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Unformatted text preview: e squares of the lengths of the diagonals of a parallelogram is equal to twice the sum of the squares of the lengths of the sides. u+v 2 = (u + v) · (u + v) = u 2 + 2u · v + v u−v 2 = (u − v) · (u − v) = u 2 − 2u · v + v , subtract to get u+v 40. 2 − u−v 2 2 and 2 = 4u · v, the result follows by dividing both sides by 4. 41. v = c1 v1 + c2 v2 + c3 v3 so v · vi = ci vi · vi because vi · vj = 0 if i = j, thus v · vi = ci vi 2 , ci = v · vi / vi 2 for i = 1, 2, 3. 42. v1 · v2 = v1 · v3 = v2 · v3 = 0 so they are mutually perpendicular. Let v = i − j + k, then v · v1 3 v · v2 1 v · v3 1 c1 = . = , c2 = = − , and c3 = = v1 2 7 v2 2 3 v3 2 21 43. (a) u = xi + (x2 + 1)j, v = xi − (x + 1)j, θ = cos−1 [(u · v)/( u v )]. Solve dθ/dx = 0 to find that the minimum value of θ occurs when x ≈ −3.136742 so the minimum angle is about 40◦ . (b) Solve u · v = 0 for x to get x ≈ −0.682328. 44. (a) u = cos θ1 i ± sin θ1 j, v = ± sin θ2 j + cos θ2 k, cos θ = u · v = ± sin θ1 sin θ2 (b) cos θ = ± sin2 45◦ = ±1/2, θ = 60◦ (c) Let θ(t) = cos−1 (sin t sin 2t); solve θ (t) = 0 for t to find that θmax ≈ 140◦ (reject, since θ is acute) when t ≈ 2.186276 and that θmin ≈ 40◦ when t ≈ 0.955317; for θmax check the endpoints t = 0, π/2 to obtain θmax = cos−1 (0) = π/2. 45. Let u = u1 , u2 , u3 , v = v1 , v2 , v3 , w = w1 , w2 , w3 . Then u · (v + w) = u1 (v1 + w1 ), u2 (v2 + w2 ), u3 (v3 + w3 ) = u1 v1 + u1 w1 , u2 v2 + u2 w2 , u3 v3 + u3 w3 = u1 v1 , u2 v2 , u3 v3 + u1 w1 , u2 w2 , u3 w3 = u · v + u · w 0 · v = 0 · v1 + 0 · v2 + 0 · v3 = 0 463 Chapter 13 EXERCISE SET 13.4 1. (a) i × (i + j + k) = i 1 1 jk 00 11 = −j + k (b) i × (i + j + k) = (i × i) + (i × j) + (i × k) = −j + k 2. (a) j × (i + j + k) = i 0 1 jk 10 11 =i−k j × (i + j + k) = (j × i) + (j × j) + (j × k) = i − k (b) k × (i + j + k) = i 0 1 jk 01 11 = −i + j k × (i + j + k) = (k × i) + (k × j) + (k × k) = j − i + 0 = −i + j 3. 7, 10, 9 4. −i − 2j − 7k −4, −6, −3 5. 6. i + 2j − 4k 7. (a) v × w = −23, 7, −1 , u × (v × w) = −20, −67, −9 (b) u × v = −10, −14, 2 , (u × v) × w = −78, 52, −26 (c) (u × v) × (v × w) = −10, −14, 2 × −23, 7, −1 = 0, −56, −392 (d) (v × w) × (u × v) = 0, 56, 392 1 1 9. u × v = (i + j) × (i + j + k) = k − j − k + i = i − j, the direction cosines are √ , − √ , 0 2 2 10. u × v = 12i + 30j − 6k, so ± √ 1 5 2 √ i+ √ j− √ k 30 6 30 −→ −→ 1 11. n = AB × AC = 1, 1, −3 × −1, 3, −1 = 8, 4, 4 , unit vectors are ± √ 2, 1, 1 6 12. A vector parallel to the yz-plane must be perpendicular to i; √ √ i × (3i − j + 2k) = −2j − k, − 2j − k = 5, the unit vectors are ±(2j + k)/ 5. 13. A = u × v = − 7i − j + 3k = √ 59 14. A = u × v = − 6i + 4j + 7k = √ 101 √ 374/2 15. A = −→ 1 1 −→ PQ × PR = 2 2 −1, −5, 2 × 2, 0, 3 = 1 2 −15, 7, 10 = 16. A = −→ 1 1 −→ PQ × PR = 2 2 −1, 4, 8 × 5, 2, 12 = 1 2 32, 52, −22 √ = 9 13 17. 80 18. 29 21. V = |u · (v × w)| = | − 16| = 16 23. (a) u · (v × w) = 0, yes 19. −3 20. 1 22. V = |u · (v × w)| = |45| = 45 (b) u · (v × w) = 0, yes (c) u · (v × w) = 245, no Exercise Set 13.4 464 24. (a) u · (w × v) = −u · (v × w) = −3 (c) w · (u × v) = u · (v × w) = 3 (e) (u × w) · v = u · (w × v) = −3 (b) (v × w) · u = u · (v × w) = 3 (d) v · (u × w) = u · (w × v) = −3 (f ) v · (w × w) = 0 because w × w = 0 √ 25. (a) V = |u · (v × w)| = | − 9| = 9 (b) A = u × w = 3i − 8j + 7k = 122 (c) v × w = −3i − j + 2k is perpendicular to the plane determined by v and w; let θ be the angle between u and v × w then cos θ = −9 u · (v × w ) = √ √ = −9/14 u v×w 14 14 so the acute angle φ that u makes with the plane determined by v and w is φ = θ − π/2 = sin−1 (9/14). 26. From the diagram, P u×v v sin θ = v v u d = u sin θ = u d θ v −→ −→ 27. (a) u = AP = −4i + 2k, v = AB = −3i + 2j − 4k, u × v = −4i − 22j − 8k; distance = u × v / v = 2 141/29 −→ −→ √ (b) u = AP = 2i + 2j, v = AB = −2i + j, u × v = 6k; distance = u × v / v = 6/ 5 1 v × w and 2 1 1 |u · (v × w)| height = projv×w u = so V = (area of base) (height) = |u · (v × w)| v×w 3 6 28. Take v and w as sides of the (triangular) base, then area of base = −→ −→ −→ 29. P Q = 3, −1, −3 , P R = 2, −2, 1 , P S = 4, −4, 3 , V= −→ −→ 1 1 −→ | P Q · (P R × P S )| = |−4| = 2/3 6 6 √ u×v 36i − 24j 12 13 (b) sin θ = = = uv 49 49 23 u·v =− 30. (a) cos θ = uv 49 (c) 232 144 · 13 2401 + = =1 2 2 49 49 492 31. From Theorems 13.3.3 and 13.4.5a it follows that sin θ = cos θ, so θ = π/4. 32. u×v 2 =u 2 v 2 sin2 θ = u 2 v 2 (1 − cos2 θ) = u 2 v 2 − (u · v)2 −→ 33. (a) F = 10j and P Q= i + j + k, so the vector moment of F about P is −→ PQ ×F = i 1 0 jk 11 10 0 √ = −10i + 10k, and the scalar moment is 10 2 lb·ft. The direction of rotation of the cube about P is counterclockwise looking along ...
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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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