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This pre-publication material is for review purposes only. Any typographical or technical errors will be corrected prior to publication. Copyright Pearson Canada Inc. All rights reserved.

Copyright Pearson Canada Inc. All rights reserved

CHAPTER Euclidean Vector Spaces CHAPTER OUTLINE. Vectors in R and R. Vectors in R n. Length and Dot Products.4 Projections and Minimum Distance.5 Cross-Products and Volumes Some of the material in this chapter will be familiar to many students, but some ideas that are introduced here will be new to most. In this chapter we will look at operations on and important concepts related to vectors. We will also look at some applications of vectors in the familiar setting of Euclidean space. Most of these concepts will later be extended to more general settings. A firm understanding of the material from this chapter will help greatly in understanding the topics in the rest of this book.. Vectors in R and R We begin by considering the two-dimensional plane in Cartesian coordinates. Choose an origin O and two mutually perpendicular axes, called the x -axis and the -axis, as shown in Figure... Then a point P in the plane is identified by the -tuple (p, p ), where p is the distance from P to the -axis, with p positive if P is to the right of this axis and negative if P is to the left. Similarly, p is the distance from P to the x -axis, with p positive if P is above this axis and negative if P is below. You have already learned how to plot graphs of equations in this plane. p Figure.. O p P = (p, p ) x Coordinates in the plane. Copyright Pearson Canada Inc. All rights reserved For applications in many areas of mathematics, and in many subjects such as physics and economics, it is useful to view points more abstractly. In particular, we will view them as vectors and provide rules for adding them and multiplying them by constants.

Chapter Euclidean Vector Spaces Definition R Definition Addition and Scalar Multiplication in R R is the set of all vectors of the form [ x the components of the vector. Mathematically, we write {[ ] } R x = x, R Remark We shall use the notation x = [ x ], where x and are real numbers called ] to denote vectors in R. Although we are viewing the elements of R as vectors, [ ] we can still interpret p these geometrically as points. That is, the vector p = can be interpreted as the point P(p, p ). Graphically, this is often represented by drawing an arrow from (, ) to (p, p ), as shown in Figure... Note, however, that the points between (, ) and (p, p ) should not be thought of as points on the vector. The representation of a vector as an arrow is particularly common in physics; force and acceleration are vector quantities that can conveniently be represented by an arrow of suitable magnitude and direction. If x = [ x ], y = [ y y Figure.. O = (, ) p = p P = (p, p ) ] [ p p x Graphical representation of a vector. ], and t R, then we define addition of vectors by x + y = [ x ] + [ y y ] [ ] x + y = + y The scalar multiplication of a vector by a factor of t, called a scalar, is defined by x tx t x = t = t The addition of two vectors is illustrated in Figure..: construct a parallelogram with vectors x and y as adjacent sides; then x + y is the vector corresponding to the vertex of the parallelogram opposite to the origin. Observe that the components really are added according to the definition. This is often called the parallelogram rule for addition. Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R and R p + q R(p + q, p + q ) EXAMPLE q O 5 Let x = and y =. Then q Q(q, q ) + 5 x + y = = + 4 r = p + q p P(p, p ) p p + q Figure.. Addition of vectors p and q. (, ) O x (, 4) x (5, ) Similarly, scalar multiplication is illustrated in Figure..4. Observe that multiplication by a negative scalar reverses the direction of the vector. It is important to note that x y is to be interpreted as x + ( ) y. (.5) d d O ( ) d Copyright Pearson Canada Inc. All rights reserved x Figure..4 Scalar multiplication of the vector d.

4 Chapter Euclidean Vector Spaces EXAMPLE EXERCISE [ ] Let u =, v =, and w =. Calculate u + v, w, and v w. Solution: We get [ ] u + v = + = 4 w = = [ ] 4 4 v w = + ( ) = + = 6 7 [ ] Let u =, v =, and w =. Calculate each of the following and illustrate with a sketch. (a) u + w (b) v (c) ( u + w) v. The vectors e = and e = play a special role in our discussion of R.We will call the set { e, e } the standard basis for R. (We shall discuss the concept of a basis further [ ] in Section..) The basis vectors e and e are important because any v vector v = can be written as a sum of scalar multiples of e v and e in exactly one way: [ ] v v = = v + v = v e + v e v Remark In physics and engineering, it is common to use the notation i = and j = instead. We will use the phrase linear combination to mean sum of scalar multiples. So, we have shown above that any vector x R can be written as a unique linear combination of the standard basis vectors. [ ] One other vector in R deserves special mention: the zero vector, =.Some important properties of the zero vector, which are easy to verify, are that for any x R, Copyright Pearson Canada Inc. All rights reserved () + x = x () x + ( ) x = () x =

EXAMPLE The Vector Equation of a Line in R Section. Vectors in R and R 5 In Figure..4, it is apparent that the set of all multiples of a vector d creates a line through the origin. We make this our definition of a line in R :aline through the origin in R is a set of the form {t d t R} Often we do not use formal set notation but simply write the vector equation of the line: x = t d, t R The vector d is called the direction vector of the line. Similarly, we define the line through p with direction vector d to be the set which has vector equation { p + t d t R} x = p + t d, t R This line is parallel to the line with equation x = t d, t R because of the parallelogram rule for addition. As shown in Figure..5, each point on the line through p can be obtained from a corresponding point on the line x = t d by adding the vector p. We say that the line has been translated by p. More generally, two lines are parallel if the direction vector of one line is a non-zero multiple of the direction vector of the other line. line x = t d + p p d + p Figure..5 The line with vector equation x = t d + p. [ ] 4 A vector equation of the line through the point P(, ) with direction vector is 5 4 x = + t, t R 5 Copyright Pearson Canada Inc. All rights reserved d x

6 Chapter Euclidean Vector Spaces EXAMPLE 4 EXERCISE EXAMPLE 5 Write the vector equation of a line through P(, ) parallel to the line with vector equation [ ] x = t, t R Solution: Since they are parallel, they have the same direction vector. Hence, the vector equation of the line is x = + t, t R Write the vector equation of a line through P(, ) parallel to the line with vector equation 4 x = + t, t R Sometimes the components of a vector equation are written separately: x = p + td x = p + t d becomes t R = p + td, This is referred to as the parametric equation of the line. The familiar scalar form of the equation of the line is obtained by eliminating the parameter t. Provided that d, d, x p = t = p d d or = p + d (x p ) d What can you say about the line if d = ord =? Write the vector, parametric, [ and] scalar equations of the line passing through the point 5 P(, 4) with direction vector. 5 Solution: The vector equation is x = + t, t R. 4 x = 5t So, the parametric equation is t R. = 4 + t, The scalar equation is = 4 5 (x ). Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R and R 7 EXAMPLE 6 Directed Line Segments For dealing with certain geometrical problems, it is useful to introduce directed line segments. We denote the directed line segment from point P to point Q by PQ. We think of it as an arrow starting at P and pointing towards Q. We shall identify directed line segments from the origin O with the corresponding vectors; we write OP = p, OQ = q, and so on. A directed line segment that starts at the origin is called the position vector of the point. For many problems, we are interested only in the direction and length of the directed line segment; we are not interested in the point where it is located. For example, in Figure.., we may wish to treat the line segment QR as if it were the same as OP. Taking our cue from this example, for arbitrary points P, Q, R in R, we define QR to be equivalent to OP if r q = p. In this case, we have used one directed line segment OP starting from the origin in our definition. O Q Figure..6 A directed line segment from P to Q. More generally, for arbitrary points Q, R, S, T in R, we define QR to be equivalent to ST if they are both equivalent to the same OP for some P. That is, if r q = p and t s = p for the same p We can abbreviate this by simply requiring that r q = t s For points Q(, ), R(6, ), S (, 4), T(, ), QR is equivalent to ST because [ ] 6 5 r q = = = = t s 4 4 S (, 4) Q(, ) O T(, ) R(6, ) Copyright Pearson Canada Inc. All rights reserved P x x

8 Chapter Euclidean Vector Spaces EXAMPLE 7 In some problems, where it is not necessary to distinguish between equivalent directed line segments, we identify them (that is, we treat them as the same object) and write PQ = RS. Indeed, we identify them with the corresponding [ ] line segment starting at the origin, so in Example 6 we write QR = ST 5 =. 4 Remark Writing QR = ST is a bit sloppy an abuse of notation because QR is not really the same object as ST. However, introducing the precise language of equivalence classes and more careful notation with directed line segments is not helpful at this stage. By introducing directed line segments, we are encouraged to think about vectors that are located at arbitrary points in space. This is helpful in solving some geometrical problems, as we shall see below. Find a vector equation of the line through P(, ), and Q(, ). Solution: The direction of the line is [ ] PQ = q p = = Hence, a vector equation of the line with direction PQ that passes through P(, ) is x = p + tpq = + t, t R O P(, ) x Q(, ) Observe in the example above that we would have the same line if we started at the second point and moved toward the first point or even if we took a direction vector in the opposite direction. Thus, the same line is described by the vector equations x = + r, r R x = + s, s R x = + t, t R In fact, there are infinitely many descriptions of a line: we may choose any point on the line, and we may use any non-zero multiple of the direction vector. Copyright Pearson Canada Inc. All rights reserved EXERCISE Find a vector equation of the line through P(, ), and Q(, ).

Definition R Vectors and Lines in R Section. Vectors in R and R 9 Everything we have done so far works perfectly well in three dimensions. We choose an origin O and three mutually perpendicular axes, as shown in Figure..7. The x - axis is usually pictured coming out of the page (or blackboard), the -axis to the right, and the x -axis towards the top of the picture. x O x Figure..7 The positive coordinate axes in R. It should be noted that we are adopting the convention that the coordinate axes form a right-handed system. One way to visualize a right-handed system is to spread out the thumb, index finger, and middle finger of your right hand. The thumb is the x -axis, the index finger is the -axis, and the middle finger is the x -axis. See Figure..8. Figure..8 O x x Identifying a right-handed system. We now define R to be the three-dimensional analog of R. R is the set of all vectors of the form where x,, x are real numbers. Mathematically, we write R = x x x x Copyright Pearson Canada Inc. All rights reserved x,, x R

Chapter Euclidean Vector Spaces Definition Addition and Scalar Multiplication in R EXAMPLE 8 If x = x x, y = y y y, and t R, then we define addition of vectors by x + y = x x + y y y x + y = x + y x + y and the scalar multiplication of a vector by a factor of t by t x = t x x tx = tx tx Addition still follows the parallelogram rule. It may help you to visualize this if you realize that two vectors in R must lie within a plane in R so that the twodimensional picture is still valid. See Figure..9. x x O Figure..9 Two-dimensional parallelogram rule in R. Let u =, v =, and w =. Calculate v + u, w and v + w u. Solution: We have v + u = + = w = = v + w u = + = + + = Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R and R EXAMPLE 9 EXERCISE 4 It is useful to introduce standard basis for R just as we did for R. Define e =, e =, e = Then any vector v = Remark v v v can be written as the linear combination v = v x + v + v x In physics and engineering, it is common to use the notation i = e, j = e and k = e instead. The zero vector = in R has the same properties as the zero vector in R. Directed line segments are the same in three-dimensional space as in the twodimensional case. A line through the point P in R (corresponding to a vector p ) with direction vector d can be described by a vector equation: x = p + t d, t R It is important to realize that a line in R cannot be described by a single scalar linear equation, as in R. We shall see in Section. that such an equation describes a plane in R. Find a vector equation of the line that passes through the points P(, 5, ), and Q(4,, ). Solution: The direction vector is d = q p = 6. Hence a vector equation of the 5 line is x = 5 + t 6, t R 5 Note that the corresponding parametric equations are x = +t, = 5 6t, x = + 5t. Copyright Pearson Canada Inc. All rights reserved Find a vector equation of the line that passes through the points P(,, ), and Q(,, ).

Chapter Euclidean Vector Spaces PROBLEMS. Practice Problems A Compute each of the following linear combinations and[ illustrate ] [ ] with a sketch. 4 (a) + (b) 4 [ ] (c) (d) 4 A Compute each of the following linear combinations. 4 (a) + (b) 4 5 [ ] 4 (c) (d) + 6 /4 (e) (f) [ ] [ ] + / 6 A Compute each of the following linear combinations. 5 (a) (b) + 4 6 4 4 5 (c) 6 5 (d) + 6 / (e) / + (f) + π A4 Let v =, and w =. Determine (a) v w (b) ( v + w) + 5 v (c) u such that w u = v (d) u such that u v = u 5 A5 Let v =, and w =. Determine (a) v + w (b) ( v + w) ( v w) (c) u such that w u = v (d) u such that u + v = w A6 Consider the points P(,, ), Q(,, ), R(, 4, ), S ( 5,, 5). Determine PQ, PR, PS, QR, SRand verify that PQ + QR = PR = PS + SR. A7 Write a vector equation of the line passing through the given points [ with] the given direction vector. 5 (a) P(, 4), d = [ ] 4 (b) P(, ), d = 6 4 (c) P(,, 5), d = (d) P(4,, 5), d = A8 Write a vector equation for the line that passes through the given points. (a) P(, ), Q(, ) (b) P(4, ), Q(, ) (c) P(,, 5), Q(,, ) (d) P(,, ), Q(4,, ) (e) P (, 4, ), Q ( ),, A9 For each of the following lines in R, determine a vector equation and parametric equations. (a) = x + (b) x + = 5 A (a) A set of points in R n is collinear if they all lie on the same line. By considering directed line segments, give a general method for determining whether a given set of three points is collinear. (b) Determine whether the points P(, ), Q(4, ), and R( 5, 4) are collinear. Show how you decide. (c) Determine whether the points S (,, ), T(,, ), and U(, 4, ) are collinear. Show how you decide. Copyright Pearson Canada Inc. All rights reserved

Section. Exercises Homework Problems B Compute each of the following linear combinations and[ illustrate ] [ with ] a sketch. 4 (a) + (b) [ ] 5 (c) (d) B Compute each of the following linear combinations. 6 6 6 6 (a) + (b) [ ] 4 (c) (d) 5 (e) [ ] + [ ] [ ] 6 / B Compute each of the following linear combinations. 5 4 (a) 4 (b) + (c) 4 5 (d) / (e) / + (f) ( + ) 5 B4 Let v = 5, and w =. Determine (a) v w (b) ( v w) w (c) u such that w u = v (d) u such that u + w = v 4 B5 Let v =, and w =. Determine (a) v w (b) v + 4 w (c) u such that v + u = v (d) u such that u w = v B6 (a) Consider the points P(, 4, ), Q(4,, ), R(, 4, ), S (8, 6, 5). Determine PQ, PR, PS, QR, SRand verify that PQ + QR = PR = PS + SR. (b) Consider the points P(,, ), Q(, 7, ), R(,, 5), S (, 4, ). Determine PQ, PR, PS, QR, SRand verify that PQ + QR = PR = PS + SR. B7 Write a vector equation of the line passing through the given points with [ the ] given direction vector. 4 (a) P(, 4), d = (b) P(, ), d = (c) P(,, ), d = 4 8 (d) P(,, ), d = B8 Write a vector equation for the line that passes through the given points. (a) P(, ), Q(, ) (b) P(,, ), Q(,, ) (c) P(, 6, ), Q(, 5, ) (d) P ( ) (,,, Q,, ) B9 For each of the following lines in R, determine a vector equation and parametric equations. (a) = x + (b) x + = B (You will need the solution from Problem A (a) to answer this.) (a) Determine whether the points P(,, ), Q(,, ), and R(4,, ) are collinear. Show how you decide. (b) Determine whether the points S (,, ), T(6,, ), and U( 4,, ) are collinear. Show how you decide. Copyright Pearson Canada Inc. All rights reserved

4 Chapter Euclidean Vector Spaces Computer Problems 6 C Let v = 46, v =, v = 8, and 7 46 47 v 4 = 9. 4 Conceptual Problems D Let v =, and w =. (a) Find [ ] real numbers t and t such that t v +t w =. Illustrate with a sketch. (b) Find [ ] real numbers t and t such that t v +t w = x for any x, R. (c) Use your result in part (b) to find [ real ] numbers t and t such that t v + t v =. π D Let P, Q, R be points in R corresponding to vectors p, q, r respectively.. Vectors in R n Use computer software to evaluate each of the following. (a) 7 v + 5 v v + 4 v 4 (b) 44 v 4 v 99 v + 669 v 4 (a) Explain in terms of directed line segments why PQ + QR + RP = (b) Verify the equation of part (a) by expressing PQ, QR, RP in terms of p, q, r. D Let p and d be vectors in R. Prove that x = p + t d, t R, is a line in R passing through the origin if and only if p is a scalar multiple of d. D4 Let x and y be vectors in R and t R be a scalar. Prove that t( x + y ) = t x + t y We now extend the ideas from the previous section to n-dimensional Euclidean space R n. Students sometimes do not see the point in discussing n-dimensional space because it does not seem to correspond to any physical realistic geometry. But, in a number of instances, more than three dimensions are important. For example to discuss the motion of a particle, an engineer needs to specify its position ( variables) and its velocity ( more variables); she therefore has 6 variables. A scientist working in string theory works with dimensional space-time variables. An economist seeking to model the Canadian economy uses many variables: one standard model has more than 5 variables. Of course, calculations in such huge models are carried out by computer. Even so, understanding the ideas of geometry and linear algebra is necessary to decide which calculations are required and what the results mean. Copyright Pearson Canada Inc. All rights reserved

Definition R n Definition Addition and Scalar Multiplication in R n Theorem Section. Vectors in R n 5 Addition and Scalar Multiplication of Vectors in R n R n is the set of all vectors of the form., where x i R. Mathematically, x x x n x R n =. x,...,x n R If x =., y =., and t R, then we define addition of vectors by x n y y n x n x y x + y x + y =. +. =. x n y n x n + y n and the scalar multiplication of a vector by a factor of t by For all w, x, y R n and s, t R we have x tx t x = t. =. x n tx n () x + y R n (closed under addition) () x + y = y + x (addition is commutative) () ( x + y ) + w = x + ( y + w) (addition is associative) (4) There exists a vector R n such that z + = z for all z R n (zero vector) (5) For each x R n there exists a vector x R n such that x + ( x ) = (additive inverses) (6) t x R n (closed under scalar multiplication) (7) s(t x ) = (st) x (scalar multiplication is associative) (8) (s + t) x = s x + t x (a distributive law) (9) t( x + y ) = t x + t y (another distributive law) () x = x (scalar multiplicative identity) Copyright Pearson Canada Inc. All rights reserved Proof: We will prove properties () and () from Theorem and leave the other proofs to the reader.

6 Chapter Euclidean Vector Spaces For (), by definition, x + y x + y =. Rn x n + y n since x i + y i R for i n. For (), x + y y + x x + y =. =. = y + x x n + y n y n + x n EXERCISE Prove properties (5), (6), and (7) from Theorem. Definition Subspace Observe that properties (), (), (7), (8), (9), and () from Theorem refer only to the operations of addition and scalar multiplication, while the other properties, (), (4), (5), and (6), are about the relationship between the operations and the set R n. These facts should be clear in the proof of Theorem. Moreover, we see that the zero vector of R n is the vector =., and the additive inverse of x is x = ( ) x. Note that the zero vector satisfies the same properties as the zero vector in R and R. Students often find properties () and (6) a little strange. At first glance, it seems obvious that the sum of two vectors in R n or the scalar multiple of a vector in R n is another vector in R n. However, these properties are in fact extremely important. We now look at subsets of R n that have both of these properties. Subspaces A non-empty subset S of R n is called a subspace of R n if for all vectors x, y S and t R: () x + y S (closed under addition) () t x S (closed under scalar multiplication) The definition requires that a subspace be non-empty. A subspace always contains at least one vector. In particular, it follows from () that if we let t = that every subspace of R n contains the zero vector. This fact provides an easy method for disqualify any subsets that do not contain the zero vector as subspaces. For example, a line in R cannot be a subspace if it does not pass through the origin. Thus, when checking to determine if a set S is non-empty, it makes sense to first check if S. It is easy to see that the set { } consisting of only the zero vector in R n is a subspace of R; this is called the trivial subspace. Additionally, R n is a subspace of itself. We will see throughout the text that other subspaces arise naturally in linear algebra. Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R n 7 EXAMPLE EXAMPLE EXERCISE Show that S = x x x + x = is a subspace of R. Solution: We observe that, by definition, S is a subset of R and that = Let x = since taking x =, =, and x = satisfies x + x =. x x, y = y y y S S. Then they must satisfy the condition of the set, so x + x = and y y + y =. To show that S is closed under addition, we must show that x + y satisfies the condition of S.Wehave x + y x + y = x + y x + y and (x + y ) ( + y ) + (x + y ) = x + x + y y + y = + = Hence, x + y S. tx Similarly, for any t R, wehavet x = tx and tx (tx ) (t ) + (tx ) = t(x + x ) = t() = So, S is closed under scalar multiplication. Therefore, S is a subspace of R. Show that T = {[ x ] } x = is not a subspace of R. Solution: To show that T is not a subspace, we just need to give one example showing that T does not satisfy the definition of a subspace. We will show that T is not closed under [ ] addition. [ ] [ ] Observe that x = and y = are both in T, but x + y = T, since (). Thus, T is not a subspace of R. Show that S = {[ x not a subspace of R. ] } x = is a subspace of R and T = {[ x ] } x + = is Copyright Pearson Canada Inc. All rights reserved

8 Chapter Euclidean Vector Spaces Theorem Definition Span Spanning Set EXAMPLE Spanning Sets and Linear Independence One of the main ways that subspaces arise is as the set of all linear combinations of some spanning set. We next present an easy theorem and a bit more vocabulary. If { v,..., v k } is a set of vectors in R n and S is the set of all possible linear combinations of these vectors, then S is a subspace of R n. S = {t v + + t k v k t,...,t k R} Proof: By properties () and (6) of Theorem, t v + + t k v k R n, so S is a subset of R n. Taking t i = for i k, we get = v + + v k S, so S is non-empty. Let x, y S. Then, for some real numbers s i and t i, i k, x = s v + + s k v k and y = t v + + t k v k. It follows that x + y = s v + + s k v k + t v + + t k v k = (s + t ) v + + (s k + t k ) v k so, x + y S since (s i + t i ) R. Hence, S is closed under addition. Similarly, for all t R, t x = t(s v + + s k v k ) = (ts ) v + (ts k ) v k S So, S is closed under scalar multiplication. Therefore, S is a subspace of R n. If S is the subspace of R n consisting of all linear combinations of the vectors v,..., v k R n, then S is called the subspace spanned by B = { v,..., v k }, and we say that the set B spans S. The set B is called a spanning set for the subspace S. We denote S by S = Span{ v,..., v k } = Span B Let v R with v and consider the line L with vector equation x = t v, t R. Then L is the subspace spanned by { v}, and { v} is a spanning set for L. We write L = Span{ v}. Similarly, for v, v R, the set M with vector equation x = t v + t v is a subspace of R with spanning set { v, v }. That is, M = Span{ v, v }. Copyright Pearson Canada Inc. All rights reserved If v R with v, then we can guarantee that Span{ v} represents a line in R that passes through the origin. However, we see that the geometrical interpretation of Span{ v, v } depends on the choices of v and v. We demonstrate this with some examples.

Section. Vectors in R n 9 EXAMPLE 4 EXAMPLE 5 The set S = Span {}, has vector equation [ ] x = t + t = t, t R Hence, S is a line in R{[ that ] [ passes ]} through the origin. The set S = Span, has vector equation x = t + t = (t t ) = t where t = t t R. Hence, S represents the same line as S. That is, {[ ]} S = Span = S {} The set S = Span, has vector equation x = t + t = [ t t ], t, t R Hence, S = R. That is, S spans the entire two-dimensional plane. From these examples, we observe that { v, v } is a spanning set for R if and only if neither v nor v is a scalar multiple of the other. This also means that neither vector can be the zero vector. We now look at this in R. The set S = Span,, has vector equation Hence, x = t + t + t = ( t + t + t ), t, t, t R S = Span Copyright Pearson Canada Inc. All rights reserved

Chapter Euclidean Vector Spaces EXAMPLE 5 (continued) Theorem The set S = Span,, has vector equation x = t + t + t, t, t, t R which can be written as x = t + t + t + t = (t + t ) + (t + t ) So, S = Span,. We extend this to the general case in R n. Let v,..., v k be vectors in R n. If v k can be written as a linear combination of v,..., v k, then Span{ v,..., v k } = Span{ v,..., v k } Proof: We are assuming that there exists t,...,t k R such that t v + + t k v k = v k Let x Span{ v,..., v k }. Then, there exists s,...,s k R such that x = s v + + s k v k + s k v k = s v + + s k v k + s k (t v + + t k v k ) = (s + s k t ) v + + (s k + s k t k ) v k Thus, x Span{ v,..., v k }. Hence, Span{ v,..., v k } Span{ v,..., v k }. Clearly, we have Span{ v,..., v k } Span{ v,..., v k } and so as required. Span{ v,..., v k } = Span{ v,..., v k } In fact, any vector which can written as a linear combination of the other vectors in the set can be removed which changing the spanned set. It is important in linear algebra to identify when a spanning set can be simplified by removing a vector that can be written as a linear combination of the other vectors. We will call such sets linearly dependent. If a spanning set is as simple as possible, then we will call the set linearly independent. To identify whether a set is linearly dependent or linearly independent, we require a mathematical definition. Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R n Definition Linearly Dependent Linearly Independent Theorem 4 EXAMPLE 6 Assume that the set { v,..., v k } is linearly dependent. Then one of the vectors, say v i, is equal to a linear combination of some (or all) of the other vectors. Hence, we can find scalars t,...t k R such that t i v i = t v + + t i v i + t i+ v i+ + + t k v k where t i. Thus, a set is linearly dependent if the equation = t v + + t i v i t i v i + t i+ v i+ + + t k v k has a solution where at least one of the coefficients is non-zero. On the other hand, if the set is linearly independent, then the only solution to this equation must be when all the coefficients are. For example, if any coefficient is non-zero, say t i, then we can write t i v i = t v + + t i v i + t i+ v i+ + + t k v k Thus, v i Span{ v,..., v i, v i+,..., v n }, and so the set can be simplified by using Theorem. We make this our mathematical definition. A set of vectors { v,..., v k } is said to be linearly dependent if there exist coefficients t,...,t k not all zero such that = t v + + t k v k A set of vectors { v,..., v k } is said to be linearly independent if the only solution to = t v + + t k v k is t = t = = t k =. This is called the trivial solution. If a set of vectors { v,..., v k } contains the zero vector, then it is linearly dependent. Proof: Assume v i =. Then we have v + + v i + v i + v i+ + + v k = Hence, the equation = t v + + t k v k has a solution with one coefficient, t i, that is non-zero. So, by definition, the set is linearly dependent. 7 Show that the set 4, 5, is linearly dependent. 6 5/4 Solution: We consider 7 t 4 + t 5 + t = 6 5/4 Copyright Pearson Canada Inc. All rights reserved

Chapter Euclidean Vector Spaces EXAMPLE 6 (continued) EXERCISE Definition Basis EXAMPLE 7 Using operations on vectors, we get 7t t t 4t + 5t 6t + 5 4 t = + t Since vectors are equal only if their corresponding entries are equal, this gives the three equations in three unknowns 7t t t = 4t + 5t = 6t + 5 4 t + t = Solving using substitution and elimination, we find that there are in fact infinitely many possible solutions. One is t = 7, t = 5, t =. Hence, the set is linearly dependent. Determine whether,, is linearly dependent or linearly independent. Remark Observe that determining whether a set { v,..., v k } in R n is linearly dependent or linearly independent requires determining solutions of the vector equation t v + + t k v k =. However, this equation actually represents n equations (one for each entry of the vectors) in k unknowns t,...,t k. In the next chapter, we will look at how to efficiently solve such systems of equations. What we have derived above is that the simplest spanning set for a subspace S is one that is linearly independent. Hence, we make the following definition. If { v,..., v k } is a spanning set for a subspace S of R n and { v,..., v k } is linearly independent, then { v,..., v k } is called a basis for S. Let v =, v =, v = 4 and let S be the subspace of R given by S = Span { } v, v, v. Find a basis for S. Solution: Observe that { } v, v, v is linearly dependent, since 4 = Copyright Pearson Canada Inc. All rights reserved

Section. Vectors in R n EXAMPLE 7 (continued) In particular, we can write v as a linear combination of v and v. Hence, by Theorem, Definition Standard Basis for R n EXAMPLE 8 S = Span{ v, v, v } = Span{ v, v } Moreover, observe that the only solution to t + t = is t = t = since neither v nor v is a scalar multiple of the other. Hence, { v, v } is linearly independent. Therefore, { v, v } is linearly independent and spans S and so is a basis for S. Bases (the plural of basis) will be extremely important throughout the remainder of the book. At this point, however, we just define the following very important basis. In R n, let e i represent the vector whose i-th component is and all other components are. The set { e,..., e n } is called the standard basis for R n. Observe that this definition matches that of the standard basis for R and R given in Section.. The standard basis for R is { e, e, e } =,,. It is linearly independent since the only solution to = t + t + t = is t = t = t =. Moreover, it is a spanning set for R since every vector x = R can be written as a linear combination of the basis vectors. In particular, x x = x + + x Copyright Pearson Canada Inc. All rights reserved Remark x Compare the result of Example 8 with the meaning of point notation P(a, b, c). When we say P(a, b, c) we mean the point P having a amount in the x-direction, b amount in the y-direction, and c amount in the y-direction. So, observe that the standard basis vectors represent our usual coordinate axes. t t t x x

4 Chapter Euclidean Vector Spaces EXERCISE 4 Definition Line in R n Definition Plane in R n Definition Hyperplane in R n EXAMPLE 9 EXAMPLE State the standard basis for R 5. Prove that it is linearly independent and show that it is a spanning set for R 5. Surfaces in Higher Dimensions We can now extend our geometrical concepts of lines and planes to R n for n >. To match what we did in R and R, we make the following definition. Let p, v R n with v. Then we call the set with vector equation x = p +t v, t R a line in R n that passes through p. From the definition of the standard basis, the set { e, e } spans the entire Cartesian plane. That is, the vector equation x = t e + t e, t, t R represents a plane. We extend this definition to R n. Let v, v, p R n, with { v, v } being a linearly independent set. Then the set with vector equation x = p + t v + t v, t, t R is called a plane in R n that passes through p. Let v,..., v n, p R n, with { v,..., v n } being linearly independent. Then the set with vector equation x = p + t v + + t n v n, t i R is called a hyperplane in R n that passes through p. The set Span,, is a hyperplane since,, is linearly independent in R 4. Show that the set Span,, defines a plane in R. Solution: Observe that the set is linearly dependent since + =. Hence, the simplified vector equation of the set spanned by these vectors is Copyright Pearson Canada Inc. All rights reserved

Section. Exercises 5 EXAMPLE (continued) PROBLEMS. Practice Problems A Compute each of the following linear combinations. (a) + (b) + 5 4 (c) + A For each of the following sets, show that the set is or is not a subspace of the appropriate R n. x (a) x x x = x x x (b) x x = x x {[ ] } x (c) x x + = x (d) x x = x x x = t + t, t, t R Since the set, is linearly independent, this is the vector equation of a plane passing through the origin in R. (e) x = + t + t 4 (f) Span A Determine whether the following sets are subspaces of R 4. Explain. (a) { x R 4 x + + x + x 4 = } (b) { v }, where v. (c) { x R 4 x + x = 5, x x 4 = } (d) { x R 4 x = x x 4, x 4 = } (e) { x R 4 x = x 4, 5x = } (f) { x R 4 x + = x 4, x = } A4 Show that each of the following sets is linearly dependent. Do so by writing a non-trivial linear combination of the vectors that equals the zero vector. (a),, (b),, 4 (c),, { [ ]} (d),, Copyright Pearson Canada Inc. All rights reserved

6 Chapter Euclidean Vector Spaces A5 Determine if the following sets represent lines, planes, or hyperplanes in R 4. Give a basis for each. (a) Span, (b) Span,, 6 (c) Span,, Homework Problems B Compute each of the following linear combinations. 4 (a) (b) + + 7 (c) + 6 4 7 B For each of the following sets, show that the set is or is{[ not ] a subspace of } the appropriate R n. x (a) x x + = x (b) x x + x x (c) x x = x, x + = x x (d) x x = x x x (e) x = + t + t (d) Span,, A6 Let p, d R n. Prove that x = p + t d, t R is a subspace of R n if and only if p is a scalar multiple of d. A7 Suppose that B = { v,..., v k } is a linearly independent set in R n. Prove that any non-empty subset of B is linearly independent. (f) Span, B Determine whether the following sets are subspaces of R 4. Explain. (a) { } x R 4 x 5x 4 = 7, = x 4 (b) { x R 4 + x = x + } x 4 (c) { } x R 4 x + + x 4 =, x = x 4 (d) { x R 4 x + x =, x x 4 = } (e), (f) Span, B4 Show that each of the following sets is linearly dependent by writing a non-trivial linear combination of the vectors that equals the zero vector. (a), 6 (b),, 4 4 8 5 (c), 4, 5 { [ ]} (d),, Copyright Pearson Canada Inc. All rights reserved

B5 Determine if the following sets represent lines, planes, or hyperplanes in R 4. Give a basis for each. (a) Span, (b) Span,, Computer Problems. 4. 9.6 4.6.4. C Let v =, v. =, v =,..4.7.99.. and v 4 =...89 Conceptual Problems D Prove property (8) from Theorem. D Prove property (9) from Theorem. D Let U and V be subspaces of R n. (a) Prove that the intersection of U and V is a subspace of R n. (b) Give an example to show that the union of two subspaces of R n does not have to be a subspace of R n. (c) Define U + V = { u + v u U, v V}. Prove that U + V is a subspace of R n. D4 Pick vectors p, v, v, v in R 4 such that the vector equation x = p + t v + t v + t v (a) Is a hyperplane not passing through the origin (b) Is a plane passing through the origin (c) Is the point (,,, ) (d) Is a line passing through the origin D5 Let B = { v,..., v k } be a linearly independent set of vectors in R n. Prove that every vector in Span B can be written as a unique linear combination of the vectors in B. (c) Span,, (d) Span,, Section. Exercises 7 Use computer software to evaluate each of the following. (a) v v + 5 v v 4 (b).4 v. v + v v 4. D6 Let B = { v,..., v k } be a linearly independent set of vectors in R n and let x Span B. Prove that { v,..., v k, x } is linearly independent. D7 Let v, v R n and let s and t be fixed real numbers. Prove that Span{ v, v } = Span{ v, s v + t v } D8 Let v, v, v R n. State whether each of the following statements is true or false. If the statement is true, explain briefly. If the statement is false, give a counterexample. (a) If v = t v for some real number t, then { v, v } is linearly dependent. (b) If v is not a scalar multiple of v, then { v, v } is linearly independent. (c) If { v, v, v } is linearly dependent, then v can be written as a linear combination of v and v. (d) If v can be written as a linear combination of v and v, then { v, v, v } is linearly dependent. (e) { v } is not a subspace of R n. (f) Span{ v } is a subspace of R n. Copyright Pearson Canada Inc. All rights reserved

8 Chapter Euclidean Vector Spaces Definition Length in R If x = Definition If x = Length in R. Length and Dot Products In many physical applications, we are given measurements in terms of angles and magnitudes. We must convert these data into vectors so that we can apply the tools of linear algebra to solve problems. For example, we may need to find a vector representing the path (and speed) of a plane flying northwest at km/h. To do this, we need to identify the length of a vector and the angle between two vectors. In this section, we see how we can calculate both of these quantities with the dot product function. Length and Dot Products in R, and R The length of a vector in R is defined by the usual distance formula (that is, Pythagoras Theorem), as in Figure... [ x ] R, its length is defined to be O x = + x x p P(x, ) Figure.. Length in R. For vectors in R, the formula for the length can be obtained from a two-step calculation using the formula for R, as shown in Figure... Consider the point X(x,, x ) and let P be the point P(x,, ). Observe that OPX is a right triangle, so that x = OP + PX = ( + x ) + x x x R, its length is defined to be x = + x + x Copyright Pearson Canada Inc. All rights reserved One immediate application of this formula is to calculate the distance between two points. In particular, if we have points P and Q, then the distance between them is the length of the directed line segment PQ. x

Section. Length and Dot Products 9 x O + x + x X(x,, x ) x + x P(x,, ) Figure.. Length in R. EXAMPLE Find the distance between the points P(,, 4), Q(, 5, ) in R. Solution: We have PQ = points is ( ) 5 = 8. Hence, the distance between the two 4 PQ = + ( 8) + ( ) = 8 Angles and the Dot Product Determining the angle between two vectors in R leads to the important idea of the dot product of two vectors. Consider Figure... The Law of cosines gives Substituting OP = p = simplifying gives PQ = OP + OQ OP OQ cos θ (.) [ p p ], OQ = q = [ q q p q + p q = p q cos θ For vectors in R, a similar calculation gives p q + p q + p q = p q cos θ Observe that if p = q, then θ = radians, and we get p + p + p = p ] [ ], PQ p q = p q = into (.) and p q Copyright Pearson Canada Inc. All rights reserved

Chapter Euclidean Vector Spaces EXAMPLE O Figure.. P(p, p ) θ p q p q Q(q, q ) PQ = OP + OQ OP OQ cos θ. This matches our definition of length in R above. Thus, we see that the formula on the left-hand side of these equations defines the angle between vectors and also the length of a vector. x y Thus, we define the dot product of two vectors x =, y = in R by Similarly, the dot product of vectors x = x y = x y + y x x and y = x y = x y + y + x y y y y y x in R is defined by Thus, in R and R, the cosine of the angle between vectors x and y can be calculated by means of the formula x y cos θ = (.) x y where θ is always chosen to satisfy θ π. Find the angle in R between v = 4, w =. 4 Solution: We have Hence, v w = () + 4( ) + ( )(4) = 9 v = + 6 + 4 = w = 9 + + 6 = 6 cos θ = 9.856 6 Copyright Pearson Canada Inc. All rights reserved So θ.966 radians. (Note that since cos θ is negative, θ is between π and π.)

Section. Length and Dot Products EXAMPLE EXERCISE Definition Dot Product Theorem Find the angle in R between v = and w =. Solution: We have v w = () + ( )() =. Hence, cos θ = =. Thus, θ = π v w radians. That is, v and w are perpendicular to each other. Find the angle in R between v = and w =. Length and Dot Product in R n O q = x [ ] p = We now extend everything we did in R and R to R n. We begin by defining the dot product. x Let x =., y =. be vectors in Rn. Then the dot product of x and y is Remark x n y y n x y = x y + y + + x n y n The dot product is also sometimes called the scalar product or the standard inner product. From this definition, some important properties follow. Let x, y, z R n and t R. Then, () x x and x x = if and only if x = () x y = y x () x ( y + z) = x y + x z (4) (t x ) y = t( x y ) = x (t y ) Copyright Pearson Canada Inc. All rights reserved [ ] Proof: We leave the proof of these properties to the reader. Because of property (i), we can now define the length of a vector in R n. The word norm is often used as a synonym for length when we are speaking of vectors.

Chapter Euclidean Vector Spaces Definition Norm EXAMPLE 4 EXERCISE Theorem x Let x =. x n. We define the norm or length of x by x = x x = / / Let x = and y =. Find x and y. / Solution: We have x = + + + ( ) = 5 + + x n y = (/) + ( /) + + ( /) = /9 + 4/9 + + 4/9 = Let x = and let y = x. Determine x and y. x We now give some important properties of the norm in R n. Let x, y R n and t R. Then () x and x = if and only if x = () t x = t x () x y x y, with equality if and only if { x, y } is linearly dependent (4) x + y x + y Proof: Property () of Theorem follows immediately from property (i) of Theorem. () t x = (tx ) + + (tx n ) = t + + x n = t x. () Suppose that { x, y } is linearly dependent. Then, either y = or x = t y. If y is the zero vector, then both sides are equal to. If x = t y, then Copyright Pearson Canada Inc. All rights reserved x y = x (t x ) = t( x x ) = t x = x t x = x y Suppose that { x, y } is linearly independent. Then t x + y for any t R. Therefore, by property (),we have (t x + y ) (t x + y ) > for any t R. Use (iii) to expand, and we obtain ( x x )t + ( x y )t + ( y y ) >, for all t R (.)

Section. Length and Dot Products Note that x x > since x. Now a quadratic expression At + Bt + C with A > is always positive if and only if the corresponding equation has no real roots. From the quadratic formula, this is true if and only if B 4AC <. Thus, inequality (.) implies that 4( x y ) 4( x x )( y y ) < which can be simplified to the required inequality. (4) Observe that the required statement is equivalent to x + y ( x + y ) The squared form will allow us to use the dot product conveniently. Thus, we consider x + y ( x + y ) = ( x + y ) ( x + y ) ( x + x y + y ) = x x + x y + y x + y y ( x x + x y + y y ) = x y x y by() Remark EXERCISE Prove that the vector ˆx = Definition Unit Vector Definition Orthogonal Property () is called the Cauchy-Schwarz Inequality (or Cauchy-Schwarz- Buniakowski). Property (4) is the Triangle Inequality. x x is parallel to x and satisfies ˆx =. A vector x R n such that x = is called a unit vector. We will see that unit vectors can be very useful. Thus, we often want to find a unit vector that has the same direction as a given vector x. Using the result in Exercise, we see that this is the vector ˆx = x x We could now define the angle between vectors x and y in R n by matching equation (.). However, in linear algebra we are generally interested only in whether two vectors are perpendicular. To agree with the result of Example, we make the following definition. Copyright Pearson Canada Inc. All rights reserved Two vectors x and y in R n are orthogonal to each other if and only if x y =. Notice that this definition implies that is orthogonal to every vector in R n.

4 Chapter Euclidean Vector Spaces EXAMPLE 5 Let v =, w =, and z =. Show that v is orthogonal to w but v is not orthogonal to z. Solution: We have v w = () + () + () + ( )() =, so they are orthogonal. v z = ( ) + ( ) + () + ( )() =, so they are not orthogonal. The Scalar Equation of Planes and Hyperplanes We saw in Section. that a plane can be described by the vector equation x = p + t v + t v, t, t R, where { v, v } is linearly independent. In many problems, it is more useful to have a scalar equation that represents the plane. We now look at how to use the dot product to find such an equation. Suppose that we want to find the equation of a plane that passes through the point P(p, p, p ). Suppose that we can find a vector n, called the normal vector of the plane, that is orthogonal to any directed line segment PQ lying in the plane. (That is, n is orthogonal to PQ for any point Q in the plane; see Figure...) To find the equation of this plane, let X(x,, x ) be any point on the plane. Then n is orthogonal to PX, so = n PX = n ( x p ) = n (x p ) + n ( p ) + n (x p ) This equation, which must be satisfied by the coordinates of a point X in the plane, can be written in the form n x + n + n x = d, where d = n p + n p + n p = n p This is the standard equation of this plane. For computational purposes, the form n ( x p ) = is often easiest to use. O x Q Copyright Pearson Canada Inc. All rights reserved P n Q Figure.. x The normal n is orthogonal to every directed line segment lying in the plane.

Section. Length and Dot Products 5 EXAMPLE 6 Find the scalar equation of the plane that passes through the point P(,, ) and has normal vector n = 4. Solution: The equation is x n ( x p ) = 4 x = x + or x 4 + x = () + ( 4) + ( )() = It is important to note that we can reverse the reasoning that leads to the equation of the plane in order to identify the set of points that satisfies an equation of the form n x + n + n x = d. Ifn, this can be written as x d/n n x d + n + n x =, or n x = where n = n n n. This equation describes a plane through the point P(d/n,, ), with normal vector n. If n, we could combine the d with the term and find that the plane passes through the point P(, d/n, ). In fact, we could find a point in the plane in many ways, but the normal vector will always be a non-zero scalar multiple of n. EXAMPLE 7 Describe the set of points in R that satisfies 5x 6 + 7x =. Solution: We wish to rewrite the equation in the form n ( x p ) =. Using our work above, we get (5x ) 6 + 7x = or ( 5 x ) 6( ) + 7(x ) = 5 5 Thus, we identify the set as a plane with normal vector n = 6, passing 7 through the point (/5,, ). Alternatively, if x = x =, we find that = /6, so the plane passes through (, /6, ). Copyright Pearson Canada Inc. All rights reserved Two planes are defined to be parallel if the normal vector to one plane is a nonzero scalar multiple of the normal vector of the other plane. Thus, for example, the plane x + y z = is parallel to the plane x + 4y z = 7. x

6 Chapter Euclidean Vector Spaces EXAMPLE 8 EXAMPLE 9 EXERCISE 4 Two planes are orthogonal to each other if their normal vectors are orthogonal. For example, the plane x + + x = is orthogonal to the plane x + x = since = Find a scalar equation of the plane that contains the point P(, 4, ) and is parallel to the plane x + 5x = 6. Solution: An equation of the plane must be of the form x + 5x = d since the planes are parallel. The plane must pass through P, so we find that a scalar equation of the plane is x + 5x = () + (4) + ( 5)( ) = Find a scalar equation of a plane that contains the point P(,, ) and is orthogonal to the plane x + 4x =. Solution: The normal vector must be orthogonal to, so we pick. Thus, an 4 equation of this plane must be of the form + x = d. Since the plane must pass through P, we find that a scalar equation of this plane is + x = () + ( ) + ()() = Find a scalar equation of the plane that contains the point P(,, ) and is parallel to the plane x x = 5. The Scalar Equation of a Hyperplane Repeating what we did above we can find the scalar equation of a hyperplane in R n. In particular, if we have a vector m that is orthogonal to any directed line segment PQ lying in the hyperplane, then for any point X(x,...,x n ) in the hyperplane, we have As before, we can rearrange this as = m PX Copyright Pearson Canada Inc. All rights reserved = m (x p, p, x p ) = m x m vp m x + + m n x n = m p Thus, we see that a single scalar equation in R n represents a hyperplane in R n.

Section. Exercises 7 EXAMPLE PROBLEMS. Practice Problems A Calculate [ ] the lengths of the given vectors. [ ] / 9 (a) (b) 5 5/ 9 (c) (d) / (e) /5 (f) / / (g) (h) A Find[ a unit ] vector in the direction of (a) 4 (c) (e) A Determine the distance from P to Q for (a) P(, ), and Q( 4, ) (b) P(,, ), and Q(,, ) (c) P(4, 6, ), and Q(, 5, ) (d) P(,,, 5), and Q(4, 6,, ) Find the scalar equation of the hyperplane in R 4 that has normal vector m = and passes through the point P(,,, ). Solution: The equation is x + x + x 4 = () + () + ( )() + ( ) = [ ] (b) (d) (f) A4 Verify the triangle inequality and the Cauchy- Schwarz inequality if 4 (a) x =, and y = 5 (b) x =, and y = 4 A5 Determine whether each pair of vectors is orthogonal. (a), (b), 7 4 (c), 4 4 (d), x / / / (e), (f), x / / x 4 A6 Determine all values of k for which each pair of vectors [ ] is[ orthogonal. ] k (a), (b), k k k (c), k k (d), k k 4 Copyright Pearson Canada Inc. All rights reserved

8 Chapter Euclidean Vector Spaces A7 Find a scalar equation of the plane that contains the given point with the given normal vector. (a) P(,, ), n = 4 (b) P(, 5, 4), n = 5 (c) P(,, ), n = 4 4 (d) P(,, ), n = A8 Determine a scalar equation of the hyperplane that passes through the given point with the given normal vector. (a) P(,, ), n = 4 (b) P(,,, ), n = Homework Problems B Calculate [ ] the lengths of the following vectors. [ ] (a) (b) 5/ (c) 6 (d) / / / / (e) / (f) / B Find[ a unit ] vector in the direction of (a) 4 / (c) / / (e) [ ] (b) (d) (f) 4 (c) P(,,, ), n = 5 (d) P(,,,, ), n = A9 Determine a normal vector for the hyperplane. (a) x + = inr (b) x + x = 7inR (c) 4x + 5x 6 = inr (d) x + x x 4 = 5inR 4 (e) x + x + x 4 x 5 = inr 5 A Find an equation for the plane through the given point and parallel to the given plane. (a) P(,, ), x + 5x = 7 (b) P(,, 4), = (c) P(,, ), x + x = 5 A Consider the statement If u v = u w, then v = w. (a) If the statement is true, prove it. If it is false, provide a counterexample. (b) If we specify u, does this change the result? B Determine the distance from P to Q for (a) P(, ), and Q(, ) (b) P(,, 5), and Q( 4,, 4) (c) P(,, ), and Q(, 4, 5) (d) P(5,,, 6), and Q(, 5, 4, ) B4 Verify the triangle inequality and the Cauchy- Schwarz inequality if (a) x = 6, and y = 4 5 4 (b) x =, and y = 5 Copyright Pearson Canada Inc. All rights reserved B5 Determine whether each pair of vectors is orthogonal. 4 (a), (b), 6 4 (c) 4, (d), 4

(e), 5 (f), B6 Determine all values of k for which each pair of vectors [ ] [ is] orthogonal. k k (a), (b), k k k k (c), k (d), k 4 B7 Find a scalar equation of the plane that contains the given point with the given normal vector. (a) P(,, ), n = 4 7 4 (b) P(6,, 5), n = (c) P(,, ), n = (d) P(,, ), n = B8 Determine a scalar equation of the hyperplane that passes through the given point with the given normal vector. Computer Problems...... C Let v = 7., v =.45, v =., and 4.5..4 6.... v 4 =...5.4 k (a) P(,,, 5), n = 5 4 (b) P(,,, 7), n = (c) P(,,, ), n = (d) P(,,,, ), n = Section. Exercises 9 B9 Determine a normal vector for the hyperplane. (a) x + = inr (b) x + 5x = 7inR (c) x + 4 x 4 = inr 4 (d) x + + x x 4 = 5inR 4 (e) x x 5 = inr 5 (f) x + x + x 4 x 5 = inr 5 B Find an equation for the plane through the given point and parallel to the given plane. (a) P(,, 7), 5x x = 6 (b) P(,, 5), + x = 7 (c) P(,, ), x + x = 7 (d) P(,, ), x 5 + x = Use computer software to evaluate each of the following. (a) v v (b) v (c) v v 4 (d) v + v 4 Copyright Pearson Canada Inc. All rights reserved

4 Chapter Euclidean Vector Spaces Conceptual Problems D (a) Using geometrical arguments, what can you say about the vectors p, n, and d if the line with vector equation x = p + t d and the plane with scalar equation n x = k have no point of intersection? (b) Confirm your answer in part (a) by determining when it is possible to find a value of the parameter t that gives a point of intersection. D Prove that, as a consequence of the triangle inequality, x y x y. (Hint: x = x y + y ). D Let v and v be orthogonal vectors in R n. Prove that v + v = v + v. D4 Determine the equation of the set of points in R that are equidistant from points P and Q. Explain why the set is a plane and determine its normal vector. D5 Find the scalar equation of the plane such that each point of the plane is equidistant from the points P(,, 5), and Q(, 4, ) in two ways. (a) Write and simplify the equation PX = QX. (b) Determine a point on the plane and the normal vector by geometrical arguments. D6 Let u R n. Prove that the set of all vectors orthogonal to u is a subspace of R n. D7 Let S be any set of vectors in R n. Let S be the set of all vectors that are orthogonal to every vector in S. That is, S = { w R n v w = for all v S } Show that S is a subspace of R n. D8 Let { v,..., v k } be a set of non-zero vectors in R n such that all of the vectors are mutually orthogonal. That is, v i v j = for all i j. Prove that { v,..., v k } is linearly independent. D9 (a) Let n be a unit vector in R. Let α be the angle between n and the x -axis, let β be the angle between n and the -axis, and let γ be the angle between n and the x -axis. Explain why cos α n = cos β cos γ (Hint: Take the dot product of n with the standard basis vectors.) Because of this equation, the components n, n, n are sometimes called the direction cosines. (b) Explain why cos α + cos β + cos γ =. (c) Give a two-dimensional version of the direction cosines and explain the connection to the identity cos θ + sin θ =..4 Projections and Minimum Distance The idea of a projection is one of the most important applications of the dot product. Suppose that we want to know how much of a given vector y is in the direction of some other given vector x (see Figure.4.4). In elementary physics, this is exactly what is required when a force is resolved into its components along certain directions (for example, into its vertical and horizontal components). When we define projections, it is helpful to think of examples in two or three dimensions, but the ideas do not really depend on whether the vectors are in R, R,orR n. Copyright Pearson Canada Inc. All rights reserved Projections First, [ let] us consider the case where x = e in R. How much of an arbitrary [ vector ] y y y = points along x? Clearly, the part of y that is in the direction of x is = y y e = ( y x ) x. This will be called the projection of y onto x and is denoted proj x y.

Section.4 Projections and Minimum Distance 4 EXAMPLE O a b ( proj a b ) O x x b proj a ( b ) Figure.4.4 proj x y is a vector in the direction of x. Next, consider the case where x R has arbitrary direction and is a unit vector. First, draw the line through the origin with direction vector x. Now, draw the line perpendicular to this line that passes through the point (y, y ). This forms a righttriangle, as in Figure.4.4. The projection of y onto x is the portion of the triangle that lies on the line with direction x. Thus, the resulting projection is a scalar multiple of x. Using trigonometry to calculate the length of the projection gives Hence, proj x y = y cos θ = y x cos θ = y x proj x y = ( y x ) x [ ] [ ] / Find the projection of u = onto the unit vector v = /. Solution: We have x [ ] proj v u = ( u v) v u = / v = ( = + )[ ] / / / = [ ] / / [ ] [ ] proj v u = = If x R is an arbitrary non-zero vector, then the unit vector in the direction of x is ˆx = x. Hence, we find that the projection of y onto x is x ( ) x x y x proj x y = ( y ˆx)ˆx = y = x x x To match this result, we make the following definition for vectors in R n. Copyright Pearson Canada Inc. All rights reserved Definition Projection For any vectors y, x in R n, with x, we define the projection of y onto x by y x proj x y = x a x

4 Chapter Euclidean Vector Spaces EXAMPLE EXAMPLE EXERCISE 4 Let v =, and u = 5. Determine proj v u and proj u v. Solution: We have u v ( )(4) + 5() + ( ) proj v u = v = v = 4 4 8/ v 4 + + ( ) 6 = 6/ / v u (4)( ) + (5) + ( )() proj u v = u = v = 4 4/9 5 u ( ) + 5 + 8 = /9 6/9 Remarks. This example illustrates that, in general, proj x y proj y x. Of course, we should not expect equality because proj x y is in the direction of x, whereas proj y x is in the direction of y.. Observe that for any x R n, we can consider proj x a function whose domain and codomain are R n. To indicate this, we can write proj x : R n R n. Since the output of this function is a vector, we call it a vector-valued function. / Let v = / 5 / and u =. Find proj v u. Solution: Since v =, we get u v proj v u = v v = ( u v) v = 6 / / / = Let v = and u =. Determine proj v u and proj u v. Copyright Pearson Canada Inc. All rights reserved

Definition Perpendicular of a Projection The Perpendicular Part Section.4 Projections and Minimum Distance 4 When you resolve a force in physics, you often not only want the component of the force in the direction of a given vector x, but also the component of the force perpendicular to x. In R, we could find the perpendicular part of the force by finding a vector in the direction perpendicular to x, using the same approach as above. However, to follow a similar strategy in R n for n > would require a lot more mathematical machinery. We now describe a much simpler approach. We begin by restating the problem. In R n, given a fixed vector x and any other y, express y as the sum of a vector parallel to x and a vector orthogonal to x. That is, write y = w + z, where w = c x for some c R and z x =. If this is possible, what can we say about w and z? To find out, we employ a very useful and common trick take the dot product of y with x : x y = x ( w + z) = x (c x + z) = c( x x ) + x z = c x + y x Therefore, c = x, so in fact, w = c x = proj x y, as we might have expected. One bonus of approaching the problem this way is that it is now clear that this is the only way to choose w to satisfy the problem. Next, since y = proj x y + z, it follows that z = y proj x y. Is this z really orthogonal to x? To check, calculate z x = ( y proj x y ) x ( ) y x = y x x x x ( ) y x = y x ( x x ) x ( ) y x = y x x x = y x y x = So, z is orthogonal to x, as required. Since it is often useful to construct a vector z in this way, we introduce a name for it. For any vectors x, y R n, with x, define the projection of y perpendicular to x to be perp x y = y proj x y Notice that perp x y is again a vector-valued function on R n. Also observe that y = proj x y + perp x y. See Figure.4.5. perp x ( y ) Copyright Pearson Canada Inc. All rights reserved θ y x ( ) proj x y O x Figure.4.5 perp x ( y ) is perpendicular to x, and proj x ( y ) + perp x ( y ) = y.

44 Chapter Euclidean Vector Spaces EXAMPLE 4 EXERCISE EXERCISE Let v =, u = 5. Determine proj v u and perp v u. Solution: u v proj v u = v v = 8 8/ 6 = 4/ 4/ 8/ 5/ perp v u = u proj v u = 5 4/ = / 4/ / Let v =, u =. Determine proj v u and perp v u. Some Properties of Projections Projections will appear several times in this book, and some of their special properties are important. Two of them are called the linearity properties: (L) proj x ( y + z) = proj x y + proj x z for all y, z R n (L) proj x (t y ) = t proj x y for all y R n and all t R Verify that properties (L) and (L) are true. It follows that perp x also satisfies the corresponding equations. We shall see that proj x and perp x are just two cases amongst the many functions that satisfy the linearity properties. proj x and perp x also have a special property called the projection property. We write it here for proj x, but it is also true for perp x : Minimum Distance proj x (proj x y ) = proj x y, for all y in R n Copyright Pearson Canada Inc. All rights reserved What is the distance from a point Q(q, q ) to the line with vector equation x = p +t d? In this and similar problems, distance always means the minimum distance. Geometrically, we see that the minimum distance is found along a line segment perpendicular

Section.4 Projections and Minimum Distance 45 EXAMPLE 5 to the given line through a point P on the line. A formal proof that minimum distance requires perpendicularity by using Pythagoras theorem. (See Problem D.) To answer the question, take any point on the line x = p + t d. The obvious choice is P(p, p ) corresponding to p. From Figure.4.6, we see that the required distance is the length perp d PQ. O P proj d PQ PQ Q perp d PQ x = (, ) + t(, ) d = (, ) Figure.4.6 The distance from Q to the line x = p + t d is perp d PQ. Find the distance from Q(4, ) to the line x = + t, t R. Solution: We pick [ the ] point [ ] P(, ) on the line. Then, PQ 4 = =. So, the distance is [ ] [ ] perp d PQ = PQ proj d PQ Q [ ] ( )[ = + P + ] = [ ] + = = O x Notice that in this problem and similar problems, we take advantage of the fact that the direction vector d can be thought of as starting at any point. When perp d AB is calculated, both vectors are located at point P. When projections were originally defined, it was implicitly assumed that all vectors were located at the origin. Now, it is apparent that the definitions make sense as long as all vectors in the calculation are located at the same point. We now want to look at the similar problem of finding the distance from a point Q(q, q, q ) to a plane in R with normal vector n. If P is any point in the plane, then proj n PQ is the directed line segment from the plane to the point Q that is perpendicular Copyright Pearson Canada Inc. All rights reserved x

46 Chapter Euclidean Vector Spaces to the plane. Hence, proj n PQ is the distance from the Q to the plane. Moreover, perp n PQ is a directed line segment lying in the plane. In particular, it is the projection of PQ onto the plane. See Figure.4.7. O x proj n PQ P x n PQ perp n PQ Figure.4.7 proj n PQ and perp n PQ, where n is normal to the plane. EXAMPLE 6 What is the distance from Q(q, q, q ) to a plane in R with equation n x + n + n x = d? Solution: Assuming that n, we pick P(d/n,, ) to be our point in the plane. Thus, the distance is proj n PQ = ( q p ) n n n = ( q p ) n n n = ( q p ) n n = (p d/n )n + p n + p n n + n + n = p n + p n + p n d n + n + n Copyright Pearson Canada Inc. All rights reserved This is a standard formula for this distance problem. However, the lengths of projections along or perpendicular to a suitable vector can be used for all of these problems. It is better to learn to use this powerful and versatile idea, as illustrated in the problems above, than to memorize complicated formulas. R Q

Section.4 Exercises 47 Finding the Nearest Point In some applications, we need to determine the point in the plane that is nearest to the point Q. Let us call this point R, as in Figure.4.7. Then we can determine R by observing that OR = OP + PR = OP + perp n PQ However, we get an easier calculation if we observe from the figure that OR = OQ + QR = OQ + proj n QP Notice that we need QP here instead of PQ. Problem D4 asks you to check that these two calculations of OR are consistent. If the plane in this problem passes through the origin, then we may take P = O, and the point in the plane that is closest to Q is given by perp n q. EXAMPLE 7 Find the point on the plane x + x = 5 that is closest to the point Q(,, ). PROBLEMS.4 Practice Problems A For each given pair of vectors v, u, determine proj v u and perp v u. Check your results by verifying that proj v u + perp v u = u and v perp v u =. (a) v =, u = 5 /5 4 (b) v =, u = 4/5 6 (c) v =, u = 5 / 4 (d) v = /, u = / (e) v =, u = Solution: We pick P(,, ) to be the point on the plane. Then QP =, and we find that the point on the plane closest to Q is OR = q + proj n QP QP = q + n n n = + 7/ 9 = / 5/ (f) v =, u = A Determine [ ] proj v u [ and ] perp v u where (a) v =, u = 4 (b) v =, u = 5 (c) v =, u = 4 (d) v =, u = (e) v =, u = Copyright Pearson Canada Inc. All rights reserved

48 Chapter Euclidean Vector Spaces A Consider the force represented by the vector F = 8 and let u = 6. 6 (a) Determine a unit vector in the direction of u. (b) Find the projection of F onto u. (c) Determine the projection of F perpendicular to u. A4 Follow the same instructions as in Problem A but with F = and u =. A5 For the given point and line, find by projection the point on the line that is closest to the given point and use perp to find the distance from the point to the line. (a) Q(, ), line x = + t, t R 4 (b) Q(, 5), line x = + t, t R 7 4 Homework Problems B For each given pair of vectors v, and u, determine proj v u and perp v u. Check your results by verifying that proj [ ] v u + [ perp ] v u = u and v perp v u =. 4 (a) v =, u = 4/5 (b) v =, u = /5 5 (c) v =, u = 4 7 (d) v =, u = (e) v =, u = 4 (f) v =, u = B Determine [ ] proj v [ u and ] perp v u where (a) v = u = (c) Q(,, ), line x = + t, t R (d) Q(,, ), line x = + t 4, t R A6 Use a projection to find the distance from the point to the plane. (a) Q(,, ), plane x + 4x = 5 (b) Q(,, ), plane x 5x = 5 (c) Q(,, ), plane x x = 5 (d) Q(,, ), plane x x = 4 A7 For the given point and hyperplane in R 4, determine by a projection the point in the hyperplane that is closest to the given point. (a) Q(,,, ), hyperplane x + x + x 4 = (b) Q(,,, ), hyperplane x + x = (c) Q(, 4,, 4), hyperplane x + 4x + x 4 = (d) Q(,,, ), hyperplane x + +x x 4 = 4 (b) v =, u = 5 (c) v =, u = 4 (d) v =, u = 4 6 (e) v =, u = 6 (f) v =, u = B Consider the force represented by the vector F = 5 and let u = 8. 4 (a) Determine a unit vector in the direction of u. (b) Find the projection of F onto u. (c) Determine the projection of F perpendicular to u. Copyright Pearson Canada Inc. All rights reserved

Section.4 Exercises 49 B4 Follow the same instructions as in Problem A but 5 with F =, u = 4. B5 For the given point and line, find by projection the point on the line that is closest to the given point and use perp to find the distance from the point to the line. (a) Q(, 5), line x = + t, t R 4 4 (b) Q(,, ), line x = + t, t R (c) Q(,, ), line x = + t, t R 4 (d) Q(,, ), line x = + t, t R 4 Computer Problems...... C Let v = 7., v =.45, v =., v 4 = 4.5.. 6...4....5.4 Conceptual Problems D (a) Given u, v in R with u and v, verify that the composite map C : R R defined by C( x ) = proj u (proj v x ) also has the linearity properties (L) and (L). (b) (b) Suppose that C( x ) = for all x R, where C is defined as in part (a). What can you say about u and v? Explain. D By linearity property (L), we know that B6 Use a projection to find the distance from the point to the plane. (a) Q(,, ), plane x 5x = 7 (b) Q(,, ), plane x + 4x = 5 (c) Q(,, ), plane x x = 6 (d) Q(,, ), plane x + x = 5 B7 For the given point and hyperplane in R 4, determine by a projection the point in the hyperplane that is closest to the given point. (a) Q(,,, ), hyperplane x x + x 4 = (b) Q(,,, ), hyperplane x + x + x 4 = (c) Q(,,, 6), hyperplane x x + x 4 = (d) Q(5,,, 7), hyperplane x + + 4x + x 4 = 4 Use computer software to evaluate each of the following. (a) proj v v (b) perp v v (c) proj v v (d) proj v 4 v proj u ( x ) = proj u x. Check, and explain geometrically, that proj u x = proj u x. D (a) (Pythagoras theorem) Use the fact that x = x x to prove that x + y = x + y if and only if x y =. (b) Let l be the line in R n with vector equation x = t d and let P be any point that is not on l. Prove that for any point Q on the line, the smallest value of p q is obtained when Copyright Pearson Canada Inc. All rights reserved

5 Chapter Euclidean Vector Spaces q = projd ( p ) (that is, when p q is perpendicular to d). (Hint: Consider p q = p projd ( p ) + proj d ( p ) q.) D4 By using the definition of perp n and the fact that PQ = QP, show that OP + perp n PQ = OQ + proj n QP D5 (a) Let u = and x = 5. Show that proj u (perp u ( x )) =..5 Cross-Products and Volumes Given a pair of vectors u and v inr, how can we find a third vector w that is orthogonal to both u and v? This problem arises naturally in many ways. For example, to find the scalar equation of a plane whose vector equation is x = r u + s v, we must find the normal vector n orthogonal to u and v. In physics, it is observed that the force on an electrically charged particle moving in a magnetic field is in the direction orthogonal to the velocity of the particle and to the vector describing the magnetic field. Cross-Products Let u, v R.If w is orthogonal to both u and v, it must satisfy the equations u w = u w + u w + u w = v w = v w + v w + v w = In Chapter, we shall develop systematic methods for solving such equations for w, w, w. For the present, we simply give a solution: u v u v w = u v u v u v u v EXERCISE Verify that w u = and w v =. Definition Cross-Product Also notice from the form of the equations that any multiple of w would also be orthogonal to both u and v. The cross-product of vectors u = u u u and v = v v v u v u v u v = u v u v u v u v is defined by Copyright Pearson Canada Inc. All rights reserved

Section.5 Cross-Products and Volumes 5 EXAMPLE EXERCISE Calculate the cross-product of and. 5 6 5 Solution: = 5 4 = 9. 5 ( ) 5 Remarks. Unlike the dot product of two vectors, which is a scalar, the cross-product of two vectors is itself a new vector.. The cross-product is a construction that is defined only in R. (There is a generalization to higher dimensions, but it is considerably more complicated, and it will not be considered in this book.) The formula for the cross-product is a little awkward to remember, but there are many tricks for remembering it. One way is to write the components of u in a row above the components of v: u u u v v v Then, for the first entry in u v, we cover the first column and calculate the difference of the products of the cross-terms: u u v v u v u v For the second entry in u v, we cover the second column and take the negative of the difference of the products of the cross-terms: u u v v (u v u v ) Similarly, for the third entry, we cover the third column and calculate the difference of the products of the cross-terms: u u v v u v u v Note carefully that the second term must be given a minus sign in order for this procedure to provide the correct answer. Since the formula can be difficult to remember, we recommend checking the answer by verifying that it is orthogonal to both u and v. Calculate the cross-product of and. 7 Copyright Pearson Canada Inc. All rights reserved

5 Chapter Euclidean Vector Spaces EXERCISE Theorem By construction, u v is orthogonal to u and v, so the direction of u v is known except for sign: does it point up or down? The general rule is as follows: the three vectors u, v, and u v, taken in this order, form a right-handed system. Let us see how this works for simple cases. Let e, e, e be the standard basis vectors in R. Verify that but e e = e, e e = e, e e = e e e = e, e e = e, e e = e Check that in every case, the three vectors taken in order form a right-handed system. These simple examples also suggest some of the general properties of the crossproduct. For x, y, z R and t R, we have () x y = y x () x x = () x ( y + z) = x y + x z (4) (t x ) y = t( x y ) Proof: These properties follow easily from the definition of the cross-product and are left to the reader. One rule we might expect does not in fact hold. In general, x ( y z) ( x y ) z This means that the parentheses cannot be omitted in a cross-product. (There are formulas available for these triple-vector products, but we shall not need them. See Problem F4 in Further Problems at the end of this chapter.) The Length of the Cross-Product Given u and v, the direction of their cross-product is known. What is the length of the cross-product of u and v? We give the answer in the following theorem. Theorem Let u, v R and θ be the angle between u and v, then u v = u v sin θ. Copyright Pearson Canada Inc. All rights reserved Proof: We give an outline of the proof. We have u v = (u v u v ) + (u v u v ) + (u v u v )

Section.5 Cross-Products and Volumes 5 EXAMPLE Expand by the binomial theorem and then add and subtract the term (u v +u v +u v ). The resulting terms can be arranged so as to be seen to be equal to Thus, and the result follows. (u + u + u )(v + v + v ) (u v + u v + u v ) u v = u v ( u v) = u v u v cos θ = u v ( cos θ) = u v sin θ To interpret this formula, consider Figure.5.8. Assuming that u v, the vectors u and v determine a parallelogram. Take the length of u to be the base of the parallelogram; the altitude is the length of perp u v. From trigonometry, we know that this length is v sin θ, so that the area of the parallelogram is (base) (altitude) = u v sin θ = u v θ v height v sin θ length u Figure.5.8 The area of the parallelogram is u v sin θ. Find the area of the parallelogram determined by u = and v =. Solution: By Theorem, we find that the area is Copyright Pearson Canada Inc. All rights reserved u v = (,, ) = u x

54 Chapter Euclidean Vector Spaces EXAMPLE EXERCISE 4 EXAMPLE 4 Find the area of the parallelogram determined by u = and v =. Solution: By Theorem, the area is u v = (,, ) = 4 Find the area of the parallelogram determined by u = and v =. Some Problems on Lines, Planes, and Distances The cross-product allows us to answer many questions about lines, planes, and distances in R. Finding the Normal to a Plane In Section., the vector equation of a plane was given in the form x = p + s u + t v, where { u, v} is linearly independent. By definition, the normal vector n must be perpendicular to u and v. Therefore, it will be given by n = u v. The lines x = + s and x = + t must lie in a common plane since they have the point (,, ) in common. Find a scalar equation of the plane that contains these lines. Solution: The normal to the plane is 4 n = = Therefore, since the plane passes through P(,, ), we find that an equation of the plane is 4x + x = ( 4)() + ( )() + () = 9 Copyright Pearson Canada Inc. All rights reserved

Section.5 Cross-Products and Volumes 55 EXAMPLE 5 Find a scalar equation of the plane that contains the three points P(,, ), Q(,, ), R(4,, ). Solution: Since P, Q, R lie in the plane, then so do the directed line segments PQ, PR. Hence, the normal to the plane is given by 6 n = PQ PR = = 6 Since the plane passes through P, we find that an equation of the plane is 6x 6 + x = (6)() + ( 6)( ) + () =, or x + x = 7 Finding the Line of Intersection of Two Planes Unless two planes in R are parallel, their intersection will be a line. The direction vector of this line lies in both planes, so it is perpendicular to both of the normals. It can therefore be obtained as the cross-product of the two normals. Once we find a point that lies on this line, we can write the vector equation of the line. EXAMPLE 6 Find a vector equation of the line of intersection of the two planes x + x = EXERCISE 5 and x + x = 6. Solution: The normal vectors of the planes are and. Hence, the direction vector of the line of intersection is d = = 7 One easy way to find a point on the line is to let x = and then solve the remaining equations x + = and x = 6. The solution is x = and =. Hence, a vector equation of the line of intersection is x = + t 7, t R Find a vector equation of the line of intersection of the two planes x + x = and x + x =. Copyright Pearson Canada Inc. All rights reserved The Scalar Triple Product and Volumes in R The three vectors u, v, and w in R may be taken to be the three adjacent edges of a parallelepiped

56 Chapter Euclidean Vector Spaces EXAMPLE 7 (see Figure.5.9). Is there an expression for the volume of the parallelepiped in terms of the three vectors? To obtain such a formula, observe that the parallelogram determined by u and v can be regarded as the base of the solid of the parallelepiped. This base has area u v. With respect to this base, the altitude of the solid is the length of the amount of w in the direction of the normal vector n = u v to the base. That is, w n w ( u v) altitude = proj n w = = n u v Thus, to get the volume of the parallelepiped, multiply this altitude by the area of the base to get w ( u v) volume of the parallelepiped = u v = w ( u v) u v The product w ( u v) is called the scalar triple product of w, u, and v. Notice that the result is a real number (a scalar). Figure.5.9 u v w v altitude proj n w O u base area u v The parallelepiped with adjacent edges u, v, w has volume given by w ( u v). The sign of the scalar triple product also has an interpretation. Recall that the ordered triple of vectors { u, v, u v} is right-handed; we can think of u v as the upwards normal vector to the plane with vector equation x = s u + t v. Some other vector w is then upwards, and { u, v, w} (in that order) is right-handed, if and only if the scalar triple product is positive. If the triple scalar product is negative, then { u, v, w} is a left-handed system. It is often useful to note that w ( u v) = u ( v w) = v ( w u) This is straightfoward but tedious to verify. Find the volume of the parallelepiped determined by the vectors,,. Solution: The volume V is V = 4 = 7 = Copyright Pearson Canada Inc. All rights reserved

PROBLEMS.5 Practice Problems A Calculate the following cross-products. 4 (a) 5 (b) 5 5 7 (c) 4 (d) 5 4 (e) (f) 6 A Let u = 4, v = and w =. Check by calculation that the following general properties hold. (a) u u = (b) u v = v u (c) u w = ( u w) (d) u ( v + w) = u v + u w (e) u ( v w) = w ( u v) (f) u ( v w) = v ( u w) A Calculate the area of the parallelogram determined by each pair of vectors. (a), (b), 4 4 (c), 5 (d), (Hint: For (d), think of these vectors as and 4 in R.) A4 Determine the scalar equation of the plane with vector equation 4 (a) x = 4 + s + t 7 (b) x = + s + t (c) x = + s + t (d) x = s + t 4 Section.5 Exercises 57 A5 Determine the scalar equation of the plane that contains each set of points. (a) P(,, 5), Q(4,, ), R(, 6, ) (b) P(,, 4), Q(,, ), R(, 4, ) (c) P(, 4, ), Q(,, ), R(,, ) (d) P(,, ), Q(,, ), R(,, ) A6 Determine a vector equation of the line of intersection of the given planes. (a) x + x = 5 and x 5 + x = 7 (b) x x = 7 and + x = 4 A7 Find the volume of the parallelepiped determined by each set of vectors. (a),, 4 (b), 5, 6 (c),, 5 (d) 5,, 4 (e), 4, Copyright Pearson Canada Inc. All rights reserved

58 Chapter Euclidean Vector Spaces Homework Problems B Calculate the following cross-products. 4 6 4 (a) (b) 4 4 6 6 5 (c) (d) 5 (e) 4 (f) / 4 B Let u =, v = 4, and w =. Check by calculation that the following general properties hold. (a) u u = (b) u v = v u (c) u w = ( u w) (d) u ( v + w) = u v + u w (e) u ( v w) = w ( u v) (f) u ( v w) = v ( u w) B Calculate the area of the parallelogram determined by each pair of vectors. 5 (a), (b), 5 4 5 (c), (d), 5 5 (Hint: For (d), think of these vectors as and in R.) B4 Determine the scalar equation of the plane with vector equation 4 (a) x = + s + t 4 5 5 Conceptual Problems D Show that if X is a point on the line through P and Q, then x ( q p ) = p q, where x = OX, p = OP, and q = OQ. D Consider the following statement: If u, and u v = u w, then v = w. If the statement is true, prove it. If it is false, give a counterexample. (b) x = + s + t (c) x = 4 + s + t 5 (d) x = s + t B5 Determine the scalar equation of the plane that contains each set of points. (a) P(5,, ), Q(,, 4), R(8,, 6) (b) P(5,, ), Q(,, 4), R(,, ) (c) P(,, ), Q(,, ), R(,, ) (d) P(, 5, ), Q(, 6, ), R(,, ) B6 Determine a vector equation of the line of intersection of the given planes. (a) x + 4 + x = 5 and x 7 x = 6 (b) x x = 4 and x + + x = B7 Find the volume of the parallelepiped determined by each set of vectors. 5 (a), 4, 6 (b),, 4 4 5 (c),, 4 6 7 (d),, 5 5 4 Copyright Pearson Canada Inc. All rights reserved D Explain why u ( v w) must be a vector that satisfies the vector equation x = s v + t w. D4 Give an example of distinct vectors u, v, and w in R such that (a) u ( v w) = ( u v) w (b) u ( v w) ( u v) w

CHAPTER REVIEW Suggestions for Student Review Organizing your own review is an important step towards mastering new material. It is much more valuable than memorizing someone else s list of key ideas. To retain new concepts as useful tools, you must be able to state definitions and make connections between various ideas and techniques. You should also be able to give (or, even better, create) instructive examples. The suggestions below are not intended to be an exhaustive checklist; instead, they suggest the kinds of activities and questioning that will help you gain a confident grasp of the material. Find some person or persons to talk with about mathematics. There s lots of evidence that this is the best way to learn. Be sure you do your share of asking and answering. Note that a little bit of embarrassment is a small price for learning. Also, be sure to get lots of practice in writing answers independently. Draw pictures to illustrate addition of vectors, subtraction of vectors, and multiplication of a vector by a scalar (general case). (Section.) Explain how you find a vector equation for a line and make up examples to show why the vector equation of a line is not unique. (Albert Einstein once said, If you can t explain it simply, you don t understand it well enough. ) (Section.) 4 State the definition of a subspace of R n. Give examples of subspaces in R that are lines, planes, and all of R. Show that there is only one subspace in R that does not have infinitely many vectors in it. (Section.) 5 Show that the subspace spanned by three vectors in R can either be a point, a line, a plane, or all of R, Chapter Quiz Note: Your instructor may have different ideas of an appropriate level of difficulty for a test on this material. Determine a vector equation of the line passing through points P(,, 4), and Q(5,, ). Determine the scalar equation of the plane that contains the points P(,, ), Q(,, ), and R( 4,, 6). Section.5 Chapter Review Student Review 59 by giving examples. Explain how this relates with the concept of linear independence. (Section.) 6 Let { v, v } be a linearly independent spanning set for a subspace S of R. Explain how you could construct other spanning sets and other linearly independent spanning sets for S. (Section.) 7 State the formal definition of linear independence. Explain the connection between the formal definition of linear dependence and an intuitive geometric understanding of linear dependence. Why is linear independence important when looking at spanning sets? (Section.) 8 State the relation (or relations) between the length in R and the dot product in R. Use examples to illustrate. (Section.) 9 Explain how projection onto a vector v is defined in terms of the dot product. Illustrate with a picture. Define the part of a vector x perpendicular to v and verify (in the general case) that it is perpendicular to v. (Section.4) Explain with a picture how projections help us to solve the minimum distance problem. (Section.4) Discuss the role of the normal vector to a plane in determining the scalar equation of the plane. Explain how you can get from a scalar equation of a plane to a vector equation for the plane and from a vector equation of the plane to the scalar equation. (Sections. and.5) State the important algebraic and geometric properties of the cross-product. (Section.5) {} Show that, is a basis for R. x 4 Prove that S = x R a + a + a x = d Copyright Pearson Canada Inc. All rights reserved x is a subspace of R for any real numbers a, a, a if and only if d =.

6 Chapter Euclidean Vector Spaces 5 Determine the cosine of the angle between v = and each of the coordinate axes. 6 Find the point on the line x = t, t R that is closest to the point P(,, 4). Illustrate your method of calculation with a sketch. 7 Find the point on the hyperplane x + + x + x 4 = that is closest to the point P(,,, ) and determine the distance from the point to the plane. 8 Determine a non-zero vector that is orthogonal to both and. 9 Prove that the volume of the parallelepiped determined by u, v, and w has the same volume as the parallelepiped determined by ( u + k v), v, w. Further Problems These problems are intended to be a little more challenging than the problems at the end of each section. Some explore topics beyond the material discussed in the text. F Consider the statement If u, and both u v = u w and u v = u w, then v = w. Either prove the statement or give a counterexample. F Suppose that u and v are orthogonal unit vectors in R. Prove that for every x R, perp u v ( x ) = proj u ( x ) + proj v ( x ) F In Problem.5 D, you were asked to show that u ( v w) = s v + t w for some s, t R. (a) By direct calculation, prove that u ( v w) = ( u w) v ( u v) w. (b) Prove that u ( v w)+ v ( w u)+ w ( u v) =. F4 Prove that (a) u v = 4 u + v 4 u v (b) u + v + u v = u + v Each of the following statements to be interpreted in R. Determine whether each statement is true, and if so, explain briefly. If false, give a counterexample. (i) Any three distinct points lie in exactly one plane. (ii) The subspace spanned by a single non-zero vector is a line passing through the origin. (iii) The set Span{ v,..., v k } is linearly dependent. (iv) The dot product of a vector with itself cannot be zero. (v) For any vectors x and y, proj x y = proj y x. (vi) For any vectors x and y, the set {proj x y, perp x y } is linearly independent. (vii) The area of the parallelogram determined by u and v is the same as the area of the parallelogram determined by u and ( v + u). (c) Interpret (a) and (b) in terms of a parallelogram determined by vectors u, and v. F5 Show that if P, Q, and R are collinear points and OP = p, OQ = q, and OR = r, then ( p q ) + ( q r) + ( r p ) = F6 In R, two lines fail to have a point of intersection only if they are parallel. However, in R, a pair of lines can fail to have a point of intersection even if they are not parallel. Two such lines in R are called skew. (a) Observe that if two lines are skew, then they do not lie in a common plane. Show that two skew lines do lie in parallel planes. (b) Find the distance between the skew lines x = 4 +s, s R and x = +t, t R Copyright Pearson Canada Inc. All rights reserved Visit the text s website at www.pearsoncanada.ca/norman for practice quizzes, additional applications, and an essay on linearity and superposition in physics.

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