# A permute x and y in exercise 31a x2 z 2 f y 2 b

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Unformatted text preview: os θ)2 + (a sin φ sin θ)2 + (a cos φ)2 = a2 sin2 φ cos2 θ + a2 sin2 φ sin2 θ + a2 cos2 φ = a2 sin2 φ(cos2 θ + sin2 θ) + a2 cos2 φ = a2 sin2 φ + a2 cos2 φ = a2 (sin2 φ + cos2 φ) = a2 EXERCISE SET 13.2 y 1. (a–c) y (d–f ) -5i + 3j 〈2, 5〉 x x 〈2, 0〉 3i - 2j 〈–5, –4〉 -6j y 2. (a–c) y (d–f ) 〈- 3, 7 〉 4i + 2 j x x 〈6, - 2 〉 4i -2 i - j 〈0, - 8 〉 z 3. (a–b) z (c–d) – i + 2j + 3k 〈1, -2, 2〉 y y 〈2, 2, –1〉 x x z 4. (a–b) 2i + 3j – k z (c–d) 〈-1, 3, 2 〉 i - j + 2k y y 〈3, 4, 2 〉 2j - k x x 453 Chapter 13 5. (a) 4 − 1, 1 − 5 = 3, −4 (b) y 0 − 2, 0 − 3, 4 − 0 = −2, −3, 4 z –2 i – 3j + 4k x 3i – 4 j y x 6. (a) −3 − 2, 3 − 3 = −5, 0 (b) 0 − 3, 4 − 0, 4 − 4 = −3, 4, 0 y z – 3i + 4 j x y –5 i x 7. (a) 2 − 3, 8 − 5 = −1, 3 8. (a) −4 − (−6), −1 − (−2) = 2, 1 (b) 0 − 7, 0 − (−2) = −7, 2 (b) −1, 6, 1 (c) −3, 6, 1 (c) 5, 0, 0 9. (a) Let (x, y ) be the terminal point, then x − 1 = 3, x = 4 and y − (−2) = −2, y = −4. The terminal point is (4, −4). (b) Let (x, y, z ) be the initial point, then 5 − x = −3, −y = 1, and −1 − z = 2 so x = 8, y = −1, and z = −3. The initial point is (8, −1, −3). 10. (a) Let (x, y ) be the terminal point, then x − 2 = 7, x = 9 and y − (−1) = 6, y = 5. The terminal point is (9,5). (b) Let (x, y, z ) be the terminal point, then x + 2 = 1, y − 1 = 2, and z − 4 = −3 so x = −1, y = 3, and z = 1. The terminal point is (−1, 3, 1). 11. (a) −i + 4j − 2k (d) 40i − 4j − 4k (b) 18i + 12j − 6k (e) −2i − 16j − 18k (c) −i − 5j − 2k (f ) −i + 13j − 2k 12. (a) (c) (e) (f ) 1, −2, 0 (b) 28, 0, −14 + 3, 3, 9 = 31, 3, −5 3, −1, −5 (d) 3( 2, −1, 3 − 28, 0, −14 ) = 3 −26, −1, 17 = −78, −3, 51 −12, 0, 6 − 8, 8, 24 = −20, −8, −18 8, 0, −4 − 3, 0, 6 = 5, 0, −10 13. (a) (c) √ √ v = 1+1= 2 √ v = 21 (b) (d) √ √ v = 1 + 49 = 5 2 √ v = 14 14. (a) (c) √ v = 9 + 16 = 5 v =3 (b) (d) √ v = 2+7=3 √ v= 3 Exercise Set 13.2 454 √ u + v = 2i − 2j + 2k = 2 3 √ √ (c) − 2u + 2 v = 2 14 + 2 2 √ √ √ (e) (1/ 6)i + (1/ 6)j − (2/ 6)k 15. (a) √ √ 14 + 2 (b) u+v= (d) 3u − 5v + w = √ − 12j + 2k = 2 37 (f ) 1 16. If one vector is a positive multiple of the other, say u = αv with α > 0, then u, v and u + v are parallel and u + v = (1 + α) v = u + v . √ √ 17 so the required vector is −1/ 17 i + 4/ 17 j √ √ (b) 6i − 4j + 2k = 2 14 so the required vector is (−3i + 2j − k)/ 14 −→ −→ √ √ (c) AB = 4i + j − k, AB = 3 2 so the required vector is (4i + j − k)/ 3 2 17. (a) − i + 4j = √ 3 4 1 3i − 4j = 5 so the required vector is − (3i − 4j) = − i + j 5 5 5 1 2 2 (b) 2i − j − 2k = 3 so the required vector is i − j − k 3 3 3 −→ −→ 3 4 (c) AB = 4i − 3j, AB = 5 so the required vector is i − j 5 5 18. (a) 1 19. (a) − v = −3/2, 2 2 (b) 20. (a) 3v = −6i + 9j (b) √ √ √ 1 17 v = 85, so √ v = √ 7, 0, −6 has length 17 85 5 − 6 2 8 2 v = √ i− √ j− √ k v 26 26 26 √ √ 21. (a) v = v cos π/4, sin π/4 = 3 2/2, 3 2/2 (b) v = v cos 90◦ , sin 90◦ = 0, 2 √ (c) v = v cos 120◦ , sin 120◦ = −5/2, 5 3/2 (d) v = v cos π, sin π = −1, 0 √ √ √ 22. From (12), v = cos√ 6, sin π/6√= 3/2, 1/2 and w = cos 3π/4, sin 3π/4 = − 2/2, 2/2 , π/ √ √ √ √ so v + w = (( 3 − 2)/2, (1 + 2)/2, v − w = (( 3 + 2)/2, (1 − 2)/2) √ √ √ 23. From (12), √ = √ 30◦ , sin 30◦ = 3/2, 1/2 and w = cos 135◦ , sin 135◦ = − 2/2, 2/2 , so v cos √ v + w = (( 3 − 2)/2, (1 + 2)/2) √ √ 24. w = 1, 0 , and from (12), v = cos 120◦ , sin 120◦ = −1/2, 3/2 , so v + w = 1/2, 3/2 25. (a) The initial point of u + v + w is the origin and the endpoint is (−2, 5), so u + v + w = −2, 5 . –2 i + 5j (b) The initial point of u + v + w is (−5, 4) and the endpoint is (−2, −4), so u + v + w = 3, −8 . y y 5 2 x -5 5 x -5 5 3i – 8j -5 -8 455 Chapter 13 26. (a) v = −10, 2 by inspection, so u − v + w = u + v + w − 2v = −2, 5 + 20, −4 = 18, 1 . (b) v = −3, 8 by inspection, so u − v + w = u + v + w − 2v = 3, −8 + 6, −16 = 9, −24 . 27. 6x = 2u − v − w = −4, 6 , x = −2/3, 1 28. u − 2x = x − w + 3v, 3x = u + w − 3v, x = 1 (u + w − 3v) = 2/3, 2/3 3 2 1 8 1 4 5 i + j + k, v = i − j − k 30. u = −5, 8 , v = 7, −11 7 7 7 7 7 7 √ (i + j) + (i − 2j) = 2i − j = 5, (i + j − (i − 2j) = 3j = 3 29. u = 31. 32. Let A, B , C be the vertices (0,0), (1,3), (2,4) and D the fourth vertex (x, y ). For the parallelogram −→ −→ ABCD, AD = BC , x, y = 1, 1 so x = 1, y = 1 and D is at (1,1). For the parallelogram ACBD, −→ −→ AD = CB , x, y = −1, −1 so x = −1, y = −1 and D is at (−1, −1). For the parallelogram −→ −→ ABDC, AC =BD, x − 1, y − 3 = 2, 4 , so x = 3, y = 7 and D is at (3, 7). 33. (a) 5 = k v = |k | v = ±3k...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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