Linear ALlgebra a modern introduction 4th edition by poole solution manual

Linear ALlgebra A Modern Introduction 4th edition by Poole Solution Manual Link full download solution manual: https://findtestbanks.com/download/linear-algebraa-modern-introduction-4th-edition-by-poole-solution-manual/ Linear Algebra A Modern Introduction FOURTH EDITION David Poole Trent University Prepared by Roger Lipsett Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States ISBN-13: 978-128586960-5 ISBN-10: 1-28586960-5 © 2015 Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as 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of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they Printed in the United States of America 17 16 15 14 13 may not copy or distribute any portion of the Supplement to any third party You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you not accept these conditions, you must return the Supplement unused within 30 days of receipt All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool Contents Vectors 1.1 The Geometry and Algebra of Vectors 1.2 Length and Angle: The Dot Product 10 Exploration: Vectors and Geometry 25 1.3 Lines and Planes 27 Exploration: The Cross Product 41 1.4 Applications .44 Chapter Review 48 Systems of Linear Equations 53 2.1 Introduction to Systems of Linear Equations 53 2.2 Direct Methods for Solving Linear Systems 58 Exploration: Lies My Computer Told Me 75 Exploration: Partial Pivoting 75 Exploration: An Introduction to the Analysis of Algorithms 77 2.3 Spanning Sets and Linear Independence 79 2.4 Applications .93 2.5 Iterative Methods for Solving Linear Systems 112 Chapter Review 123 Matrices 129 3.1 Matrix Operations 129 3.2 Matrix Algebra 138 3.3 The Inverse of a Matrix 150 3.4 The LU Factorization 164 3.5 Subspaces, Basis, Dimension, and Rank 176 3.6 Introduction to Linear Transformations 192 3.7 Applications 209 Chapter Review 230 Eigenvalues and Eigenvectors 235 4.1 Introduction to Eigenvalues and Eigenvectors 235 4.2 Determinants 250 Exploration: Geometric Applications of Determinants 263 4.3 Eigenvalues and Eigenvectors of n × n Matrices 270 4.4 Similarity and Diagonalization 291 4.5 Iterative Methods for Computing Eigenvalues 308 4.6 Applications and the Perron-Frobenius Theorem 326 Chapter Review 365 CONTENTS Orthogonality 371 5.1 Orthogonality in Rn 371 5.2 Orthogonal Complements and Orthogonal Projections 379 5.3 The Gram-Schmidt Process and the QR Factorization 388 Exploration: The Modified QR Process 398 Exploration: Approximating Eigenvalues with the QR Algorithm 402 5.4 Orthogonal Diagonalization of Symmetric Matrices 405 5.5 Applications 417 Chapter Review 442 Vector Spaces 451 6.1 Vector Spaces and Subspaces 451 6.2 Linear Independence, Basis, and Dimension 463 Exploration: Magic Squares 477 6.3 Change of Basis 480 6.4 Linear Transformations 491 6.5 The Kernel and Range of a Linear Transformation 498 6.6 The Matrix of a Linear Transformation 507 Exploration: Tiles, Lattices, and the Crystallographic Restriction 525 6.7 Applications 527 Chapter Review 531 Distance and Approximation 537 7.1 Inner Product Spaces 537 Exploration: Vectors and Matrices with Complex Entries 546 Exploration: Geometric Inequalities and Optimization Problems 553 7.2 Norms and Distance Functions 556 7.3 Least Squares Approximation 568 7.4 The Singular Value Decomposition 590 7.5 Applications 614 Chapter Review 625 Codes 633 8.1 Code Vectors 633 8.2 Error-Correcting Codes 637 8.3 Dual Codes 641 8.4 Linear Codes 647 8.5 The Minimum Distance of a Code 650 Chapter Vectors 1.1 The Geometry and Algebra of Vectors (–2, 3} (2, 3} (3, 0} –2 –1 –1 (3, –2} –2 Since Σ Σ Σ Σ Σ Σ + = , −3 −3 Σ Σ Σ Σ Σ Σ 2 + = , −3 Σ Σ Σ Σ Σ Σ −2 + = , −3 plotting those vectors gives –1 c b –2 a –3 d –4 –5 Σ Σ Σ Σ Σ Σ + = , −3 −2 −5 CHAPTER VECTORS 2z c b –2 –1 0 y2 –1 a x –1 d –2 Since the heads are all at (3, 2, 1), the tails are at 3 − = 0, 1 3 − = 0, 1 2 − −2 = , 1 #» The four vectors AB are c d a –1 b –2 In standard position, the vectors are #» (a) AB = [4 − 1, − (− 1)] = [3, 3] #» (b) AB = [2 − 0, −1 − (−2)] = [2, 1] Σ Σ Σ #» Σ (c) AB = − 2, − = − , 2 2 #» Σ6 Σ Σ3 Σ (d) AB = − , − = − , a c b d –1 3 − −1 −1 −2 = 1.1 THE GEOMETRY AND ALGEBRA OF VECTORS Recall the notation that [a, b] denotes a move of a units horizontally and b units vertically Then during the first part of the walk, the hiker walks km north, so a = [0, 4] During the second part of the walk, the hiker walks a distance of km northeast FromΣthe components, we get √ √ Σ 5 b = [5 cos 45◦ , sin 45◦ ] = , 2 Thus the net displacement vector is c=a+b= Σ √ √ Σ ,4+ 7.a + b = Σ Σ Σ Σ Σ Σ Σ Σ 3+2 + = = +3 a +b b a Σ Σ Σ Σ Σ Σ Σ Σ −2 − (−2) − = = 8.b −c = 3−3 3 —c b b—c Σ Σ Σ Σ Σ Σ −2 9.d − c = − = −2 −5 2 3 d –1 –2 d —c –3 —c –4 10.a + d = Σ Σ −2 ΣΣ Σ Σ Σ Σ 3 3+3 + = = −2 + (−2) –5 a d –1 a +d –2 11 2a + 3c = 2[0, 2, 0] + 3[1, −2, 1] = [2 · 0, · 2, · 0] + [3 · 1, · (−2), · 1] = [3, −2, 3] 12 3b − 2c + d = 3[3, 2, 1] − 2[1, −2, 1] + [−1, −1, −2] = [3 · 3, · 2, · 1] + [−2 · 1, −2 · (−2), −2 · 1] + [−1, −1, −2] = [6, 9, −1] CHAPTER VECTORS 13.u = [cos 60◦ , sin 60◦ ] = Σ , √ Σ Σ u+v = Σ √ Σ , and v = [cos 210◦ , sin 210◦ ] = − 32, − 2, so that Σ √ √ 3 − , − , 2 2 Σ u−v = Σ √ √ 3 + , + 2 2 #» AB = b − a #» #» #» #» (b) Since OC = AB, we have BC = OC − b = (b − a) − b = − a #» (c) AD = −2a #» #» #» (d) CF = −2OC = −2AB = −2(b − a) = 2(a − b) # » # » #» (e) AC = AB + BC = (b − a) + (−a) = b − 2a #» #» #» #» (f ) Note that F A and OB are equal, and that DE = −AB Then 14.(a) #» #» #» #» #» BC + DE + F A = −a − AB + OB = −a − (b − a) + b = 15 2(a − 3b) + 3(2b + a) property e distributivity = (2a − 6b) + (6b + 3a) property b associativity = (2a + 3a) + (−6b + 6b) = 5a 16 property e distributivity −3(a − c) + 2(a + 2b) + 3(c − b) = (−3a + 3c) + (2a + 4b) + (3c − 3b) property b associativity = (−3a + 2a) + (4b − 3b) + (3c + 3c) = −a + b + 6c 17.x − a = 2(x − 2a) = 2x − 4a ⇒ x − 2x = a − 4a ⇒ −x = −3a ⇒ x = 3a 18 x + 2a − b = 3(x + a) − 2(2a − b) = 3x + 3a − 4a + 2b x − 3x = −a − 2a + 2b + b ⇒ −2x = −3a + 3b ⇒ 3 x = a − b 2 ⇒ 19 We have 2u + 3v = 2[1, −1] + 3[1, 1] = [2 · + · 1, · (−1) + · 1] = [5, 1] Plots of all three vectors are w v –1 v u –1 u –2 v v 1.1 THE GEOMETRY AND ALGEBRA OF VECTORS 20 We have −u − 2v = −[−2, 1] − 2[2, −2] = [−(−2) − · 2, −1 − · (−2)] = [−2, 3] Plots of all three vectors are —u —v —u w —v u –2 –1 v –1 –2 21 From the diagram, we see that w = −2u + 4v —u —u w v v v v –1 u –1 22 From the diagram, we see that w = 2u + 3v u w u v u v v –2 –1 23 Property (d) states that u + ( −u) = The first diagram below shows u along with − u Then, as the diagonal of the parallelogram, the resultant vector is Property (e) states that c(u + v) = cu + cv The second figure illustrates this CHAPTER VECTORS cv c(u + v) cu v cu u u u +v u cv v —u 24 Let u = [u1, u2, , un] and v = [v1, v2, , vn], and let c and d be scalars in R Property (d): u + (−u) = [u1, u2, , un] + (−1[u1, u2, , un]) = [u1, u2, , un] + [−u1, −u2, , −un] = [u1 − u1, u2 − u2, , un − un] = [0, 0, , 0] = Property (e): c(u + v) = c ([u1, u2, , un] + [v1, v2, , vn]) = c ([u1 + v1, u2 + v2, , un + vn]) = [c(u1 + v1), c(u2 + v2), , c(un + vn)] = [cu1 + cv1, cu2 + cv2, , cun + cvn] = [cu1, cu2, , cun] + [cv1, cv2, , cvn] = c[u1, u2, , un] + c[v1, v2, , vn] = cu + cv Property (f): (c + d)u = (c + d)[u1, u2, , un] = [(c + d)u1, (c + d)u2, , (c + d)un] = [cu1 + du1, cu2 + du2, , cun + dun] = [cu1, cu2, , cun] + [du1, du2, , dun] = c[u1, u2, , un] + d[u1, u2, , un] = cu + du Property (g): c(du) = c(d[u1, u2, , un]) = c[du1, du2, , dun] = [cdu1, cdu2, , cdun] = [(cd)u1, (cd)u2, , (cd)un] = (cd)[u1, u2, , un] = (cd)u 25.u + v = [0, 1] + [1, 1] = [1, 0] 26.u + v = [1, 1, 0] + [1, 1, 1] = [0, 0, 1] 34 CHAPTER VECTORS x y be a normal vector for the desired plane P Since P is perpendicular to the z xy-plane, their normal vectors must be orthogonal Thus (b) Let n = x y z · = x · + y · + z · = z = x Thus z = 0, so the normal vector is of the form n = y But the normal vector is also perpendicular to the plane in question, by definition Since that plane contains both the origin and (1, 1, 1), the normal vector is orthogonal to (1, 1, 1) − (0, 0, 0): x y · 1 = x · + y · + z · = x + y = Thus x + y = 0, so that y = −x So finally, a normal vector to P is given by n = −x x for − Since the plane passes through (0, 0, 0), we let p = Then substituting in n · x = n · p gives any nonzero x We may as well choose x = 1, giving n = x −1 · y = −1 · , z 0 or x − y = Thus the general equation for the plane perpendicular to the xy-plane and containing the diagonal from the origin to (1, 1, 1) is x − y = (c) As in Example 1.22 (Figure 1.34) in Section 1.2, use u = [0, 1, 1] and v = [1, 0, 1] as two vectors x z in the required plane If n = y is a normal vector to the plane, then n · u = = n · v: z n·u = x y z · 1 = y + z = ⇒ y = −z, n·v = x y z · = x + z = ⇒ x = −z −z z for any z Taking z = gives n = − − z −1 Now, the side diagonals pass through (0, 0, 0), so set p = Then n · x = n · p yields Thus the normal vector is of the form n = 1 −1 −1 · x y z z = 1 · , −1 or x + y − z = 0 The general equation for the plane containing the side diagonals is x + y − z = 26 Finding the distance between points A and B is equivalent to finding d(a, b), where a is the vector from the origin to A, and similarly for b Given x = [x, y, z], p = [1, 0, −2], and q = [5, 2, 4], we want to solve d(x, p) = d(x, q); that is, √ √ d(x, p) = (x − 1)2 + (y − 0)2 + (z + 2)2 = (x − 5)2 + (y − 2)2 + (z − 4)2 = d(x, q) 1.3 LINES AND PLANES 35 Squaring both sides gives (x − 1)2 + (y − 0)2 + (z + 2)2 = (x − 5)2 + (y − 2)2 + (z − 4)2 ⇒ x − 2x + + y + z + 4z + = x − 10x + 25 + y − 4y + + z − 8z + 16 8x + 4y + 12z = 40 ⇒ 2x + y + 3z = 10 2 2 2 Thus all such points (x, y, z) lie on the plane 2x + y + 3z = 10 27 To calculate d(Q, A) = |ax0 +by0 −c| √ Σ , we first put A into general form With d = a2 +b2 Σ −1 since then n · d = Then we have n·x=n·p⇒ ⇒ , we get n = Σ Σ 1 Σ Σ Σ Σ Σ ΣΣ Σ x −1 · = = 1 y Thus x + y = and thus a = b = c = Since Q = (2, 2) = (x0, y0), we have √ |1 · + · − 1| 3 √ d(Q, A) = 12 + 12 = √ = −2 As suggested by #» Figure 1.63, we need to calculate the length of RQ, where R is the point on the line at the foot of the # » perpendicular from Q So if v = P Q, then #» #» P R = projd v, RQ = v − projd v 28 Comparing the given equation to x = p + td, we get P = (1, 1, 1) and d = #» Now, v = P Q = q − p = − proj v = d 1 = d ·v −1 , so that −1 Σ d= −2 · (−1) + · (−1) d·d −2 · (−2) + · v − projd v = −1 − Thus −1 Σ −2 13 = − 133 = 13 − 3 13 −1513 − 10 13 Then the distance d(Q, A) from Q to A is "v − projd v" = +by +cz −d| 13 = √ 13 32 + 22 = √ 13 13 29 To calculate d(Q, P ) = |ax0√ 20 2 , we first note that the plane has equation x + y − z = 0, so a +b +c that a = b = 1, c = −1, and d = Also, Q = (2, 2, 2), so that x0 = y0 = z0 = Hence √ |1 · + · − · − 0| 2 d(Q, P) = √ = √ = 3 12 + 12 + (−1)2 36 CHAPTER VECTORS +by0 +cz0 −d| 30 To calculate d(Q, P ) = |ax0√ , we first note that the plane has equation x − 2y + 2z = 1, so that a = 1, b = −2, c = 2, and d = Also, Q = (0, 0, 0), so that x0 = y0 = z0 = Hence a2 +b2 +c2 |1 · − · + · − 1| √ 12 + (−2)2 + 22 d(Q, P) = = # » 31 Figure 1.66 suggests that we let v = PΣQ; Σ then w = P#R» = projd v Comparing Σ Σ Σ the Σ given Σ Σ line A to −1 # » x = p + td, we get P = (−1, 2) and d = − = Next, Then v = P Q = q − p = −1 2 Σ ΣΣ Σ Σ Σ d·v · + (−1) · 1 w = proj v = d= = d −1 −1 d·d · + (−1) · (−1) So Σ Σ Σ Σ ΣΣ # » −1 = + r = p + P R = p + projd v = p + w = −2 2 Σ So the point R on A that is closest to Q is 2, #» #» 32 Figure 1.66 suggests that we let v = P Q; then P R = projd v Comparing the given line A to x = p+td, −2 #» Then v = P Q = q − p = 3 we get P = (1, 1, 1) and d = proj v = d d ·v Σ d= 1 − = 1 Σ −2 −2 · (−1) + · (−1) 13 = (−2)2 + 32 d·d −1 Next, −1 13 − 3 So #» r = p + P R = p + projd v = + So the point R on A that is closest to Q is #» 15 13 , 1, Σ 10 13 13 = 15 13 − 13 10 13 #» 33 Figure 1.67 suggests we let v = P Q, where P is some point on the plane; then QR = projn v The 1 Setting y = shows that P = (1, 0, 1) is a point −1 equation of the plane is x + y − z = 0, so n = on the plane Then #» v = PQ = q −p = so that Σ 1 − = 2, 1 Σ = projn v = n·v n= n·n 1·1 + ·1 −1 ·1 12 + 12 + (−1)2 −1 Finally, #» #» 1 −3 1.3 LINES AND PLANES r = p + P Q + P R = p + v − projn v = + − Therefore, the point R in P that is closest to Q is 4 , 3 Σ ,3 = 3 −3 3 37 38 CHAPTER VECTORS # Q, » where P is some point on the plane; then #QR » = proj v The 34 Figure 1.67 suggests we let v = P n equation of the plane is x − 2y + 2z = 1, so n = point on the plane Then −2 Setting y = z = shows that P = (1, 0, 0) is a #» v=PQ=q−p= − so that −1 , Σ Σ = −19 n·v projn v = n·n · (−1) n= 12 + (−2)2 + 22 −9 −2 = Finally, #» 2 #» −1 r = p + P Q + P R = p + v − projn v = + −9 − − 9 − 92 = − 92 − Σ Therefore, the point R in P that is closest to Q is − 19 , 29, − 29 35 Since the given lines A1 and A2 are parallel, choose arbitrary points Q on A1 and P on A2, say Q = (1, 1) and P = (5, 4) The direction vector of A2 is d = [−2, 3] Then Σ Σ Σ Σ Σ Σ # » −4 , v = P Q = q − p= − = −3 so that Σ ΣΣ Σ Σ Σ −2 −2 proj v = d · v d = −2 · (−4) + · (−3) = − d 3 13 d·d (−2)2 + 32 Then the distance between theΣlines Σ is given by Σ Σ Σ Σ Σ Σ −4 + 18 18 √ −2 −54 "v − proj v" = = = = 13 d −3 13 13 −36 13 13 36 Since the given lines A1 and A2 are parallel, choose arbitrary points Q on A1 and P on A2, say Q = (1, 0, −1) and P = (0, 1, 1) The direction vector of A2 is d = [1, 1, 1] Then − −1 #» v=PQ=q−p= so that projd v = d·v Σ 1 = −1 , −2 · + · (−1) + · (−2) d·d d= 12 + 12 + 12 Σ 1 1 =− Then the distance between the lines is given by "v − projd v" = −1 + 1 √ = − − 34 = 1√ + (−1)2 + (−4)2 = 42 1.3 LINES AND PLANES −2 −2 3 3 39 37 Since P1 and P2 are parallel, we choose an arbitrary point on P1, say Q = (0, 0, 0), and compute d(Q, P2 ) Since the equation of P2 is 2x + y − 2z = 5, we have a = 2, b = 1, c = −2, and d = 5; since Q = (0, 0, 0), we have x0 = y0 = z0 = Thus the distance is d(P1, P2) = d(Q, P2) = |ax0 + by0 + cz0 − d| √ a2 + b2 + c2 |2 · + · − · − 5| √ = 22 + 12 + 22 = 40 CHAPTER VECTORS 38 Since P1 and P2 are parallel, we choose an arbitrary point on P1 , say Q = (1, 0, 0), and compute d(Q, P2) Since the equation of P2 is x + y + z = 3, we have a = b = c = and d = 3; since Q = (1, 0, 0), we have x0 = 1, y0 = 0, and z0 = Thus the distance is √ |1 · + · + · − 3| 2 |ax0 + by0 + cz0 − d| √ √ = 12 + 12 + 12 Σ Σ a +by −c| 39 We wish to show that d(B, A) = |ax√0 , where n = , n· d(P1, P2) = d(Q, P2) = a2 +b2 d(B, A) = "projn v" = n · v Σ n ·n |ax +by +cz −d| , 0 √ a2 +b2 +c2 = # » Σ Σ Σ Σ a x · − c = ax0 + by0 − c y0 b n · v = n · (b − a) = n · b − n · a = Then from Figure 1.65, we see that a = c, and B = (x0, y0) If v = AB = b b − a, then 40 We wish to show that d(B, A) = = √ a2 + b2 + c2 n = |n · v| = "n" |ax0 + by0 − c| √ a2 + b a b , n · a = d, and B = (x0, y0, z0) If c where n = # » v = AB = b − a, then a n · v = n · (b − a) = n · b − n · a = Then from Figure 1.65, we see that d(B, A) = "projn v" = b n · v Σ n ·n x0 · y0 − d = ax0 + by0 + cz0 − d c z0 n = |n · v| "n" = |ax0 + by0 + cz0 − d| √ a2 + b2 + c2 41 Choose B = (x0, y0) on A1; since A1 and Σ Σ AΣ2 are Σ parallel, the distance between them is d(B, A2) Then a x0 since B lies on A1, we have n · b = · = ax0 + by0 = c1 Choose A on A2, so that n · a = c2 Set b y0 v = b − a Then using the formula in Exercise 39, the distance is d(A1 , A2 ) = d(B, A 2) = |n · v| |n · (b − a)| |n · b − n · a| |c1 − c2| = = = "n" "n" "n" "n" 42 Choose B = (x0, y0 , z0 ) on P1; since P1 and P2 are parallel, the distance between them is d(B, P2) a x0 b · y0 = ax0 + by0 + cz0 = d1 Choose A on P2, so c z0 that n · a = d2 Set v = b − a Then using the formula in Exercise 40, the distance is Then since B lies on P1, we have n · b = d(P 1, P )2 = d(B, P ) 2= 43 Since P1 has normal vector n1 = |n · v| |n · (b − a)| |n · b − n · a| |d1 − d2| = = = "n" "n" "n" "n" 1 and P2 has normal vector n2 = , the angle θ between −2 the normal vectors satisfies cos θ = n1 · n2 · + · + · (−2) √ = √ = √ 2 2 2 "n " "n " 3 + + + + (−2) Thus θ = cos −1 √ 3 Σ ≈ 78.9 ◦ 1.3 LINES AND PLANES 41 44 Since P1 has normal vector n1 = −1 and P2 has normal vector n2 = , the angle θ between −1 the normal vectors satisfies n1 · n2 cos θ = "n " "n " · − · + · (−1) = √ 32 + (−1)2 + 22 √ 12 + 42 + (−1)2 = −√ √ = −√ 14 18 28 This is an obtuse angle, so the acute angle is π − θ = π − cos −1 −√ Σ 28 ◦ ≈ 79.1 45 First, to see that P and A intersect, substitute the parametric equations for A into the equation for P, giving x + y + 2z = (2 + t) + (1 − 2t) + 2(3 + t) = + t = 0, so that t = −9 represents the point of intersection, which is thus (2 + (−9), − 2(−9), + (−9)) = 1 (−7, 19, −6) Now, the normal to P is n = , and a direction vector for A is d = −2 So if θ is the angle between n and d, then θ satisfies the angle between n and d, then θ satisfies n·d · + · (−2) + · 1 √ cos θ = =√ = , "n" "d" 12 + 12 + 12 12 + (−2)2 + 12 so that θ = cos−1 Σ ≈ 80.4◦ Thus the angle between the line and the plane is 90◦ − 80.4◦ ≈ 9.6◦ 46 First, to see that P and A intersect, substitute the parametric equations for A into the equation for P, giving 4x − y − z = · t − (1 + 2t) − (2 + 3t) = −t − = 6, so that t = −9 represents the point of intersection, which is thus (−9, + · (−9), + · (−9)) = (−9, −17, −25) Now, the normal to P is n = −1 , and a direction vector for A is d = So if θ −1 is the angle between n and d, then θ satisfies n ·d cos θ = "n" "d" = √ 4·1 −1 ·2 −1 ·3 √ =−√ √ 42 + 12 + 12 12 + 2 + 18 14 This corresponds to an obtuse angle, so the acute angle between the two is Σ θ = π − cos−1 − √ √ ≈ 86.4◦ 18 14 Thus the angle between the line and the plane is 90◦ − 86.4◦ ≈ 3.6◦ 47 We have p = v − c n, so that c n = v − p Take the dot product of both sides with n, giving (c n) · n = (v − p) · n ⇒ c(n · n) = v · n − p · n ⇒ c(n · n) = v · n (since p and n are orthogonal) n·v c= n·n ⇒ 42 CHAPTER VECTORS n·v c = n·n another interpretation of the figure is that c n = projn v = Note that n·nΣ n·v n, which also implies that Now substitute this value of c into the original equation, giving n · vΣ p = v −cn = v − n n ·n 48.(a) A normal vector to the plane x + y + z = is n = Then 1 n·v= · n·n = 1 1 = · + · + · (−2) = −1 −2 · 1 = · + · + · = 3, so that c = − 13 Then Σ 1 3 n·v p=v− 1= n·n n= 0+ −2 −3 −1 Then (b) A normal vector to the plane 3x − y + z = is n = n · v = −1 · 1 = · − · + · (−2) = −2 n · n = −1 · −1 = · − · (−1) + · = 11, so that c = 11 Then Σ 1 n·v p=v− n·n 11 0− n= −2 (c) A normal vector to the plane x − 2z = is n = 11 −1 = 1 Then −2 − 11 23 11 1.3 LINES AND PLANES n ·v = n ·n = 0· = · + · − · (−2) = −2 −2 1 = · + · − · (−2) = 5, 0· −2 −2 43 Exploration: The Cross Product 41 so that c = Then n · v Σ p=v− n ·n − n= −2 = −2 Note that the projection is because the vector is normal to the plane, so its projection onto the plane is a single point (d) A normal vector to the plane 2x − 3y + z = is n = n · v = −3 · −3 Then 1 = · − · + · (−2) = −2 2 n · n = −3 · −3 = · − · (−3) + · = 14, 1 Note that the projection is the vector itself because the −2 vector is parallel to the plane, so it is orthogonal to the normal vector so that c = Thus p = v = Exploration: The Cross Product 1.(a)u × v= = u1v2 − u2v1 (b)u × v = u2 v3 − u3v2 u3v1 − u1v3 u1 v − u v (c)u × v = u2 v − u3 v u3 v − u1 v u1 v − u v (d)u × v = u2 v3 − u3v2 u3v1 − u1v3 u1v2 − u2v1 3 · − · (−1) u2 v3 − u3v2 u3v1 − u1v3 = ·3 −0 ·2 · (−1) − · = −3 −3 −1 · − · = −3 ·0 −3 ·1 3 · − (−1) · · (−6) − · (−4) = · − (−1) · (−6) −1 · (−4) − · = ·3−1 ·2 1 · − · = −2 = 1·2−1·1 We have 0 ·0 −0 ·1 e1 × e2 = × = · − · = =e3 0 1 ·1 −0 ·0 ·1 −0 ·0 0 e2 × e3 = × = · − · = =e1 0 ·0 −1 ·0 e3 × e1 = × 1 0 ·0 −1 ·0 = 1·1−0·0 ·0 −0 ·1 = = e2 42 CHAPTER VECTORS Two vectors are orthogonal if their dot product equals zero But u2 v3 − u3 v2 u1 u1v2 − u2v1 u3 (u × v) · u = u3v1 − u1v3 · u1v2 − u2v1 u2 u3 u3 = (u2v3 − u3v2)u1 + (u3v1 − u1v3)u2 + (u1v2 − u2v1)u3 = (u2v3u1 − u1v3u2) + (u3v1u2 − u2v1u3) + (u1v2u3 − u3v2u1) = u2 v3 − u3 v2 v1 u1v2 − u2v1 (u × v) · v = u3v1 − u1v3 · u1v2 − u2v1 v3 v2 v3 v3 = (u2v3 − u3v2)v1 + (u3v1 − u1v3)v2 + (u1v2 − u2 v1)v3 = (u2v3v1 − u2v1v3) + (u3v1v2 − u3v2v1) + (u1v2v3 − u1v3v2) = 4.(a) By Exercise 1, a vector normal to the plane is n = u × v = × −1 = 1 · − · (−1) 1·3−0·2 = · (−1) − · · (−1) − · 3 −3 So the normal form for the equation of this plane is n · x = n · p, or 3 −3 · x y z −3 = 3 · −3 −3 z = −2 This simplifies to 3x + 3y − 3z = 9, or x + y − z = #» and v = P R = #» (b) Two vectors in the plane are u = P Q = So by Exercise 1, a vector −2 normal to the plane is n = u ×v = × 1 = −2 · (−2) − · · − · (−2) = 2·3 −1 ·1 2·3 −1 ·1 −2 So the normal form for the equation of this plane is n · x = n · p, or −5 x · y z z = −5 5 · This simplifies to −5x + 5y + 5z = 0, or x − y − z = 5.(a)v × u= v2u3 − v3u2 u2 v − u3 v v3u1 − u3v1 = − u3 v1 − u1v3 u1 v − u2 v v1u2 − v2u1 = −(u × v) −1 1 = −5 5 u2 · − u3 · u1 Product Exploration: The Cross (b)u × = u2 × = u3 · − u1 · u3 u1 · − u2 · (c)u × u = (d)u × kv = = = 0 u2 u3 − u3 u2 u3u1 − u1u = = u1u2 − u2u1 u2kv3 − u3kv2 u2 v − u3 v u3kv1 − u1kv3 = k u3v1 − u1v3 u1kv2 − u2kv1 u1 v − u2 v = k(u × v) 43 44 CHAPTER VECTORS (e)u × ku = k(u × u) = k(0) = by parts (d) and (c) (f ) Compute the cross-product: u2(v3 + w3) − u3(v2 + w2) u × (v + w) = u3(v1 + w1) − u1(v3 + w3) u1(v2 + w2) − u2(v1 + w1) (u2v3 − u3v2) + (u2w3 − u3w2) = (u3v1 − u1v3) + (u3w1 − u1w3) (u1v2 − u2v1) + (u1 w2 − u2w1) u2 v − u3 v u2 w3 − u3 w2 = u3v1 − u1v3 + u3 w1 − u1 w3 u1 v2 −u2 v1 u1 w2 − u2 w1 = u × v + u × w In each case, simply compute: (a) u1 v2w3 − v3w2 u3 v1w2 − v2w1 u · (v × w) = u2 · u3 v3w1 − v1w3 = u1v2w3 − u1v3w2 + u2v3w1 − u2v1w3 + u3v1w2 − u3v2w1 = (u2v3 − u3v2)w1 + (u3v1 − u1v3)w2 + (u1v2 − u2v1)w3 = (u × v) · w (b) v2w3 − v3w2 v3w1 − v1w3 v1w2 − v2w1 u1 u × (v × w) = u2 × u3 u2(v1 w2 − v2 w1 ) − u3(v3 w1 − v w3 ) = u3(v2 w3 − v3 w2 ) − u1(v1 w2 − v2 w1 ) u1(v3 w1 − v1 w3 ) − u2(v2 w3 − v3 w2 ) (u1w1 + u2 w2 + u3 w3)v1 − (u1v1 + u2v2 + u3 v3)w1 = (u1w1 + u2w2 + u3 w3)v2 − (u1v1 + u2v2 + u3v3)w2 (u1w1 + u2 w2 + u3 w3)v3 − (u1v1 + u2v2 + u3 v3)w3 v1 w1 = (u1w1 + u2w2 + u3 w3 ) v2 − (u1v1 + u2v2 + u3 v3 ) w2 v3 w3 = (u · w)v − (u · v)w (c) u v −u v "u × v" = u3 v − u1 v u1 v − u2 v = (u2v3 − u3v2)2 + (u3v1 − u1v3)2 + (u1v2 − u2v1)2 = (u21 + u22 + u23)2(v21+ v2 2+ v2)32 − (u1v1 + u2v2 + u3v3)2 2 = "u" "v" − (u · v)2 ... a ƒ= in Z5 have a solution because is a prime number (b) a = and a = because they have no common factors with other than (c) a and m can have no common factors other than 1; that is, the greatest... if and only if "u" = "v" It follows immediately that u + v and u − v are orthogonal if and only if "u" = "v" (b) Part (a) tells us that the diagonals of a parallelogram are perpendicular if and... office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: www.cengage.com/global Cengage Learning products are represented