Thomas Calculus 13th [Solutions]

884 Chapter 12 Vectors and the Geometry of Space
39. AB (5 a) i (1 b) j (3 c) k i 4j 2k 5 a 1, 1 b 4, and 3 c 2 a 4, b 3, and
c 15 A is the point (4, 3, 5)
40. AB ( a 2) i ( b 3) j ( c 6) k 7i 3j 8k a 2 7, b 3 3, and c 6 8 a 9, b 0, and
c 14 B is the point ( 9, 0,14)
41. 2 i j a( i j) b( i j) ( a b) i ( a b) j a b 2 and a b 1 2a 3 a 3 and b a 1 1
42. i 2 j a(2i 3 j) b( i j) (2 a b) i (3 a b) j 2a b 1 and 3a b 2 a 3 and
b 1 2a 7 u1
a(2i 3 j) 6i 9j and u2 b( i j) 7i 7j
43. 25 west of north is 90 25 115 north of east. 800 cos115 , sin115 338.095, 725.046
2
2
44. Let u x,
y represent the velocity of the plane alone, v 70 cos 60 , 70sin 60 35, 35 3 , and let the
resultant u v 500, 0 . Then x, y 35, 35 3 500,0 x 35, y 35 3 500,0
x 35 500 and y 35 3 0 x 465 and y 35 3 u 465, 35 3
2
2
| u | 465 35 3 468.9 mph, and
35 3
tan 7.4 7.4 south of east.
465
45.
3
F 1
1 F1 cos30 , F1 sin 30 F
2 1 , F
2 1 , F 1 1
2 F2 cos 45 , F2 sin 45 | F
2
2|, | F
2
2| , and
w 0, 100 . Since
3
F 1 1 1
1 F2 0, 100 2 | F1 | | F
2
2 |, 2
| F1 | | F
2
2 | 0, 100
3
| F 1
2 1| | F
2
2 | 0 and 1 | F1 | 1 | F
2 2
2 | 100. Solving the first equation for | F 2 | results in:
6
| F2 | | F
2 1 |. Substituting this result into the second equation gives us:
1 1 6
| F
2 1| | F
2 2 1 | 100
200
100 6
| F1 | 73.205 N | F
1 3
2 | 89.658 N F
1 3
1 63.397, 36.603 and F2 63.397, 63.397
46. F1 35 cos , 35 sin ,
1 3
F2 | F2 |cos 60 , | F2 |sin 60 | F
2 2|, | F
2 2| , and w 0, 50 .
3
Since F 1 1
1 F2 0, 50 35 cos | F
2 2 |, 35sin | F
2 2 | 0, 50 35 cos | F
2 2| 0 and
3
35 sin F
2 2 50. Solving the first equation for F 2 results in: F 2 70cos . Substituting this result
into the second equation gives us: 35sin 35 3 cos 50 3 cos 10 sin
7
2 100 20 2 2 100 20
2 2
3cos sin sin 3 1 sin sin sin 196sin 140sin 47 0
49 7 49 7
sin
5 6 2
14
. Since
5 6 2
0 sin 0 sin 74.42 , and F
14
2 70cos 18.81 N.
47. F1 | F1 | cos 40 , | F1
| sin 40 , F 2 100cos35 ,100sin 35 , and w 0, w . Since F1 F2 0, w
F1 cos 40 100cos35 , F1 sin 40 100sin 35 0, w F 1 cos 40 100cos35 0 and
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