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Unformatted text preview: → P Q × F = −10i + 10k toward its initial point. 465 Chapter 13 −→ (b) F = 10j and P Q= j + k, so the vector moment of F about P is −→ PQ ×F = i 0 0 jk 11 10 0 = −10i, and the scalar moment is 10 lb·ft. The direction of rotation of the cube about P is counterclockwise looking along −10i toward its initial point. −→ (c) F = 10j and P Q= j, so the vector moment of F about P is −→ PQ ×F = i 0 0 jk 10 10 0 = 0, and the scalar moment is 0 lb·ft. Since the force is parallel to the direction of motion, there is no rotation about P . −→ 1000 34. (a) F = √ (−i + k) and P Q= 2j − k, so the vector moment of F about P is 2 i jk −→ √ √ P Q × F = 500 2 0 2 −1 = 500 2(2i + j + 2k), and the scalar moment is −1 0 1 √ 1500 2 N·m. (b) The direction angles of the vector moment of F about the point P are cos−1 (2/3) ≈ 48◦ , cos−1 (1/3) ≈ 71◦ , and cos−1 (2/3) ≈ 48◦ . 35. F makes an angle 72◦ with the positive x-axis, and P Q = 0.2i + 0.03j makes an angle α = tan−1 (0.03/0.2) with the x-axis, so φ ≈ 72◦ − 8.53◦ = 63.47◦ and F = 200 N, so −→ P Q ×F = P Q F | sin 63.47◦ | = 200 (0.2i + 0.030j) | sin 63.47◦ | ≈ 36.19 N·m. Part (b) : let u = u1 , u2 , u3 , v = v1 , v2 , v3 , and w = w1 , w2 , w3 ; show that u × (v + w) and (u × v) + (u × w) are the same. Part (c) : 36. (u + v) × w = −[w × (u + v)] from part (a) = −[(w × u) + (w × v)] from part (b) = (u × w) + (v × w) from part (a) 37. Let u = u1 , u2 , u3 and v = v1 , v2 , v3 ; show that k (u × v), (k u) × v, and u × (k v) are all the same; Part (e) is proved in a similar fashion. 38. Suppose the first two rows are interchanged. Then by definition, b1 a1 c1 b2 a2 c2 b3 a3 c3 = b1 a2 c2 a3 c3 − b2 a1 c1 a3 c3 + b3 a1 c1 a2 c2 = b1 (a2 c3 − a3 c2 ) − b2 (a1 c3 − a3 c1 ) + b3 (a1 c2 − a2 c1 ), which is the negative of the right hand side of (2) after expansion. If two other rows were to be exchanged, a similar proof would hold. Finally, suppose ∆ were a determinant with two identical rows. Then the value is unchanged if we interchange those two rows, yet ∆ = −∆ by Part (b) of Theorem 13.4.1. Hence ∆ = −∆, ∆ = 0. 39. −8i − 20j + 2k, −8i − 8k 40. (a) From the first formula in Exercise 39 it follows that u × (v × w) is a linear combination of v and w and hence lies in the plane determined by them. (b) Similar to (a). Exercise Set 13.5 466 41. If a, b, c, and d lie in the same plane then a × b and c × d are parallel so (a × b) × (c × d) = 0 42. Let u and v be the vectors from a point on the curve to the points (2, −1, 0) and (3, 2, 2), respectively. Then u = (2 − x)i + (−1 − lnx)j and v = (3 − x)i + (2 − lnx)j + 2k. The area of the triangle is given by A = (1/2) u × v ; solve dA/dx = 0 for x to get x = 2.091581. The minimum area is 1.887850. −→ −→ −→ −→ −→ 43. P Q × F =P Q × F+ QQ × F =P Q × F, since F and QQ are parallel. EXERCISE SET 13.5 In many of the Exercises in this section other answers are also possible. 1. (a) L1 : P (1, 0), v = j, x = 1, y = t L2 : P (0, 1), v = i, x = t, y = 1 L3 : P (0, 0), v = i + j, x = t, y = t (b) L1 : L2 : L3 : L4 : P (1, 1, 0), v = k, x = 1, y = 1, z = t P (0, 1, 1), v = i, x = t, y = 1, z = 1 P (1, 0, 1), v = j, x = 1, y = t, z = 1 P (0, 0, 0), v = i + j + k, x = t, y = t, z = t 2. (a) L1 : x = t, y = 1, 0 ≤ t ≤ 1 L2 : x = 1, y = t, 0 ≤ t ≤ 1 L3 : x = t, y = t, 0 ≤ t ≤ 1 (b) L1 : L2 : L3 : L4 : x = 1, y = 1, z = t, 0 ≤ t ≤ 1 x = t, y = 1, z = 1, 0 ≤ t ≤ 1 x = 1, y = t, z = 1, 0 ≤ t ≤ 1 x = t, y = t, z = t, 0 ≤ t ≤ 1 −→ 3. (a) P1 P2 = 2, 3 so x = 3 + 2t, y = −2 + 3t for the line; for the line segment add the condition 0 ≤ t ≤ 1. −→ (b) P1 P2 = −3, 6, 1 so x = 5 − 3t, y = −2 + 6t, z = 1 + t for the line; for the line segment add the condition 0 ≤ t ≤ 1. −→ 4. (a) P1 P2 = −3, −5 so x = −3t, y = 1 − 5t for the line; for the line segment add the condition 0 ≤ t ≤ 1. −→ (b) P1 P2 = 0, 0, −3 so x = −1, y = 3,z = 5 − 3t for the line; for the line segment add the condition 0 ≤ t ≤ 1. 5. (a) x = 2 + t, y = −3 − 4t (b) x = t, y = −t, z = 1 + t 6. (a) x = 3 + 2t, y = −4 + t (b) x = −1 − t, y = 3t, z = 2 7. (a) r0 = 2i − j so P (2, −1) is on the line, and v = 4i − j is parallel to the line. (b) At t = 0, P (−1, 2, 4) is on the line, and v = 5i + 7j − 8k is parallel to the line. 8. (a) At t = 0, P (−1, 5) is on the line, and v = 2i + 3j is parallel to the line. (b) r0 = i + j − 2k so P (1, 1, −2) is on the line, and v = j is parallel to the line. 9. (a) (b) 10. (a) (b) x, y = −3, 4 + t 1, 5 ; r = −3i + 4j + t(i + 5j) x, y, z = 2, −3, 0 + t −1, 5, 1 ; r = 2i − 3j + t(−i + 5j + k) x, y = 0, −2 + t 1, 1 ; r = −2j + t(i + j) x, y, z = 1, −7, 4 + t 1, 3, −5 ; r = i − 7j + 4k + t(i + 3j − 5k) 11. x = −5 + 2t, y = 2 − 3t 12. x = t, y = 3 − 2t 467 Chapter 13 13. 2x + 2yy = 0, y = −x/y = −(3)/(−4) = 3/4, v = 4i + 3j; x = 3 + 4t, y = −4 + 3t 14. y = 2x = 2(−2) = −...
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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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