APPLIED MATHEMATICS 1A (ENG) Mathematics 132: Vectors and Matrices

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1 APPLIED MATHEMATICS A (ENG) Mathematics 3: Vectors and Matrices University of KwaZulu-Natal Pietermaritzburg c C. Zaverdinos, 00. All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the author.

2 Preface This course has grown out of lectures going back well over 30 years. Names which come to mind are J.Nevin, R.Watson, G.Parish and J. Raftery. I have been involved with teaching linear algebra to engineers since the early 90s and some of my ideas have been incorporated in the course. What is new in these notes is mainly my approach to the theoretical side of the subject. Several of the numerical examples and exercises come from earlier notes, as well as a some theoretical ones. I would like to thank Dr. Paddy Ewer for showing me how to use the program LatexCad for the diagrams and especially the Reverend George Parish for proof-reading and also for elucidating No.5 of Exercise. About this course The course is meant to be studied as a whole. Many examples and exercises in later Chapters refer the reader to earlier chapters. This is especially true of Chapter 3. Chapter motivates the idea of a vector through geometry and discusses lines and planes and transformations related to such geometric objects. A vector can be thought of as a displacement in space and an ordered triple of numbers. Chapter generalizes the idea of a triple to an n-tuple and motivates linear algebra through the problem of finding solutions to simultaneous linear equations in n unknowns. The coefficient matrix of such equations is known as a matrix. Simplification of such a matrix by row operations forms the major part of this chapter. Chapter 3 considers matrices in detail and looks at them dynamically in the converse sense of Chapter : A matrix defines a transformation of points in n-dimensional space. Matrix multiplication is introduced in terms of the composition of such transformations and some other key concepts such as linear independence, the rank and inverse of a matrix are discussed. Abstract vector spaces are never mentioned, but the the proof of the basic theorem in 3.4. on linear independence goes through word-for-word for such spaces which also leads to the well-known Replacement Theorem (see No.4 of Exercise 93). Chapter 4 is about determinants and the cross product (also called the vector product). The theory of determinants predates that of matrices, going back to Leibnitz in the 7th Century. One of the founders of linear algebra, the 9th Century mathematician Arthur Cayley, once remarked that many things about determinants should really come after the study of matrices, and this is the modern approach adopted by us. The cross product is used extensively in mechanics, in particular in the notes Dynamics for Mathematics 4. Algebraic properties of the cross product are derived from those of 3 3 determinants, while the exercises can serve as an introduction to some of its applications. Note Some exercises (particularly those marked with an asterisk *) are harder and, at the discretion of the instructor, can be omitted or postponed to a later stage.

3 3 Bibliography The following books can be consulted but they should be used with caution since different authors have a variety of starting points and use different notation. Many books also have a tendency to become too abstract too early. Unless you are a mature reader, books can confuse rather than help you.. A.E. Hirst: Vectors in or 3 Dimensions (Arnold). This may be useful for our Chapter since it makes 3-dimensional space its central theme.. R.B.J.T. Allenby: Linear Algebra (Arnold). This is a quite elementary text. 3. J.B. Fraleigh & R.A. Beauregard: Linear Algebra (Addison-Wesley). 4. K. Hardy: Linear Algebra for Engineers and Scientists (Prentice-Hall). 5. D. Lay: Linear Algebra and its Applications (3rd Edition, Pearson). This book treats matrix multiplication in much the same way as we do, but its treatment of geometric aspects is less thorough. It has over 560 pages, becomes abstract and advanced after p.30, but will probably be useful in later years of study. 6. H. Anton: Elementary Linear Algebra (6th Edition, John Wiley and Sons). 7. E.M. Landesman & M.R. Hestenes: Linear Algebra for Mathematics, Science and Engineering (Prentice-Hall). This is quite advanced. I recommend especially () for Chapter and (4) and (5) for later chapters of the notes. Because of its elementary nature, () is good for Chapter. Some of the above books have been used on the Durban campus and, if consulted with care, can be helpful. C. Zaverdinos Pietermaritzburg, January, 00

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5 Contents Two and Three-Dimensional Analytic Geometry. Vectors 7 Matrices and the Solution of Simultaneous Linear Equations 4 3 Linear Transformations and Matrices 65 4 Determinants and the Cross Product 3 5

6 6 CONTENTS

7 Chapter Two and Three-Dimensional Analytic Geometry. Vectors. Points in three-dimensional space The study of the geometry of lines and planes in space provides a good introduction to Linear Algebra. Geometry is a visual subject, but it also has an analytical aspect which is the study of geometry using algebra: how geometric problems can be expressed and solved algebraically. This geometry is also called Cartesian Geometry, after its founder, the 7th Century philosopher and mathematician René Descartes. From school you are already familiar with the Cartesian plane. There are an x axis and a y axis meeting at right-angles at the origin O. Every point A in the x y plane is uniquely represented by an ordered pair (a, b) of real numbers, where a is the x coordinate and b is the y coordinate as in Figure.. We write A = (a, b). Here M is the foot of the perpendicular from A to the x axis and likewise N is the the foot of the perpendicular from A to the y axis. Notice that M = (a, 0) and N = (0, b). y axis N... a A = (a, b) b b O a. M x axis Figure. In order to represent a point A in space we add the z axis which is perpendicular to both the 7

8 8CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS x and y axes as in Figure.. Here A = (a, b, c) is a typical point and a is the x coordinate, b is the y coordinate and c is z coordinate of the point A. In the diagram P is the foot of the perpendicular from A to the y z plane. Similarly, Q and R are the feet of the perpendiculars from A to the z x and x y planes respectively. z axis N a b a P Q b A = (a, b, c) c O c M a a y axis L b R x axis Figure... The Corkscrew Rule and right-handed systems You wish to open a bottle of wine using a corkscrew. You first turn the corkscrew so that it enters the cork. As observed by the opener the turning is clockwise, but observed from the underside of the bottle the sense is reversed: it is anticlockwise. The direction in which a corkscrew moves as it is turned is given by the Corkscrew Rule. In Figure., rotating from the x axis to the y axis and applying this rule produces motion along the (positive) z axis. Rotating from the y axis to the z axis and applying the same rule gives motion along the (positive) x axis. Finally, rotating from the z axis to the x axis and applying the rule produces motion along the (positive) y axis. The axes x, y, z (in that order) are said to form a right-hand system. This can be represented by the scheme x z y x z y x Have a look at yourself in the mirror while opening a bottle of wine and observe that what is clockwise for you is anticlockwise for the person in the mirror. Thus a right-hand system of axes observed in a mirror does not follow the corkscrew rule and is called a left-hand system. The usual convention is to use only right-hand systems... Displacements and vectors Consider two numbers a and b on the real line R. The difference c = b a is the displacement from a to b. Given any two numbers from a, b and c, the third is uniquely determined by the

9 .. POINTS IN THREE-DIMENSIONAL SPACE 9 equation c = b a. For example, let a = and b = 3. The displacement from a to b is c = 3 ( ) = 5. The displacement from 6 to is also 5. Let A = (a, a, a 3 ), and B = (b, b, b 3 ) be two points in space. The displacement from A to B is denoted by AB and is given by..3 Units and distance AB = (b a, b a, b 3 a 3 ) (.) The unit can be any agreed upon distance, like the meter or kilometer. Units (of length, time) are important in applications but for the most part we won t specify them. The distance of A = (a, b, c) from the origin O is given by OA = a + b + c To see this, see Figure. and use the theorem of Pythagoras: OA = OR + RA = OL + LR + RA = a + b + c The distance from A to B is written as AB and is given by AB = (b a ) + (b a ) + (b 3 a 3 ) (.) and is also called the length or magnitude of AB. Because the displacement AB has both direction and magnitude it is called a vector and pictured as an arrow going from A to B. The tail of the arrow is at A and its head is at B as in Figure.3. As an example, let A = (,.3, 4.5) m and B = (., 3.6,.5) m. Then AB = (. ( ), 3.6.3,.5 ( 4.5)) = (3.,.3,.0) m. Two displacement vectors AB and CD are equal if they are equal as displacements. Hence if C = ( 3.5,., 3.3) and D = (.4,.4, 5.3) and A, B are as above then CD = (.4,.4, 5.3) ( 3.5,., 3.3) = (3.,.3,.0) = AB. The magnitude of this vector is AB = CD = (3.) + (.3) + (.0) = 3.95 m. Various vectors (displacement, velocity, acceleration, force) play an important role in mechanics and have their own units. For example, if positions are measured in meters and time is measured in seconds, a velocity vector will have units ms. So if a particle moves in such a way that each second it displaces (,, 3) meters, we say its velocity is constantly (,, 3) ms. Depending on the geometric interpretation, the tail (head) of displacement vectors may be at any point. Vectors are often written as u = (u, u, u 3 ), v = (v, v, v 3 ), a = (a, a, a 3 ), b = (b, b, b 3 ) etc. We now have three ways of expressing the position of a point A in space. If A has coordinates a = (a, a, a 3 ), then A = (a, a, a 3 ) = a = OA (.3) Although equation (.3) identifies a point with its position, geometrically speaking we like to distinguish the point itself from its position. The vector OA is called the position vector

10 0CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS z axis AB B A O a 3 b 3 a a b b y axis x axis Figure.3 of the point A. When we say a is the position vector of point A it is understood that the tail of a is at the origin and its head is at the point A. Our convention is that we use capital letters A, B,... to denote points in space. Given the point C, there is a unique point D in space such that CD = AB (see Exercise, No.4). There is also a unique point D such that OD = AB and we may write D = AB. In that case the tail of D can only be O, and it is better to use the notation OD to bring out the vectorial nature of the displacement from O to D...4 More systematic notational convention Instead of speaking of the x, y and z axis, we also refer to these as the x, x and x 3 axis respectively. Accordingly, given vectors a, b, c,... it will be convenient to assume (unless otherwise stated) that a = (a, a, a 3 ), b = (b, b, b 3 ), c = (c, c, c 3 ),... Also by convention, points A, P, Q,... have position vectors a, p, q,..., unless otherwise specified...5 Addition of vectors Let u = (u, u, u 3 ) and v = (v, v, v 3 ) be two vectors. Their sum u + v is defined by the equation u + v = (u + v, u + v, u 3 + v 3 ) The vector v is defined as v = ( v, v, v 3 ) and the difference The zero vector is 0 = (0, 0, 0). u v = u + ( v) = (u v, u v, u 3 v 3 )..6 Basic properties of addition of vectors For all vectors u, v and w

11 .. POINTS IN THREE-DIMENSIONAL SPACE. u + (v + w) = (u + v) + w (associative law). u + v = v + u (commutative law) u = u (the vector 0 = (0, 0, 0) behaves like the number 0) 4. u + ( u) = 0 An important result for addition is the following:..7 Geometrical interpretation: The triangle law for addition of vectors AB + BC = AC To see this, let the position vectors of A, B and C be a, b and c respectively. Then AB = b a and BC = c b. Hence by the properties of addition, AB + BC = (b a) + (c b) = (b b) + (c a) = 0 + c a = c a = AC Figure.4 illustrates this geometrical result in which it is understood that the head B of AB is the tail of BC etc. The point D is chosen so that AD = BC. It follows that DC = AB (why?). We have a geometric interpretation of the above commutative law (known as the parallelogram law for addition of vectors): BC + AB = AD + DC = AC = AB + BC B BC b O. AB AC DC A D AD C AB = DC AD = BC Figure.4 AC = AB + BC..8 Multiplication of vectors by scalars. Direction of vectors. Parallel vectors In contrast to a vector, a scalar is just a real number, i.e. an element of the real number system R. We use α, β, a, b, r, s, t,... to denote scalars. Note that f (f underlined) denotes a vector while f is a scalar.

12 CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS Let u = (u, u, u 3 ) be a vector and α R be a scalar. The product αu is defined by αu = (αu, αu, αu 3 ) The vector αu is said to be a (scalar) multiple of u. We usually omit the qualification scalar...9 Properties of the product αu For all scalars α, β and vectors u and v,. α (u + v) = αu + αv. (α + β) u = αu + βu 3. α (βu) = (αβ) u 4. u = u 5. αu = 0 if, and only if, α = 0 or u = 0 6. αu = α u (the length of αu is α times the length of u) The proofs of the first four are left to you. We prove the last: αu = (αu, αu, αu 3 ) = (αu ) + (αu ) + (αu 3 ) = α u + α u + α u 3 = α (u + u + u 3 ) = α u + u + u 3 = α u + u + u 3 = α u..0 Geometric interpretation of αu. Parallel vectors Since αu = α u, multiplication of u by the scalar α multiplies the length of u by α. Let u be non-zero, so that it has a definite direction. If α > 0 the direction of αu is the same as that of u, while if α < 0 the direction of αu is opposite to that of u. Let u and v be non-zero vectors. We say they have the same direction if u = αv for some scalar α > 0 and opposite directions if u = αv for some α < 0. The vector u is parallel to v if v is a multiple of u, that is, u = αv for some scalar α. Necessarily α 0 and we write u v to express this relation. Notice that we only speak of vectors being parallel if they are non-zero... Properties of parallelism For all non-zero u, v and w,. u u. u v implies v u 3. u v and v w together imply u w.

13 .. POINTS IN THREE-DIMENSIONAL SPACE 3 These properties are geometrically evident, while analytic proofs are left as an exercise. For example, to see property () analytically, let u = αv. As α 0, v = αu and v u, as expected. Example.. Let u = AB, where A is the point (, 7, 4) and B is the point ( 4,, ), so that u = ( 3, 4, 5). Let v = CD, where C = (, 9, ) and D = ( 7,, 4), so CD = ( 9,, 5) = 3u. Hence u and v have the same direction while u and v have opposite directions but are parallel... The dot (scalar) product of two vectors Let u = (u, u, u 3 ) and v = (v, v, v 3 ) be any two vectors. Their dot product u v is defined as u v = u v + u v + u 3 v 3 (.4) Note that the dot product of two vectors is a scalar and not a vector (that is why it is also called the scalar product)...3 Properties of the dot product For all scalars α, β and vectors u, v and w,. u v = v u. α (u v) = (αu) v = u (αv) 3. (αu + βv) w = (αu w) + (βv w) 4. u u = u, so u = + u u 5. u u 0 and u u = 0 if, and only if u = Geometric interpretation of the dot product Let u = AB and v = AC be non-zero vectors. Suppose that AB and AC make an angle θ between them at the vertex A, where 0 θ 80 in degrees, or 0 θ π in radians (recall that 80 degrees = π radians). Then u v = u v cos θ (.5) To see this result, see Figure.5 and use AB + BC = AC, so that BC = AC AB = v u and BC = (v u) (v u) = v v + ( u) ( u) v u u v = v + u u v Then by the cosine rule (which applies whether θ is acute or obtuse), BC = AB + AC AB AC cos θ v + u u v = u + v u v cos θ (from the previous equation) Cancelling v + u from both sides leads to the required result (.5). Example.. Let A = (, 0, ), B = (, 0, 3), C = ( 4, 0, ) and D = (,, ). Find the angle θ between AB and AC. Decide if the angle φ between AB and AD is acute or obtuse.

14 4CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS B B u u A θ v θ acute 7 C θ A v θ obtuse C Figure.5 Solution: AB = (, 0, 3) (, 0, ) = (, 0, ) and AC = ( 4, 0, ) (, 0, ) = ( 3, 0, ). Hence ( ) AB AC cos θ = AB AC = (, 0, ) 3, 0, ( 0 ) = + and θ = 60 degrees or π 3 radians. Similarly, AD = (,, 3) and AB AD cos φ = AB AD = (, 0, ) (,, 3) 5 = ( 3) 4 Since the dot product is negative the angle φ is obtuse...5 Some further properties of the dot product. The non-zero vectors AB and AC are at right angles (are perpendicular AB AC) if, and only if, (AB) (AC) = 0. Such vectors are also called mutually orthogonal.. u v u v for any vectors u and v (Cauchy-Schwartz inequality). 3. u + v u + v for any vectors u and v (Cauchy s inequality). The first result follows from the fact that AB AC if, and only if, cos θ = 0. The second result is left as an exercise. To see the third result, use () and consider u + v = (u + v) (u + v) = u u + v v + u v u u + v v + u v = u + v + u v = ( u + v ) As u + v ( u + v ), it follows that u + v u + v.

15 .. POINTS IN THREE-DIMENSIONAL SPACE 5 Example..3 Find a non-zero vector perpendicular to both u = (,, ) and v = (3, 3, ). Solution: Let x = (x, x, x 3 ) satisfy x u and x v. Then x u = x + x + ( ) x 3 = 0 x v = 3x + 3x + ( ) x 3 = 0 By subtraction x x = 0, or x = x. Thus from the first equation, x 3 = x +x = 3x. We may let x be any non-zero number, e.g. x =. Then x = (,, 3) is perpendicular to u and to v. We will see later why it is always possible to find a vector x 0 perpendicular to two non-zero vectors. This fact will also come into our study of planes (see.3.4)...6 Unit vectors. Components. Projections A vector u of length is called a unit vector. For example, u = ( 3 5, 0, 4 5) is a unit vector. A vector c 0 defines the vector ĉ = c c which is the unique unit vector having the same direction as c. To see that ĉ has length we note ( ) ( ) ĉ = ĉ ĉ = c c c c = c c c = c c = The component of the vector f in the direction of (or along) the non-zero vector c is defined as f ĉ. Suppose that f 0 and that the angle between f and c is θ. Then To see equation (.6), use equation (.5): f ĉ = f cos θ (.6) f f sin θ (f 0) θ f cos θ c Figure.6

16 6CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS f ĉ = f c c = f c f c cos θ = = f cos θ c c (See Figure.6). Notice that if π < θ π then the component (.6) is negative (as is only natural). The projection of f on c is the vector ( f ĉ ) ĉ = f c c c (.7) Figure.7 shows two cases of the projection of f 0 on c: when cos θ > 0 and when cos θ < 0. f f (f 0) θ. ĉ. θ ĉ Heavy arrow = projection of f on c Figure.7 Let d c. Then the projection of f on c is the same as the projection of f on d. See Exercise, No.3a. The situation with components is slightly different: If d and c have the same direction then the components of f along c and d are the same. If d and c have opposite directions then the components of f along c and d are the same apart from sign - one is the negative of the other. We return to projections in subsection..5. Example..4. If we imagine that c is pointing due East and f is a displacement, then ( ) f ĉ ĉ is the Easterly displacement, while f cos θ is the Easterly component and f sin θ is the Northerly component f. You may think of f as a vector representing a force that is dragging a box placed at its tail. Then f sin θ is the lifting effect of f while f cos θ is the horizontal dragging effect of the force.. Let f = (, 3, 4) and c = (5,, ). The length of c is c = 5 + ( ) + ( ) = 30. The unit vector ĉ in the direction of c is ĉ = 30 (5,, ) The component of f along c is f ĉ = (, 3, 4) 30 (5,, ) =

17 .. POINTS IN THREE-DIMENSIONAL SPACE 7 and the projection of f on c is ( ) ( ) 5 f ĉ ĉ = 30 (5,, ) = 5 (5,, ) The vectors i, j, k The vectors i, j and k are defined as i = (, 0, 0), j = (0,, 0), and k = (0, 0, ). They have unit length and are directed along the x, y and z axis respectively. It is clear that they are also mutually orthogonal and that any vector a = (a, a, a 3 ) has a unique expression a = a i + a j + a 3 k (.8) The scalars a, a and a 3 are the components of a in the directions i, j and k respectively. See also No.5 of Exercise below...8 Linear combinations Let a and b be two vectors and λ and µ two scalars. The vector x = λa + µb is called a linear combination of a and b with coefficients λ and µ. Similarly, for scalars r, s and t, the vector ra + sb + tc is a linear combination of a, b, and c with coefficients r, s and t. Equation (.8) shows that every vector is a linear combination of i, j and k. If a = (,, 3), b = (5, 7, ), c = ( 4, 8, ) then a + 3 b + is a linear combination of a, b, and c ( 5 ) ( ) 37 c = 6 6, 3 3, 3 Exercise. In Figure. express ON, MP, OM, LP, NR, NM and OP as vectors involving a, b and c. Write down these vectors as linear combinations of i, j and k and find their lengths in terms of a, b and c.. Interpret the equation (r (,, 5)) (r (,, 5)) = 49 geometrically. 3. Let u = (, 3, ), v = (, 0, ) and w = (3, 3, 4). (a) Find the linear combinations u + 3v w, ( 5) u + v + 3w and 3 4 u v + 4 w. (b) If the tail of u is the point A = (,, 3), what is the head of u? If the head of w is B = (, 9, 0), what is the tail of w? (c) Find coefficients α, β such that w = αu + βv. (d) Is 3 (3,, 3) a linear combination of u, v and w? This means you must look for scalars p, q, r such that pu + qv + rw = 3 (3,, 3), and this involves three equations in three unknowns. 4. (Generalization of No.3b). Let u be a given vector. Show that if the tail A of u is given then the head B of u is uniquely determined. State and prove a similar result if the head of u is given. Draw a picture. 5. See Figure.8. Imagine that in each case the axes are part of a room in which you are standing and looking at the corner indicated. Decide which of the systems of mutually orthogonal axes are left-handed and which right-handed. If you are outside the room, what would your answer be? 6. Let A 0,...,A 9 be ten points in space and put a m = A m A m for m =,, 9 and a 0 = A 9 A 0. Find the sum a + + a 0.

18 8CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS y y x x z x x z y z z y Figure.8 7. Complete the proofs in statements..6,..9 and.. above. Hint 3 These properties just reflect similar properties of numbers. commutative law of addition of vectors is proved by For example, the The proof of the first item of..9 is u + v = (u + v, u + v, u 3 + v 3 ) = (v + u, v + u, v 3 + u 3 ) = v + u α(u + v) = α(u + v, u + v, u 3 + v 3 ) = (α(u + v ), α(u + v ), α(u 3 + v 3 )) = (αu + αv, αu + αv, αu 3 + αv 3 ) = αu + αv 8. Complete the proofs of subsections..3 and Deduce that a dot product (αr + βs) (γu + δv) of linear combinations multiplies out just like in ordinary arithmetic and conclude that αu + βv = α u + β v + αβu v. 0. Show that if the vectors u and v in equation (.5) are unit vectors, then cos θ = u v.. Show that the non-zero vectors u and v are parallel if, and only if, u v = u v. Solution: The vectors are parallel if, and only if, the angle between them is 0 or π radians.. Let a = (3,, ) and b = ( 7, 3, ). Find (a) The unit vectors â and b. (b) The component of a along b and the component of b along a. (c) The projection of a on b and the projection of b on a. 3. (a) If γ 0, show that the projections of f on c and f on γc are equal. (b) Show that if f 0, then the projection of f on itself is f.

19 .. THE STRAIGHT LINE 9 4. Let u and v be two vectors. The set of all linear combinations of u and v is called the span of the vectors and is denoted by sp (u, v). (This is discussed more fully in item of subsection 3.. of Chapter 3). Let u, v u, v be four vectors. Show that sp (u, v ) sp (u, v ) if, and only if that each of the vectors u and v are linear combinations of the vectors u and v. 5. Let l, m and n be three mutually orthogonal unit vectors. Placing the tails of these vectors at the origin, it is visually clear that we may use these vectors as a frame of reference in place of i, j and k. (An analytic proof of this will follow from work done in Chapter 3 - See Exercise 78, No.9). Assuming this result, any vector a can be expressed as a linear combination a = γ l + γ m + γ 3 n Show that necessarily the coefficients γ, γ, γ 3 are the components (in the above sense) of a in the directions l, m and n respectively. In fact, γ l, γ m, γ 3 n are the projections of the vector a on l, m and n respectively. Conclude that a = γ + γ + γ 3 Show that l = 6 (,, ), m = 3 (,, ) and n = (,, 0) is such a system of three mutually orthogonal unit vectors and find the coefficients γ,..., γ 3, if a = ( 3,, ). Hint 4 Take the dot product of both sides of the above equation with each of l, m and n. The coefficients for the specific example are the components a l = γ etc. A longer procedure to find the coefficients for the specific example is to use the method of 3d above. 6. What familiar result from Euclidean Geometry does b a a + b represent? 7. Let u and v have the same length: u = v. Show that (u v) (u + v) = 0. In other words, if u v and u + v are not zero they are at right angles. 8. How are our general results affected (and simplified) when one of the coordinates is restricted to be zero for all vectors under consideration?. The straight line Two distinct points determine a line. Points that lie on one and the same line they are called collinear. Let A, B and C be three distinct points. Then it is geometrically obvious that C lies on the (infinite) line through A and B if, and only if, AC is parallel to AB. We need to show that this axiom is symmetric with regard to A, B and C. In other words, that the statements AC AB, CB CA and BA BC are equivalent. It is left as an exercise to prove this equivalence algebraically. See Exercise 5, No.. This axiom for the collinearity of A, B, and C gives the clue on how to describe algebraically the line l through A and B... Generic (parametric) representations of l Let A and B be two distinct points. Then as t R varies, the point R with position vector r = r (t) = OA + tab (.9)

20 0CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS varies over the entire straight line l through A and B. As in Figure.9, from O jump onto the line at A and then go parallel to AB to R. The point R = A is obviously on l and corresponds with t = 0. If R A we have AR = tab for some t 0, so in general r = OR = OA + AR = OA + tab When t = then we are at R = B. B l. OA R tab A r = OR = OA + tab O Figure.9 As r = r (t) represents a general point on the line l, the equation (.9) is called a generic or parametric equation of the line through A and B. The variable t is called the parameter... Some general observations In what follows a = OA, b = OB, etc.. As t varies over the set R of real numbers the point r = OR = a + t (b a) of equation (.9) varies continuously over the line l. In this representation of the line l we consider b a = AB as the direction of l. In Figure.9 as t increases R moves to the right, as t decreases R moves to the left.. We remarked that t = 0 corresponds to R = A. When t = we have R = B. If t is between 0 and, say t = we get OR = a + (b a) = (a + b) Thus (a + b) is the position vector of the point midway between A and B. 3. More generally, a point r = r(t) = a + t (b a) = ( t) a + tb where 0 t lies between A and B and the set AB of these points is called the line segment joining A and B. If A B and 0 < t <, the point r(t) is said to lie strictly between A and B. The line segment AB is a set of points and it must be distinguished from the vector AB and also from the distance AB.

21 .. THE STRAIGHT LINE 4. Let u 0 define a direction and suppose a is the position vector of the point A. Then a generic equation of the line through A with direction u is A typical point a + tu on the line is called a generic point. r (t) = a + tu (.0) The set l of points R = r (t) satisfying a generic equation such as (.0) as t varies over the real numbers is the analytic definition of a straight line. 5. A generic equation is not unique. (a) If C D are two points on the line given by equation (.9), then since these points also determine the line, another generic equation (now with parameter s) is r (s) = OC + scd The parameters t and s are of course related. See Example.. of subsection..3 and No.b of Exercise 5. (b) When do two generic equations like (.0) define the same line? Let u and v be non-zero vectors and suppose that a + tu is a generic equation of line l and b + sv a generic equation of line l. Then l = l if, and only if, u v and a b is a multiple of u. (See No.5 of Exercise 5)...3 Some illustrative Examples Example.. Find two generic equations for the line l passing through A = (,, 3) and B = (,, 4). Solution: One generic equation is Another is r = (,, 3) + t ((,, 4) (,, 3)) = (,, 3) + t (,, ) = ( t, t, 3 + t) r = (,, 4) + s ((,, 3) (,, 4)) = (,, 4) + s (,, ) = ( + s, + s, 4 s) In the second representation we have used r = OB + sba, with parameter s. The relationship between s and t is s = t, as can be easily seen. Example.. Find the position vector of the point R lying on the line l of Example.. that is a distance 3 AB from A but on the side of A opposite to that of B. Solution: To find R put t = 3 in OR = OA + tab to get ( OR = OA + ) AB 3 ( = (,, 3) + ) (,, ) 3 ( 5 = 3, 7 3, 8 ) 3

22 CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS Example..3 Find the foot Q of the perpendicular from the point P = (,, 4) to the line l of Example... Hence find the shortest distance between P and l. Solution: Let Q = ( t, t, 3 + t) be the required foot of the perpendicular from P = (,, 4) to the line l. Then P Q = ( t, 3 t, + t) must be perpendicular to the direction (,, ) of l: ( t, 3 t, + t) (,, ) = 0 This gives t = 3 and so Q = q = ( 3, 5 3, ) 0 3. The shortest distance of P from l is thus ( P Q = 5 3, 8 3, ) 3 = (See Figure.0). P = (,, 4) r = ( t, t, 3 + t) Q Figure.0 Example..4 Find the foot Q of the perpendicular from a general point P = (p, p, p 3 ) to the line r = a + tu if a = (,, 3) and u = (,, ). Solution: Let t be such that Q has position vector a + tu. As before, we consider P Q = a + tu p and must solve P Q u = 0 for t and then substitute this value in a + tu to obtain Q. We have ( a + tu p ) u = 0 (.) Since P Q = ( t p, t p, 3 + t p 3 ), the condition P Q u = 0 becomes ( t p, t p, 3 + t p 3 ) (,, ) = + 6t + p + p p 3 = 0 This gives t = 6 ( p p + p 3 ). Substituting this value of t in ( t, t, 3 + t) gives q = ( p + 3 p 3 p 3, p + 6 p 6 p 3, p 6 p + ) 6 p 3

23 .. THE STRAIGHT LINE 3..4 A general result for the foot of a perpendicular to a line In Exercise 5, No. you are asked to show that the foot Q of the perpendicular from P to the line a + tu has position vector ( ) ( ) q = a + p a u u u (.) Example..5 Let us apply this formula to the above example..3 to find the foot of the perpendicular from P = (,, 4) to the line (,, 3) + t (,, ). Solution: We have u = 6 and the foot has position vector ( ) q = (,, 3) + ((,, 4) (,, 3)) (,, ) (,, ) 6 ( = 3, 5 3, 0 ) 3 This is the same result as before...5 Projections on a line through the origin When a line l passes through the origin, the foot Q of the perpendicular from P to l is called the projection of P on l. The projection of p on the line tu is just the projection of p on the vector u. This is equation (.7) with f = p and c = u: ( ) p u q = u u (.3) Example..6 Find the projection of P = (p, p, p 3 ) on the line t (,, ). Solution: In this case P Q = ( t p, t p, t p 3 ) and we can proceed as before by solving ( t p, t p, t p 3 ) (,, ) = 6t + p + p p 3 = 0 for t and substituting into t (,, ). It amounts to the same thing to use (.3) and the required projection is q = (p, p, p 3 ) (,, ) (,, ) (,, ) = p p + p 3 (,, ) 6 ( = 3 p + 3 p 3 p 3, 3 p + 6 p 6 p 3, 3 p 6 p + ) 6 p 3 We will return to this example in Chapter 3 (See Example 3.3.). (.4) Example..7 A straight line lies in the x y plane and passes through the origin. The line makes an angle of 30 with the positive x axis. Find the foot Q of the projection from P = (p, p ) on the line. Solution: Let u = ( cos π 6, sin π 6 ) = ( 3, ). The parametric equation of the line is r = t ( 3, ) and the foot Q of the perpendicular from P to the line is the projection of p = (p, p ) on the unit vector u: ( ) ( ) 3p q = + p 3, ( = (3p + ) 3p, 3p ) 4 p (.5) We shall return to this example in subsection.4 below and again in Chapter 3 Example 3...

24 4CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS A well-known result on parallelograms Example..8 Show that the diagonals of a parallelogram bisect each other. Solution: In Figure. ABCD is a parallelogram with AB = DC and BC = AD. It is understood that the parallelogram is non-degenerate in that no three of A, B, C, D are collinear. B X... C AB = DC AD = BC A D AX = XC Figure. Let X be the midpoint of AC so that AX = XC. BX = XC AB = XC DC = XC + CD = XD. Then AB + BX = AX = XC, so..6 Intersecting, parallel and skew lines Let l and l given by generic equations a + su and a + tu respectively be two straight lines. They are parallel if u u. Exactly one of the following holds (see Exercise 5, No.6* for a rigorous proof):. The lines are identical.. They are distinct and parallel. 3. They are not parallel and meet in a (unique) point. 4. They are not parallel and do not intersect (they are by definition skew). l P Q l Figure.

25 .. THE STRAIGHT LINE 5 A familiar example of skew lines is provided by a road passing under a railway bridge. A train and car can only crash at a level crossing! In the case of skew lines there will be points P on l and Q on l such that P Q is perpendicular to the directions of both lines and the distance P Q is minimum. See Figure.. Example..9 Let l be the line through A = (0, 0, 0) and B = (,, ) and suppose l is the line through C = (,, ) and D = (, 4, 3). Are the lines skew? In any case, find the shortest distance between them. Solution: Generic equations for l and l are t(,, ) and (,, )+s(, 3, ) respectively. Since (,, ) (, 3, ), the lines are not parallel (and so cannot possibly be identical). They will meet if for some t and s t(,, ) = (,, ) + s(, 3, ) This means that the three equations t = +s, t = +3s and t = +s must hold. From the first and last equation, +s = ( + s) and thus s = 3, t = 3. But then t = 3 +3 ( ) 3 and the second equation fails. It follows that l and l are skew. Let P = t(,, ) and Q = (,, ) + s(, 3, ) be the points where P Q is minimum. Then P Q = (,, ) + s(, 3, ) t(,, ) is perpendicular to both directions (,, ) and (, 3, ). Hence: ((,, ) + s(, 3, ) t(,, )) (,, ) = + 5s 6t = 0 ((,, ) + s(, 3, ) t(,, )) (, 3, ) = 6 + 4s 5t = 0 Solving these two equations gives s = 6 59 and t = 59. Hence, ( 35 P Q = (,, ) + s(, 3, ) t(,, ) = 59, 5 59, 5 ) 59 The shortest distance between the lines is ( 35 59, 5 59, ) 5 59 = Exercise 5. let A = (,, 3), B = (,, ), C = (5, 4, 3), D = ( 5,, ) and E = (,, 3). Answer the following questions. (a) Find a generic equation with parameter t of the straight line l through A and B in the form of equation (.9). (b) Is E on l? Show that C and D are on l and find a corresponding generic equation of the line with parameter s. Find the relationship between s and t. (c) Find two points on l at a distance 4 from E. (d) Find an equation describing the line segment joining A and B. In particular find the midpoint of the segment and also the three points strictly between A and B dividing the segment into four equal parts.. * Let A, B and C be three distinct points. Show that any of the conditions AB AC, BC AC, and AB BC are equivalent. Hint 6 Since A, B and C are distinct, AB = b a, AC = c a and BC = c b are all non-zero. Let p be the statement AB AC, and q the statement BC AC and r the statement AB BC. The statement p says there is s with b a = s(c a). From this we get c b = c (a + s (c a)) = c a s (c a) = ( s) (c a). We have just shown p q (p implies q). Now substitute C for A, A for B and B for C in p q. You should get q r. Making the same substitutions in q r you should get r p.

26 6CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS 3. Use the previous criterion to show that any three points satisfying equation (.0) lie on a straight line (are collinear) and hence that the equation does indeed define a straight line. Hint 7 Use the results of the previous problem. 4. Let the generic equation r(t) = a + tu represent a line l, where u 0 and suppose that P with position-vector p is a point. Show that P lies on l (a) if, and only if, p a is a multiple of u. (b) if, and only if, the equation has a unique solution for t. a + tu = p (Compare this with No.4 below for another criterion). 5. * Let u and v be non-zero vectors and suppose that a+tu and b+sv are generic equations of lines l and l respectively. Show that necessary and sufficient conditions for the generic equations to represent the same set of points (i.e. l = l ) are (i) u v and (ii) a b is a multiple of v. Using this result show that two lines sharing two distinct points are identical. 6. * Use the results of the previous exercise to show that the four conditions in subsection..6 are in fact mutually exclusive and exhaustive. 7. Find - but not by simply substituting in equation (.) - the point Q of the line l with parametric equation ( t, t, 3 + 5t) which is closest to (a) the origin, Solution. Solve ( t, t, 3 + 5t) (,, 5) = t = 0. So t = 9 30 and Q is ( t, t, 3 + 5t) = ( 4 5, 4 30, ) 6 (b) the point P = (5, 8, 4), (c) the point P = ( 3,, ). 8. Find the projections of the following points P on the line ( t, t, 5t). (a) P = (4, 6, 7) Solution: We have (( t, t, +5t) (4, 6, 7)) (,, 5) = + 30t = 0 and t = 7 0. The projection is ( t, t, +5t) = ( 7 5, 7 0, 7 ). Alternatively, the projection of P on (,, +5) is (b) P = (,, 3). (4, 6, 7) (,, +5) (,, +5) (,, +5) = 7 (,, +5) 0 Hint 8 These can be done in the same way you handled the previous question, but realize that you can use the projection of p on the vector (,, 5). 9. Check your results of 7a and 7b using equation (.). Can you see any connection between 7a and 8b and also between between 7b and 8a?

27 .. THE STRAIGHT LINE 7 0. Find the projection of the point P with position vector p = (p, p, p 3 ) on the line ( t, t, 5t).. Prove the formula (.) that finds the foot Q of the perpendicular from P = (p, p, p 3 ) onto the line a + tu. Deduce the projection formula (.3). Hint 9 Show that the solution to ( a + tu p ) u = 0 (equation (.) is t = ( a p ) u. Show that the closest the line l with parametric equation a + tu comes to the point P is u d = ( p ) (( ) ) a u p a u (.6) u 3. Show that r (t) in equation (.0) is a quadratic in t. How does your knowledge of quadratics tie up with the formula (.) and equation (.6)? 4. Deduce from equation (.6) that l passes through the point P if, and only if, (p a) u = ± p a u. Note that geometrically this is clear in view of the formula for the angle between p a and u if p a * If t is time in seconds and u is the displacement in meters per second (the constant velocity), then equation (.0) can be interpreted as the position of (say) an aircraft in meters at time t. (a) In a situation like this we would be interested in how close the aircraft comes to a certain point P for t 0. Find a condition for (.6) to hold at some time t 0. If this condition fails, what is the closest the aircraft gets to P and when does this occur? (b) How can your findings be used to find the closest distance two aircraft travelling with constant velocities get to each other, assuming that as time t 0 increases the distance between the aircraft decreases? Hint 0 Let a + tu and b + tv describe the positions of the two aircraft at time t. Consider the line b a + t (v u). You want its closest distance from the origin. By assumption, why is (b a) (v u) < 0? 6. A rhombus is a (non-degenerate) parallelogram all of whose sides have the same length. Show that the diagonals of a rhombus intersect at right angles. Hint In Example..8 (Figure.) use AB = BC as well as AC = AB + BC and AB + BD = AD, so BD = AD AB = BC AB. Now use the result of Exercise No *(Generalization of previous exercise). Let a + tu and b + sv represent two straight lines and suppose that A = a + t u, B = b + s v, C = a + t u, and D = b + s v are points on these lines with A C and B D. Show that AB + CD = BC + DA if, and only if, the lines are perpendicular. Hint Expand AB = (b + s v (a + t u)) (b + s v (a + t u)) etc. and show that AB + CD BC DA = u v(t t )(s s ).

28 8CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS 8. Let A, B and C be three points not on a line and so forming the distinct vertices of a triangle. The median from A is the line joining A to the midpoint of the opposite side BC. Show that all three medians intersect in a single point M. This point is called the centroid of the triangle and is the centre of mass of three equal masses placed at the vertices of the triangle. Hint 3 Let a, b, and c, be the position vectors of A, B and C respectively. Then the midpoint of BC has position vector (b + c). A generic equation of the line through A and the midpoint of BC is ( ) a + t (b + c) a Put t = 3 and simplify the result. You should conclude that the centroid lies two thirds of the way from any vertex along its median toward the opposite side. 9. (Cartesian equations for a straight line). Let equation (.0) define a straight line, where u = (u, u, u 3 ) and each u i 0. Show that r = (x, y, z) is on the line if, and only if, x a u = y a u = z a 3 u 3 What happens if one or two of the u i = 0? Find Cartesian equations for the line of Question a above. Find generic and Cartesian equations for the x, y and z axis. 0. Let l be the line through A = (,, 6) and B = (3,, ) and l the line through C = (, 3, ) and D = (,, 5). Solve the following problems. (a) Show that l and l are skew. (b) Find points P on l and Q on l such that P Q is perpendicular to the directions of both lines. Hence find the shortest distance between l and l.. Let l be the line through A = (,, 4) and B = (3, 0, 5) and l the line through C = (, 0, ) and D = (,, 0). Show that l and l meet and find the point of intersection.. How are our general results affected (and simplified) when one of the coordinates (say the z coordinate) is restricted to be zero? In particular what does equation (.) reduce to?.3 Planes.3. Generic Equation for a plane Our first object is to find a generic equation for the plane passing through three non-collinear points A = (a, a, a 3 ), B = (b, b, b 3 ) and C = (c, c, c 3 ). If four points lie on a plane, they are called coplanar. So if R is a point, our first question is when are A, B, C and R coplanar? The answer is suggested by Figure.3: Suppose that R lies in the same plane as A, B and C. Let the line through R parallel to AC meet the line through A and B at the point P. Then AP = sab for some scalar s. Similarly, let the line through R parallel to AB meet the line through A and C at the point Q, so that AQ = tac for some scalar t. The figure AP RQ forms a parallelogram and OR = OA + AR = OA + (AP + P R) = OA + ( AP + AQ ) = OA + sab + tac (.7) This equation describes the position vector r = OR of a general point on the plane through A, B and C and is called a generic equation of the plane through A, B and C.

29 .3. PLANES 9 B sab = AP = QR tac = AQ = P R P R A O Q C Figure.3 Example.3. Let A = (,, ), B = (,, 3) and C = (, 0, ) be three given points. They are not collinear since the vectors AB = (0, 3, ) and AC = (,, ) are not parallel. Hence a generic equation of the plane through A, B and C is r = (,, ) + s(0, 3, ) + t(,, ) In equation (.7) the vectors u = AB and v = AC are non-zero and non-parallel and are thought of as parallel to the plane described. This suggests the following definition:.3. Analytic definition of a plane Let u and v be two non-zero, non-parallel vectors and A = a a given point. Then the set Π of points R = r satisfying the generic (parametric) equation r = r(s, t) = a + su + tv (.8) describes a plane. The parameters are s and t and we call a + su + tv a generic point of the plane..3.3 Equality of planes As with lines, a generic equation for a plane is not unique. Let a + s u + t v and a + s u + t v Represent planes Π and Π respectively. Then Π = Π if, and only if, (i) and sp (u, v ) = sp (u, v ) (ii) a a is in this common span. For the meaning of span see No.4 of Exercise and compare No.5 of Exercise 5. The proof of the above statement is left as an exercise. See No. of Exercise 6 below.

30 30CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS.3.4 Cartesian Equation of a plane A vector n = (n, n, n 3 ) 0 is called a normal of the plane defined by equation (.8) if it is perpendicular to every line lying in the plane. So for a variable point R on the plane we must have n AR = 0. Thus as in Figure.4, all points on the plane with position vector r = (x, x, x 3 ) must satisfy n (r a) = 0 n plane (r a) n = 0 AR R = (x, x, x 3 ) A Figure.4 This is a Cartesian equation of the plane. More fully this reads n x + n x + n 3 x 3 (n a + n a + n 3 a 3 ) (.9) = n x + n x + n 3 x 3 + c = 0 where c = (n a + n a + n 3 a 3 ) A Cartesian equation is unique up to non-zero constant multiples of (n, n, n 3, c) since evidently multiplying n, n, n 3 and c by β 0 cannot change a solution (x, x, x 3 )..3.5 Finding a normal n If n is a normal to the plane given by equation (.8), then in particular n u and n v. In other words, simultaneously n u + n u + n 3 u 3 = 0 (.0) n v + n v + n 3 v 3 = 0 Conversely, if equation (.0) holds then n must be a normal (see Exercise 6, No.4 below). Because equation (.0) involves two equations for three unknowns, a solution n 0 can always be found. See Chapter,..5 for a proof of this. Example.3. Find a normal n 0 to the plane of Example.3. and hence its Cartesian equation.

31 .3. PLANES 3 Solution: We solve (n, n, n 3 ) (0, 3, ) = 3n + n 3 = 0 (n, n, n 3 ) (,, ) = n + n n 3 = 0 simultaneously for n, n and n 3. (We did much the same thing in Example..3). From the first equation, n 3 = 3n. Substituting this into the second equation, we get n = (n n 3 ) = n. Thus, provided n 0, the vector n = (n, n, 3n ) = n (,, 3) is perpendicular (normal) to the plane. We may take n = and n = (,, 3). With A = (,, ) we get a Cartesian equation for the plane: (,, 3) ((x, x, x 3 ) (,, )) = x + x 3x = Some remarks on planes Remark 4 The following statements are geometrically obvious, but analytic proofs can be given after we have studied the subject more deeply, in particular after we ave done Chapter 3. See No.0 in Exercise 78.. Provided n = (n, n, n 3 ) 0, an equation n x + n x + n 3 x 3 + c = 0 always represents a plane with normal n. (If n = 0 the equation represents nothing if c 0 and all of space if c = 0).. By definition, two distinct planes are parallel if their normals are parallel. Such planes are a constant positive distance apart and have no points in common. 3. Two non-parallel planes meet in a line. 4. By definition, a line is parallel to a plane if it is perpendicular to its normal (draw a picture to see that this is correct). A line not parallel to a plane meets the plane in exactly one point. 5. A line and a point not on it determine a unique plane. 6. There are an infinity of planes containing a given line. Example.3.3 Find a generic equation for the line of intersection of the planes Solution: x + x + x 3 + = 0 x + x x 3 + = 0 Note that (,, ) (,, ) so the planes are not parallel. Eliminating x gives x + 3x 3 = 0 or x = 3x 3. Substituting x = 3x 3 into the first (or second) equation gives x = x 3. Hence, with t = x 3 the parameter, is the line of intersection of the two planes. r = (x, x, x 3 ) = (t, 3t, t) Example.3.4 Find a generic equation of the plane with Cartesian equation x x + 5x 3 = 0 (.)

32 3CHAPTER. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS Solution: By far the easiest solution is r = (0 + x 5x 3, x, x 3 ) Here the independent parameters are x and x 3 and they can vary freely over R. We could of course let s = x and t = x 3, but this is only an artificial device and changes nothing. A more long-winded approach is to find three non-collinear points A, B and C on the plane and then proceed as in Example (.3.). Letting x = x 3 = 0 gives x = 0 and A = (0, 0, 0) as one point. Putting x = x = 0 gives x 3 = and B = (0, 0, ). With x = x 3 = 0 we have x = 5 and C = (0, 5, 0). The points A, B and C do not lie on a line and another generic equation of the plane is r = OA + sab + tac = (0 0s 0t, 5t, s) Remark 5 Instead of A, B and C we could have chosen any three non-collinear points satisfying equation (.). The parametric equation will be different but will represent the same plane. Example.3.5 Find two non-parallel vectors u and v both perpendicular to n = (,, 5). Solution: Two such vectors are u = AB and v = AC Example.3.6 Find the point of intersection of the line through (,, 0) and (0,, ) and the plane 3x + x + x 3 + = 0 Solution: For some t the point must be on the given plane, i.e. r = (,, 0) + t ((0,, ) (,, 0)) = (,, 0) + t (,, ) = ( t, + t, t) 3 ( t) + ( + t) + (t) + = 0 Thus t = 3 and the point of intersection is ( 5 3, 7 3, 4 3)..3.7 Perpendicular from a Point to a Plane. Projections If Π is a plane and P a point, the problem in this section is to find the foot Q of the perpendicular from P to the plane. See Figure.5. If the plane passes through the origin, the foot Q is called the projection of P on the plane. Example.3.7 Find the foot Q of the perpendicular from the point P = (4,, 3) to the plane x + y + z = 4. Hence find the shortest distance from P to the given plane. Solution: The vector (,, ) is normal to the plane so the line l defined by (4,, 3) + t (,, ) = (4 + t, + t, 3 + t) passes through P and has direction normal to the plane. Geometry tells