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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 xaxis, and P Q = 0.2i + 0.03j makes an angle
α = tan−1 (0.03/0.2) with the xaxis, 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.
 Spring '14
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