, + 4y = 2, y = b 2a (x a), or (2): + (2a b)y ac = 0, which is perpendicular to 4x + 5y = 3. = a+b -2c (x 0), or (3): 2cr + (a + b)y c(a + b) = 0. Solution. The equation of any line through the intersection is m(2x - y 1) + n(3x + 4y 2) = 0, and - (2m + + 4n) We see that in the simplified equations on the right, the third linear expres- is its slope (the slope of Ax + By + C = A/B); and -1 is the sion is the sum of the first two; or, in symbols, (3) = (1) + (2) = 0. other slope. For two lines to be perpendicular, the product of their slopes Hence Eq. (3) is that of a line in the family determined by lines (1) is and (2), that is, it goes through their intersection. 4 5 + 4n = -1 or 3m 32n = 0; This method works equally well in three dimensions. If u(x, y, = 0, v(x, y, 2) = 0 are the equations of two surfaces, then mu + = 0 is another surface through their curve of intersection, if any. m = 32, are the smallest whole number values that do this. For example: the two planes of Exercise 5, page 80, Hence the answer is: 1) 3(3x + 4y 2) = 0 or 55x - 44y = 26. + 3y 10, As a check we see the slope is the negative reciprocal of intersect in a line L. The equation and we know it goes through the desired point of intersection, whatever it is, because it came from mu + no = 0. [It happens to be m(3x 2y + 4z 2) + n(2x + 3y 10) = 0 If we had used + kv = 0 for the family in this problem, k would gives every plane through L. If we wish this plane (Fig. 3-48) to go through (3, -2, 4) we have 27m - 14n = 0, or m = 14, n = 27, from EXAMPLE 2. To prove that the three medians of any triangle are con- which we get current. 2y + 2) + 27 + 3y 10) = 0 Solution. Choosing the coordinate axes as in Fig. 3-47, we find the or midpoints of the three sides to be + 53y + 29z = 298 as the answer to Exercise 5.84 THE STRAIGHT LINE [CHAP. 3 3-9] SUMMARY AND REVIEW 85 EXERCISE GROUP 3-9 information, such as the direction of a line or the normal to a plane. We have learned how to do certain chores like finding midpoints, distances, Draw schematic figures. slopes, points of intersection, angle between two lines, and rotating co- 1. Find by the above method the equation of the plane through: ordinate axes. We have made liberal use of vector methods. Learning to (a) the line think in terms of vectors and to use the important tools of dot products + 3y 4z = 1, and cross products is a very valuable accomplishment. y + 2z = 3, RÉSUMÉ and the point (4, 2, 3). (b) the line 3x - y = 2z + x - 1 = 4y and perpendic- I. STANDARD EQUATIONS TO BE REMEMBERED ular to the plane 4x + 3y 2z = 7. (c) the line = 2 + 3t, y = 1 + 4t, = and parallel to A. Straight lines in two dimensions 2j + 4k. 1. Point slope equation: 2. In the family (or pencil) of lines determined by 2x + 3y = 9, 3x - y = 6, find the line which: = = tan = difference of y's (Fig. 3-49). difference of (a) contains (4, -3). (b) is parallel to 2x - 5y = 1. 2. Lines parallel to coordinate axes: y = constant: parallel to (c) is perpendicular to 4x + 3y = 12. x-axis; x = constant: parallel to y-axis (Fig. 3-50). (d) goes through the origin. (e) is vertical. 3. Prove by the method similar to that in the text, that in any triangle: = B A (Fig. 3-51). (a) the three altitudes are concurrent. (b) the three perpendicular bisectors are concurrent. 4. Similarly prove that in any trapezoid, the line joining the midpoints y Slope À of the parallel sides and (a) the nonparallel sides are concurrent. (b) the diagonals are concurrent. (a,b) 5. Find the circle through (4, 1) and the two intersections of the two circles y² = 4, + 4x = 5. I 6. Find the two members of the family of straight lines determined by 0 3x - 8y 0 and 4x+y=5 = that cut from the first quadrant tri- angles of area 21 square units. Use the pencil of lines. FIGURE 3-49 y y 3-9 Summary and review. We pause now to look back over the last two chapters. They are full of important concepts and information that we shall be using constantly. y = constant Here is a list of the major developments in these chapters. We have a set of standard equations of straight lines in both rectangular and polar 0 coordinates, and of planes and straight lines in three dimensions, indicat- I 0 - constant ing what is needed to write the equations of these geometric configurations. Also from an equation or equations, we have learned how to extract certain FIGURE 3-50 FIGURE 3-51THE STRAIGHT LINE [CHAP. 3 3-9] SUMMARY AND REVIEW 87 86 4. cos α) = h any line not through 0, of distance h # - constant from 0, h at angle a to 0A (Fig. 3-52). h 5. = constant: ray (Fig. 3-53). 0 A 0 A B. Straight lines in three dimensions FIGURE 3-52 FIGURE 3-53 1. Point, direction vector: x-a = y b = m, n 0 (Fig. 3-54). (a,b,c) m n L + mj + nk 2. Parametric form: - = a + It, y r = r₀ + 0 2 = + nt, 3. As intersection of two planes (Fig. 3-55): + B₁y + + D₁ = 0, FIGURE 3-54 + B₂y + + D₂ = 0. Its direction vector is the cross product + + + = 0 N₁ X N₂ of the two normals + + + D₂ = 0 = A₂i + + C. The plane 1. Point-normal equation (Fig. 3-56): if N = Ai + Bj + Ck, L = X FIGURE 3-55 a) + B(y b) + C(z c) = 0 or Ax + By + = constant, determined by (a,b,c). II. MISCELLANEOUS INFORMATION N M 1. Midpoint coordinates (Fig. 3-57): y M: + 2 , + 2 + 2 or each coordinate = the sum of the corresponding endpoint FIGURE 3-56 FIGURE 3-57 coordinates.