Vectors and the geometry of space (página 2)

Partes: 1, 2

JJJ
AB 2, 1, 1 BA 2, 1, 1
JJJK JJJ
2 , 2 , 1 2 , 3 , 1
JJJK JJJ
1 , 1 , 2 1 , 1 , 2
JJJ JJJK
AB ? AC 5 2 1 ! 0
JJJ JJJ
BA ? BC 1 1 2 ! 0
JJJ JJJ
CA ? CB 4 4 2 ! 0
The triangle has three acute angles, so it is an acute
triangle.

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5
3
1

1
K K
K
KK
K
K K
K K
2
2
2
9 4
1
9
9
4
9
4
2
2
2
2
2
2
2
1
Chapter 11
Vectors and the Geometry of Space
30. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1
JJJ JJJ
AB 3, 12, 5 BA 3, 12, 5
JJJK JJJ
AC 2, 13, 4 CA 2, 13, 4
JJJ JJJ
BC 5, 1, 9 CB 5, 1, 9
JJJ JJJK
AB ? AC 6 156 20 ! 0
JJJ JJJ
BA ? BC 15 12 45 ! 0
JJJ JJJ
CA ? CB 10 13 36 ! 0
The triangle has three acute angles, so it is an acute
32. u 5, 3, 1 u 35
cos D
35
cos E
35
cos J
35
cos2 D cos2 E cos2 J
25 9 1
35 35 35
triangle.
33. u
0, 6, 4 , u
52
2 13
31. u
i 2 j 2k , u
3
cos D
0
cos D
1
3
cos E
3
13
cos E 3
cos J 3
cos2 D cos2 E cos2 J

1
cos J
13
cos2 D cos2 E cos2 J
0

13 13
1
34. u
a, b, c , u
a 2 b2 c 2
cos D

cos E

cos J
a
a 2 b2 c 2
b
a 2 b2 c 2
c
a 2 b2 c 2
cos2 D cos2 E cos2 J
a 2
a b c
b 2
a b c
c 2
a b 2 c 2
35. u
3, 2, 2
u
17
cos D

cos E

cos J
3
17
2
17
2
17
? D | 0.7560 or 43.3q

? E | 1.0644 or 61.0q

? y | 2.0772 or 119.0q
36. u
4, 3, 5
u
50
5 2
cos D

cos E
4
5 2
3
5 2
? D | 2.1721 or 124.4q

? E | 1.1326 or 64.9q
cos J
5
5 2
1
2
? J |
S
4
or 45q
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K
2
2
50
80
800 2
N
¨
©
¹
v
100
¨
v
©
¹
cos D |
230.239
Section 11.3
The Dot Product of Two Vectors
37. u
1, 5, 2
u
30
JJJ
41. OA
0, 10, 10
cos D
1
30
? D | 1.7544 or 100.5q
cos D
0
0 102 102
0 ? D
90q
cos E
5
30
? E | 0.4205 or 24.1q
cos E
cos J
10
0 102 102
cos J

38. u
2
30

2, 6, 1
? J | 1.1970 or 68.6q

41
u
42. F1
C1 0, 10, 10 .
1
2
? E
J
45q
cos D

cos E

cos J
2
41
6
41
1
41
? D | 1.8885 or 108.2q

? E | 0.3567 or 20.4q

? J | 1.4140 or 81.0q
F1
F1

F2
F3
F
200 C110 2 ? C1
0, 100 2, 100 2

C2 4, 6, 10
C3 4, 6, 10
0, 0, w
10 2 and
39. F1: C1 | 4.3193
F1
F2 : C2 | 5.4183
F2
F F1 F2
| 4.3193 10, 5, 3 5.4183 12, 7, 5
F F1 F2 F3 0
4C2 4C3 0 ? C2 C3

100 2 6C2 6C3 0 ? C2 C3

W 10C2 10C3 100 2
3
25 2
3
108.2126, 59.5246, 14.1336
F | 124.310 lb
43. u
6, 7 , v
1, 4
cos D |

cos E |
108.2126
F
59.5246
F
? D | 29.48q

? D | 162.02q
F
(b) w 2
u w1
30
1, 3
10
9, 7
3, 9

3, 9
6, 2
cos E |

cos J |
36.062
F
65.4655
F
? E | 98.57q

? J | 74.31q
© 2010 Brooks/Cole, Cengage Learning

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26
¨
2 ¸
¨ v
¸
5, 1
5 1
,
¨
(a) w1
projv u
©
¹
¨
v
©
¹

u ? v
.
¨
v
©
¹
1, 1, 1
6
v1 v2 v3
¨
v
©
¹
§ u ? v ·
¨ v 2 ¸ v
¸
(b) ¨
©
¹
18
¨
©
¹
0, 3, 4
v
u
S
2
S
2
33 44
8 6
S
2
Chapter 11
Vectors and the Geometry of Space
45. u
2i 3j
2, 3 , v
5i j
5, 1
50. u
i 4k
1, 0, 4
(a) w1 projv u v

(b) w 2 u w1 2, 3 ,

46. u 2i 3j 2, 3 , v 3i 2 j
§ u ? v ·
© ¹
2 5 3 1
52 1
13 5 1
5, 1 ,
26 2 2
2 2
1 5
2 2

3, 2
v 3i 2k 3, 0, 2

(b) w 2 u w1 1, 0, 4
§ u ? v ·
¨ v 2 ¸ v
1 3 4 2
32 22
11
3, 0, 2
13
3, 0, 2

33 22
, 0,
13 13
33 22
, 0,
13 13
(a) w1

(b) w 2
projv u

u w1
§ u ? v ·
¨ v 2 ¸
2 3 3 2
32 22
0 3, 2 0, 0
2, 3
3, 2
20 30
, 0,
13 13

51. u ? v u1 , u2 , u3 ? v1 , v2 , v3 u1v1 u2v2 u3v3

52. The vectors u and v are orthogonal if u ? v 0. The
angle T between u and v is given by cos T
u v
47. u
0, 3, 3 , v
1, 1, 1
53. (a) and (b) are defined. (c) and (d) are not defined
(a) w1
projv u
§ u ? v ·
¨ v 2 ¸
0 1 3 1 3 1
1 1 1
1, 1, 1 2, 2, 2
3
because it is not possible to find the dot product of a
scalar and a vector or to add a scalar to a vector.

54. See page 786. Direction cosines of v v1 , v2 , v3 are
cos D , cos E , cos J . D , E , and J
v v v
are the direction angles. See Figure 11.26.
(b) w 2
u w1
0, 3, 3 2, 2, 2
2, 1, 1
55. See figure 11.29, page 787.
48. u 8, 2, 0 , v

(a) w1 projv u
2, 1, 1
§ u ? v ·
¨ v 2 ¸
8 2 2 1 0 1
22 1 1
2, 1, 1
56. (a) ¨
© ¹
§ u ? v ·
¨ v 2 ¸ v
orthogonal.
u ? u

0 ? u ? v
cv ? u and v are parallel.

0 ? u and v are
(b) w 2
u w1
6
2, 1, 1 6, 3, 3

8, 2, 0 6, 3, 3
2, 1, 3
57. Yes,
u ? v
v 2
v
v ? u
u 2
u
49. u
v
2i j 2k
3j 4k
2, 1, 2
0, 3, 4
u ? v
v
v 2
1
v ? u

1
u
u 2
(a) w1
projv u
§ u ? v ·
¨ v 2 ¸ v
2 0 1 3 2 4
32 42
11 33 44
0, 3, 4 0, ,
25 25 25
v
u

58. (a) Orthogonal, T

(b) Acute, 0 T
(b) w 2
u w1
2,1, 2 0, ,
25 25
2, ,
25 25
(c) Obtuse,
T S
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v
v
v
K
K
K

¨ 2
j¸
¹
u
W
1 3
u
0.
v
W
v
W
F
v
w
Section 11.3
The Dot Product of Two Vectors
59. u
3240, 1450, 2235
1.35, 2.65, 1.85
71. (a) Gravitational Force F

cos 10qi sin 10q j
48,000 j
u ? v
3240 1.35 1450 2.65 2235 1.85
$12,351.25
w1
F ? v
v 2
v
F ? v v
This represents the total amount that the restaurant
earned on its three products.
48,000 sin 10q v
| 8335.1 cos 10qi sin 10q j
60. u
3240, 1450, 2235
w1 | 8335.1 lb
1.35, 2.65, 1.85
Increase prices by 4%: 1.04v
New total amount: 1.04 u ? v
1.04 12,351.25
(b) w 2
F w1
48,000 j 8335.1 cos 10qi sin 10q j
8208.5i 46,552.6 j
$12,845.30
w 2 | 47,270.8 lb
61. (a)–(c) Programs will vary.
JJJ
72. OA
10, 5, 20 , v
0, 0, 1
62.
u | 9.165
v | 5.745
T 90q
JJJ
projv OA
JJJ
projv OA
20
12
20
0, 0, 1
0, 0, 20
63. Programs will vary.

21 63 42
64. , ,
26 26 13
73. F

v
W
§ 1
85¨ i
©
10i
F ? v
3 ·
2 ¸

425 ft-lb
65. Because u and v are parallel, projv u
74. F
25 cos 20qi sin 20q j
66. Because u and v are perpendicular, projv u

67. Answers will vary. Sample answer:
0
v
50i
F ? v
1250 cos 20q | 1174.6 ft-lb
i j. Want u ? v
4 2
75. F
v
1600 cos 25q i sin 25q j
2000i
12i 2 j and v
12i 2 j are orthogonal to u.
F ? v
1600 2000 cos 25q
68. Answers will vary. Sample answer:
u 9i 4 j. Want u ? v 0.
| 2,900,184.9 Newton meters (Joules)
| 2900.2 km-N
4i 9 j and v 4i 9 j
are orthogonal to u.

69. Answers will vary. Sample answer:
JJJK
76. PQ
40i
100 cos 25qi
JJJK
F ? PQ 4000 cos 25q | 3625.2 Joules
u
3, 1, 2 . Want u ? v
0.
77. False.
0, 2, 1 and v
0, 2, 1 are orthogonal to u.
For example, let u
1, 1 , v
2, 3 and
70. Answers will vary. Sample answer:
1, 4 . Then u ? v
2 3
5 and
u
4, 3, 6 . Want u ? v
0
u ? w
1 4
5.
v 0, 6, 3 and v
are orthogonal to u.
0, 6, 3
78. True
w ? u v w ? u w ? v
u v are orthogonal.
0 0
0 so, w and
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v
c
c
§ 1 ·
© 3 ¹
z
.
c
c
1
y
c c
1
x
z
2
-1
y
x
x
6
c
c
1 1
6
3
10 10
cos T .
1
3
.
c
c
r
cc
c
c
1
5
1
c
1
1
c
y
1
x
Chapter 11
Vectors and the Geometry of Space
79. Let s
length of a side.
82. (a) The graphs y1
x3 and y2
x1 3 intersect at
s, s, s
1, 1 , 0, 0 and 1, 1 .
v

cos D
s 3

cos E
cos J
s
s 3
1
3
1
(b) y1 3x 2 and y2 .
3x 2 3
At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is
D
E
J
arcos¨ ¸ | 54.7q
tangent to y2 .
At 1, 1 , y1
3 and y2
1
3
s
v
s
s
r 1, 3 is tangent to y1 , r
10
to y2 .
At 1, 1 , y1 3 and y2
1
3
1
10

.
3, 1 is tangent
80.
v1
s, s, s
r
10
to y2 .
1, 3 is tangent to y1 , r
1
10
3, 1 is tangent
v1
v 2
s 3
s, s, 0
y
y = x 1/3
v 2

cos T
s 2
2 2
2 3
6
3
v1

v2
(s, s, s)
1
(0, 0)
-2 -1
(-1, -1)
(1, 1)

1
y = x 3
2
(s, s, 0)
T arcos | 35.26q
3

81. (a) The graphs y1 x 2 and y2 x1 3 intersect at
0, 0 and 1, 1 .
1
(b) y1 2 x and y2 .
3x 2 3
At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is
tangent to y2 .
-2

(c) At 0, 0 , the vectors are perpendicular 90q .
At 1, 1 ,
1, 3 ? 3, 1
1 1 10 5
T | 0.9273 or 53.13q
By symmetry, the angle is the same at 1, 1 .
At 1, 1 , y1
2 and y2
83. (a) The graphs of y1
1 x 2 and y 2
x 2 1
1
5
1, 2 is tangent to y1 , r
1
10
3, 1 is tangent
intersect at 1, 0 and 1, 0 .
(b) y1 2 x and y2 2 x.
to y2 .
(c) At 0, 0 , the vectors are perpendicular 90q .
At 1, 0 , y1
2 and y2
2. r
1
5
1, 2 is
At 1, 1 ,
tangent to y1 , r
1, 2 is tangent to y2 .
cos T

T
1
5

45q
1, 2 ?
1 1
10
3, 1
5
50
1
2
At 1, 0 , y1

tangent to y1 , r
5
2 and y2 2. r 1, 2 is
5
1, 2 is tangent to y2 .
2
y = x 2
(c) At 1, 0 , cos T
1
5
1, 2 ?
1
5
1, 2
3
5
.
(1, 1)
T | 0.9273 or 53.13q
-1
(0, 0) 1
y = x 1/3
2
By symmetry, the angle is the same at 1, 0 .
-1
© 2010 Brooks/Cole, Cengage Learning

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2
2
2
1
y
–
v
z
2
k
x
k
2
1
1
2

60q
© 2 ¹
k k k
, ,
1
2
K
1 1 5
S

4
1
3
§ k ·
© 2 ¹
1
90q.
2
.
u ? v
u v
2
0
Section 11.3
The Dot Product of Two Vectors
84. (a) To find the intersection points, rewrite the second
equation as y 1 x3 . Substituting into the first
equation
y 1 x ? x6 x ? x 0, 1.
There are two points of intersection, 0, 1 and
1, 0 , as indicated in the figure.

y = x 3-1
86. If u and v are the sides of the parallelogram, then the
diagonals are u v and u v, as indicated in the
figure.
the parallelogram is a rectangle.
? u ? v 0
? 2u ? v 2u ? v
? u v ? u v u v ? u v
? u v u v
? The diagonals are equal in length.
-1
(0, -1)
(1, 0)
1
2
x
u
u +
u
v
v
x = (y +1) 2
-2
87. (a)
(b) First equation:
y 1 x ? 2 y 1 yc
1 ? yc
1
2 y 1
(k, 0, k)
(0, k, k)
At 1, 0 , yc
1
2
.
k
k
y
Second equation: y
1, 0 , yc 3.
x3 1 ? yc
3x 2 . At
(k, k, 0)

(b) Length of each edge:
k 2 k 2 02
r 2, 1 unit tangent vectors to first curve,
5
r 1, 3 unit tangent vectors to second curve
10
At 0, 1 , the unit tangent vectors to the first curve
are r 0, 1 , and the unit tangent vectors to the
(c) cos T

J
(d) r1
k
k 2 k 2
§ 1 ·
arccos¨ ¸

k , k , 0 , ,
2 2 2
k k k
2 2 2
second curve are r 1, 0 .
(c) At 1, 0 ,

cos T 2, 1 ? 1, 3
5 10 50

T | or 45q
4
At 0, 1 the vectors are perpendicular, T
88. u
JK k k k k k k
r2 0, 0, 0 , , , ,
2 2 2 2 2 2
k 2
cos T 2
¨ ¸ ? 3
T 109.5q
cos D , sin D , 0 , v cos E , sin E , 0
85. In a rhombus, u
u v.
u v ? u v
v . The diagonals are u v and

u v ? u u v ? v
u ? u v ? u u ? v v ? v
The angle between u and v is D E . Assuming that
D ! E . Also,
cos D E
u 2 v
So, the diagonals are orthogonal.
u – v
cos D cos E sin D sin E
1 1
cos D cos E sin D sin E .
u
u + v
v

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2
2
u
u
2
2
2
u
u
u
2
i
Chapter 11
Vectors and the Geometry of Space
89.
u v
u
v ? u v
91.
u v
u
v ? u v
v ? u u v ? v
u ? u v ? u u ? v v ? v
v ? u u v ? v
= u ? u v ? u u ? v v ? v
u
u ? v u ? v v
u 2 2u ? v v
u
2
v 2 2u ? v
d u 2 2 u
v v
2
d

u v

2
90.
u ? v
u ? v
u
v cos T
v cos T
So, u v d u v .

92. Let w1
projvu, as indicated in the figure. Because
v cos T
w1 is a scalar multiple of v, you can write
d u
v because cos T d 1.
w1 w 2
cv w 2 .
Taking the dot product of both sides with v produces
u ? v
cv
w 2 ? v
cv ? v w 2 ? v
c v 2, because w 2 and v are orthogonol.
So, u ? v
c v
? c
u ? v
v 2
and
w1
projv u
cv
u ? v
v 2
v.
w2
u
? v
w1

Section 11.4 The Cross Product of Two Vectors in Space
i
j k
i
j k
1. j u i
0
1 0
k
3. j u k
0 1 0
1 0
z

1
0
0 0
z

1
1
k
j
j
x
1
i
-1
– k
1
y
x
1
i
-1
1
y
i
j k
i
j k
2. i u j
1 0
0
k
4. k u j
0 0
1
i
0 1 0
z

1
k
j
0
z

1
k
1 0

– i
j
1
i
1
1
1
x
-1
y
x
-1
y
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31
i
1
i
1
j
z
1
k
x
y
i
0
1
5
0
5
i
0
Section 11.4
The Cross Product of Two Vectors in Space
j k
11. u
12, 3, 0 , v
2, 5, 0
5. i u k
1 0
0
j
i
j k
0 0
1
u u v
12 3
0
54k
0, 0, 54
z
2
5
0
– 1
– j
k
u ? u u v
12 0 3 0 0 54
0 ? u A u u v
x
1
-1
1
y
v ? u u v
2 0 5 0 0 54
0 ? v A u u v
12. u
1, 1, 2 , v
0, 1, 0
i
j k
i
j k
6. k u i
0 0
u u v
1 1 2
2i k
2, 0, 1
1 0
0
0 1 0
u ? u u v
1 2 1 0 2 1
0 ? u A u u v
v ? u u v
0 2 1 0 0 1
1
i
j
1
0 ? v A u u v
-1
13. u
2, 3, 1 , v
i
j k
1, 2, 1
7. (a) u u v
j k
2 4
20i 10 j 16k
u u v
2 3
1 2
1
i j k
1, 1, 1
(b) v u u
(c) v u v
3 2
u u v

i
j
k
20i 10 j 16k
u ? u u v

v ? u u v
2 1 3 1 1 1
0 ? u A u u v
1 1 2 1 1 1
0 ? v A u u v
8. (a) u u v
3 0
15i 16 j 9k
14. u
10, 0, 6 , v
5, 3, 0
2 3 2
j k
(b) v u u
(c) v u v
u u v
0
15i 16 j 9k
u u v 10 0 6 18i 30 j 30k

u ? u u v 10 18 0 30 6 30
5 3 0
18, 30, 30

0
9. (a) u u v
i j k
7 3 2
1 1 5
17i 33j 10k
? u A u u v
v ? u u v
? v A u u v
5 18 3 30 0 30
(b) v u u
(c) v u v
u u v
0
17i 33j 10k
15. u
i j k , v
i
j
k
2i j k
10. (a) u u v
i j k
3 2 2
1 5 1
8i 5 j 17k
u u v 1 1 1 2i 3j k
2 1 1
u ? u u v 1 2 1 3 1 1
2, 3, 1
(b) v u u
(c) v u v
u u v
0
8i 5 j 17k
v ? u u v

v u u
0 ? u A u u v
2 2 1 3 1 1
0 ? v A u u v
v u u u u v
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32
u
6
5

,
,
,
Chapter 11
Vectors and the Geometry of Space
16. u
i 6 j, v
i
2i j k
j k
22.
v
8, 6, 4
10, 12, 2
u u v 1 6 0 6i j 13k
2 1 1
u ? u u v 1 6 6 1 0 ? u A u u v
u u v
u u v
u u v
60, 24, 156
1
60, 24, 156
36 22
17.
v ? u u v

z
2 6 1 1 1 13
0 ? v A u u v
5 2 13
, ,
3 22 3 22 3 22
23.
u
3, 2, 5 , v
0.4, 0.8, 0.2
4
3
2
1
4
3
2
1
v

u
4
6
y
u u v
u u v
u u v
3.6, 1.4, 1.6
1.8
4.37
0.7
4.37
,
0.8
4.37
x
18.
z
24.
u
0, 0,
7
10
, v
3
2
, 0,
31
5
6
5
4
u u v
0,
21
20
,0
1
3
2
1
v
u u v
u u v
0, 1, 0
x
4
3
2
u
4
6
y
25. Programs will vary.
19.
6
z
26.
u u v 50, 40, 34
u u v | 72.498
5
4
3
v
27.
u
j
x
4
3
2
1
2
1
u
4
6
y
v

u u v
j k
i j k
0 1 0
i
0 1
1
20.
6
z
A
u u v
i
1
5
1
4
3
2
1
v
28.
u
v
i j k
j k
x
4
3
2
u
4
6
y
u u v
i j k
1 1 1
j k
0 1
1
21.
u
u u v
4, 3.5, 7 , v
73.5, 5.5, 44.75
2.5, 9, 3
A
u u v
j k
2
u u v
u u v

2.94
11.8961
0.22
11.8961
1.79
11.8961
29.
u
v
3, 2, 1
1, 2, 3
i
j
k
u u v
3 2 1
1 2
3
8, 10, 4
A
u u v
8, 10, 4
180
6 5
© 2010 Brooks/Cole, Cengage Learning

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33
K
K
K
2
1
5
1
9 5
K
K
K K
K
K
K
K
K
K
2
K
K
K
1 cos 40q j sin 40qk
z
K
K
KK
K
K
K
K
1
1
z
F
y
K
K
Section 11.4
The Cross Product of Two Vectors in Space
30. u
v
2, 1, 0
1, 2, 0
34. A 2, 3, 4 , B 0, 1, 2 , C 1, 2, 0
JJJ JJJK
AB 2, 4, 2 , AC 3, 5, 4
u u v
i j k
2 1 0
0, 0, 3
JJJ JJJK
AB u AC
i j k
2 4 2
6i 2 j 2k
A
1
u u v
2
0
0, 0, 3
3
A
1
2
3 5 4
JJJ JJJK
AB u AC 1 44
11
31. A 0, 3, 2 , B 1, 5, 5 , C 6, 9, 5 , D 5, 7, 2
JJJ
AB 1, 2, 3
JJJK
DC 1, 2, 3
JJJ
BC 5, 4, 0
JJJK
AD 5, 4, 0
JJJ JJJK JJJ JJJK
Because AB DC and BC AD, the figure ABCD is
a parallelogram.
JJJ JJJK
AB and AD are adjacent sides
i j k
JJJ JJJK
AB u AD 1 2 3 12, 15, 6
5 4 0
JJJ JJJK
A AB u AD 144 225 36
35. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1
JJJ JJJK
AB 3, 12, 5 , AC 2, 13, 4
i j k
JJJ JJJK
AB u AC 3 12 5 113, 2, 63
2 13 4
JJJ JJJK
A 2 AB u AC 1 16,742

36. A 1, 2, 0 , B 2, 1, 0 , C 0, 0, 0
JJJ JJJK
AB 3, 1, 0 , AC 1, 2, 0
i j k
JJJ JJJK
AB u AC 3 1 0 5k
1 2 0
JJJ JJJK
A 2 AB u AC 2
32. A 2, 3, 1 , B 6, 5, 1 , C 7, 2, 2 , D 3, 6, 4
JJJ
AB 4, 8, 2
JJJK
DC 4, 8, 2
JJJ
BC 1, 3, 3
JJJK
AD 1, 3, 3
JJJ JJJK JJJ JJJK
Because AB DC and BC AD, the figure ABCD is
a parallelogram.
JJJ JJJK
AB and AD are adjacent sides
37. F 20k
JJJK
PQ

JJJK
PQ u F 0 cos 40q 2 sin 40q 2

JJJK
PQ u F 10 cos 40q | 7.66 ft-lb
i j k

PQ
0 0 20
10 cos 40qi
i j k
JJJ JJJK
AB u AD 4 8 2 18, 14, 20
1 3 3
JJJ JJJK
A AB u AD 324 196 400
2 230
x
1
2
ft
40°
F
y
33. A 0, 0, 0 , B 1, 0, 3 , C 3, 2, 0
JJJ JJJK
AB 1, 0, 3 , AC 3, 2, 0
38. F 2000 cos 30q j sin 30qk
JJJK
PQ 0.16k
1000 3j 1000k
i j k
JJJ JJJK
AB u AC 1 0 3
3 2 0
JJJ JJJK
A 2 AB u AC 2
6, 9, 2

36 81 4
11
2
JJJK
PQ u F

JJJK
PQ u F
i j k
0 0 0.16
0 1000 3 1000
160 3i
160 3 ft-lb
PQ
0.16 ft
60°
x

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34
JJJ
K
K
3 3 3
4
4
K
¬ ¼
¬ ¼
K
¬ ¼
¬ ¼
¨ 2
«
©
¬
sin T ¸ 42¨
¹»
¹
©
¼
K
K
F
5
K
C
A
K
4
K
© 2 ¹
K
4
K
Chapter 11
Vectors and the Geometry of Space
39. (a) Place the wrench in the xy-plane, as indicated in the figure.
The angle from AB to F is 30q 180q T 210q T
JJJ
OA 18 inches 1.5 feet
y
A
30
B
JJJ
OA

F
1.5ªcos 30q i sin 30q jº i j
56ªcos 210q T i sin 210q T jº
O
18
30
in.
F
x
i
j
k
JJJ
OA u F
3 3 3
0
4 4
56 cos 210q T 56 sin 210q T 0
100
y = 84 sin ?
ª42 3 sin 210q T 42 cos 210q T ºk
0
0
180
ª42 3 sin 210q cos T cos 210q sin T 42 cos 210q cos T sin 210q sin T ºk
JJJ
OA u F
ª § 1 3
«42 3¨ cos T
2
84 sin T , 0 d T d 180q
· § 3 1 ·º
¸ ¨ 2 cos T 2 sin T ¸»k
84 sin T k
(b) When T
JJJ
45q, OA u F
84
2
2
42 2 | 59.40
(c) Let T
84 sin T
dT
dT
84 cos T
0 when T
90q.
This is reasonable. When T
90q, the force is perpendicular to the wrench.
40. (a) AC
15 inches
5
4
feet
BC
JJJ
AB
12 inches
j k
4
1 foot
B
F
180 cos T j sin T k
12 in.
(b)
JJJ
AB u F
i j
0 5
0 180 cos T
k
1
180 sin T
15 in.
JJJ
AB u F
225 sin T 180 cos T i
225 sin T 180 cos T
(c) When T
JJJ
30q, AB u F
§ 1 · § 3 ·
225¨ ¸ 180¨ 2 ¸ | 268.38
© ¹
(d) If T 225 sin T 180 cos T , T 0 for 225 sin T

For 0 T 141.34, T c T 225 cos T 180 sin T
180 cos T ? tan T ? T | 141.34q.

0 ? tan T ? T | 51.34q. AB and F are perpendicular.
4
5
5 JJJ
(e)
400
0
180
0
From part (d), the zero is T | 141.34q, when the vectors are parallel.

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35
6
0
2
c
V
V
¬ ¼ ¬ ¼ ¬ ¼
Section 11.4
The Cross Product of Two Vectors in Space
1 0 0
48. u
0, 4, 0
41. u ? v u w
0 1 0
0 0 1
1
v
w
3, 0, 0
1, 1, 5
42. u ? v u w
1 1 1
2 1 0
0 0 1
1
0 4 0
u ? v u w 3 0 0
1 1 5
V u ? v u w 60
4 15
60
43. u ? v u w
2 0 1
0 3 0
0 0 1
49. u u v
u ? v
0 ? u and v are parallel.
0 ? u and v are orthogonal.
So, u or v (or both) is the zero vector.
44. u ? v u w
2 0 0
1 1 1
0 2 2
50. (a) u ? v u w
v u w ? u b
w ? u u v u u v ? w c
v ? w u u ux w ? v d
1 1 0
45. u ? v u w 0 1 1
1 0 1
V u ? v u w 2
(e) u ? w u v

So, a b
v ? w u u w u u ? v h
w ? v u u f
w ? v u u u u v ? w g
d h and e f g
46. u ? v u w
1 3 1
0 6 6
4 0 4
72
51. u u v
u1 , u2 , u3 ? v1 , v2 , v3
u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k
u ? v u w
72
52. See Theorem 11.8, page 794.
47. u
v
w
3, 0, 0
0, 5, 1
2, 0, 5
53. The magnitude of the cross product will increase by a
factor of 4.
54. From the vectors for two sides of the triangle, and
compute their cross product.
u ? v u w
3 0 0
0 5 1
2 0 5
75
x2 x1 , y2 y1 , z2 z1 u x3 x1 , y3 y1 , z3 z1

55. False. If the vectors are ordered pairs, then the cross
u ? v u w
75
product does not exist.

56. False. In general, u u v
v u u
57. False. Let u
Then, u u v
1, 0, 0 , v
u u w
1, 0, 0 , w
0, but v z w.
1, 0, 0 .
58. True
59. u
u1 , u2 , u3 , v
i
v1 , v2 , v3 , w
j
w1 , w2 , w3
k
u u v w
u1
u2
u3
v1 w1 v2 w2
v3 w3
ªu2 v3 w3 u3 v2 w2 º i ªu1 v3 w3 u3 v1 w1 º j ªu1 v2 w2 u2 v1 w1 ºk
u2v3 u3v2 i u1v3 u3v1 j u1v2 u2 v1 k u2 w3 u3 w2 i u1w3 u3 w1 j u1w2 u2 w1 k
u u v u u w

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36
¬ ¼
u
u
¬ ¼ ¬ ¼
¬ ¼
¬ ¼ ¬ ¼
¬ ¼
Chapter 11
Vectors and the Geometry of Space
60. u
u1 , u2 , u3 , v
v1 , v2 , v3 , c is a scalar:
i
j
k
cu u v
cu1 cu2
cu3
v1
v2
v3
cu2v3 cu3v2 i cu1v3 cu3v1 j cu1v2 cu2 v1 k
c ª u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k º

61. u = u1 , u2 , u3
c u u v
i
j
k
u u u
u1 u2
u3
u2u3
u3u2 i u1u3 u3u1 j u1u2 u2u1 k
0
u1 u2
u3
u1
u2
u3
62. u ? v u w
v1
w1
v2
w2
v3
w3
w1
w2
w3
u u v ? w
w ? u u v
u1
v1
u2
v2
u3
v3
w1 u2v3 v2u3 w2 u1v3 v1u3 w3 u1v2 v1u2
u1 v2 w3 w2v3 u2 v1w3 w1v3 u3 v1w2 w1v2
u ? v u w
63.
u u v
u u v ? u
u u v ? v
u2v3
u2v3
u2v3
u3v2 i u1v3 u3v1 j u1v2 u2 v1 k
u3v2 u1 u3v1 u1v3 u2 u1v2 u2v1 u3
u3v2 v1 u3v1 u1v3 v2 u1v2 u2 v1 v3
0
0
So, u u v A u and u u v A v.
64. If u and v are scalar multiples of each other, u
cv for some scalar c.
u u v
cv u
v
c v u v
c 0
0
If u u v
0, then u
v sin T
0. Assume u z 0, v z 0. So, sin T
0, T
0, and u and v are parallel. So,
cv for some scalar c.
65.
u u v
u
v sin T
If u and v are orthogonal, T
S 2 and sin T
1. So, u u v
v .
66. u
a1 , b1 , c1 , v
i
a2 , b2 , c2 , w
j
k
a3 , b3 , c3
v u w
a2
b2
c2
b2c3
b3c2 i a2c3 a3c2 j a2b3 a3b2 k
a3
b3
c3
i
j
k
u u v u w
b2c3
a1
b3c2
a3c2
b1
a2 c3
a2b3
c1
a3b2
u u v u w
ªb1 a2b3 a3b2 c1 a3c2 a2c3 º i ªa1 a2b3 a3b2 c1 b2c3 b3c2 º j
ªa1 a3c2 a2c3 b1 b2c3 b3c2 ºk
ªa2 a1a3 b1b3 c1c3 a3 a1a2 b1b2 c1c2 º i ªb2 a1b3 b1b3 c1c3 b3 a1a2 b1b2 c1c2 º j
ªc2 a1a3 b1b3 c1c3 c3 a1a2 b1b2 c1c2 ºk
a1a3 b1b3 c1c3 a2 , b2 , c2 a1a2 b1b2 c1c2 a3 , b3 , c3 u ? w v u ? v w

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37
t
3
3

3 3

2
5
z
5
5 5
x 3 y 7
1 3 1 7
t
2,
JJJK
2.
3
1
3

3
x
0
5
t
z
2t
5
y
2
z
Section 11.5 Lines and Planes in Space
Section 11.5
Lines and Planes in Space
1. x
(a)
1 3t , y
z
2 t , z
2 5t
x

y
(b) When t
JJJK
PQ
0, P
9, 3, 15
1, 2, 2 . When
3, Q
10, 1, 17 .
JJJK
The components of the vector and the coefficients of t are proportional because the line is parallel to PQ.
(c) y
0 when t
2. So, x
7 and z
12.
Point: 7, 0, 12
x
0 when t
1. So, y
7
3
and z
1.
Point: 0, 7 , 1
0 when t
5 . So, x
1 and y
12 .
Point: 1 , 12 , 0
2. x
(a)
2
3t , y
z
2, z
1 t
4.
2 8
z 2
(a) 7, 23, 0 : Substituting, you have
7 3
2
23 7
8
0 2
x

(b) When t 0, P
Q 4, 2, 1 .
JJJK
PQ 6, 0, 2
y

2, 2, 1 . When
2 2 2
Yes, 7, 23, 0 lies on the line.
(b) 1, 1, 3 : Substituting, you have
3 2
2 8
1 1 1
The components of the vector and the coefficients of
t are proportional because the line is parallel to PQ.
Yes, 1, 1, 3 lies on the line.
(c) z 0 when t
Point: 1, 2, 0
1. So, x
1 and y
5. Point: 0, 0, 0
Direction vector: 3, 1, 5
x
0 when t
2 . So,
y
2 and z
Direction numbers: 3, 1, 5
3. x
Point: 0, 2, 1

2 t , y

3t , z
4 t
(a) Parametric: x
x
(b) Symmetric:
3
3t , y

y
z
5
t , z
5t
(a)
0, 6, 6 : For
2 t , you have
6. Point: 0, 0, 0
2. Then y 3 2 6 and
4 2 6. Yes, 0, 6, 6 lies on the line.
Direction vector: v
2, , 1
2
(b)
2, 3, 5 : For x
t 4. Then y
2
3 4
2 t , you have
12 z 3. No, 2, 3, 5 does
Direction numbers: 4, 5, 2
(a) Parametric: x 4t , y
5t , z
not lie on the line.
(b) Symmetric:
x
4
© 2010 Brooks/Cole, Cengage Learning

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7. Point: 2, 0, 3
Vectors and the Geometry of Space
13. Points: 7, 2, 6 , 3, 0, 6
Direction vector: v
2, 4, 2
Direction vector: 10, 2, 0
Direction numbers: 2, 4, 2
Direction numbers: 10, 2, 0
(a) Parametric: x
2 2t , y
4t , z
3 2t
(a) Parametric: x
7 10t , y
2 2t , z
(b) Symmetric:

8. Point: 3, 0, 2
x 2
2
z 3
2
(b) Symmetric: Not possible because the direction
number for z is 0. But, you could describe the
line as 6.
10 2
Direction vector: v 0, 6, 3
Direction numbers: 0, 2, 1
(a) Parametric: x 3, y

(b) Symmetric: z 2, x
2
2t , z

3
2 t
14. Points: 0, 0, 25 , 10, 10, 0
Direction vector: 10, 10, 25
Direction numbers: 2, 2, 5
(a) Parametric: x 2t , y
2t , z
25 5t
9. Point: 1, 0, 1
Direction vector: v 3i 2 j k
Direction numbers: 3, 2, 1
(b) Symmetric:

15. Point: 2, 3, 4
z 25
5
(a) Parametric: x
1 3t , y
2t , z
1 t
Direction vector: v
(b) Symmetric:
x 1
3
y
2
z 1
1
Direction numbers: 0, 0, 1
Parametric: x 2, y 3, z
4 t
10. Point: 3, 5, 4
Directions numbers: 3, 2, 1
16. Point: 4, 5, 2
Direction vector: v
(a) Parametric: x
3 3t , y
5 2t , z
4 t
Direction numbers: 0, 1, 0
(b) Symmetric:
3 2
11. Points: 5, 3, 2 , ¨ , , 1¸
z 4
Parametric: x 4, y

17. Point: 2, 3, 4
Direction vector: v
5 t , z

3i 2 j k
Direction vector: v
17
3
i j 3k
3
Direction numbers: 3, 2, 1
Parametric: x 2 3t , y
3 2t , z
4 t
Direction numbers: 17, 11, 9
(a) Parametric:
x 5 17t , y 3 11t , z
2 9t
18. Point 4, 5, 2
Direction vector: v
i 2 j k
(b) Symmetric:
x 5
17
y 3
11
z 2
9
Direction numbers: 1, 2, 1
Parametric: x 4 t , y
5 2t , z
2 t
12. Points: 0, 4, 3 , 1, 2, 5
Direction vector: 1, 2, 2
Direction numbers: 1, 2, 2
(a) Parametric: x t , y

(b) Symmetric: x
2
4 2t , z
z 3
2
3 2t
19. Point: 5, 3, 4
Direction vector: v 2, 1, 3
Direction numbers: 2, 1, 3
Parametric: x 5 2t , y 3 t , z

20. Point: 1, 4, 3
4 3t
Direction vector: v
5i j
Direction numbers: 5, 1, 0
Parametric: x
1 5t , y
4 t , z
3
© 2010 Brooks/Cole, Cengage Learning

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39
2
8
t
v
v
v
1
2
y
y
z
z
7
s
Section 11.5
Lines and Planes in Space
21. Point: 2, 1, 2
30. L1: v
2, 1, 2
3, 2, 2 on line
Direction vector: 1, 1, 1
Direction numbers: 1, 1, 1
Parametric: x 2 t , y
1 t , z
2 t
L2 : v
L3: v
L4 : v
4, 2, 4
1, 1 , 1
2, 4, 1
1, 1, 3 on line
2, 1, 3 on line
3, 1, 2 on line
22. Point: 6, 0, 8
Direction vector: 2, 2, 0
Direction numbers: 2, 2, 0
Parametric: x 6 2t , y
2t , z
L1 , L2 and L3 have same direction.
3, 2, 2 is not on L2 nor L3
1, 1, 3 is not on L3
So, the three lines are parallel, not identical.
23. Let t

v
0: P
1, 2, 0
3, 1, 2 other answers possible
any nonzero multiple of v is correct
31. At the point of intersection, the coordinates for one line
equal the corresponding coordinates for the other line.
So,
(i) 4t 2 2s 2, (ii) 3 2s 3, and
24. Let t

v
0: P
4, 1, 3
0, 5, 4 other answers possible
any nonzero multiple of v is correct
(iii) t 1 s 1.
From (ii), you find that s
(iii), t 0. Letting s
0 and consequently, from
0, you see that equation (i)
25. Let each quantity equal 0:
is satisfied and so the two lines intersect. Substituting
zero for s or for t, you obtain the point 2, 3, 1 .
P
7, 6, 2 other answers possible
any nonzero multiple of v is correct
4, 2, 1
u
4i k
2i 2 j k
First line
Second line
26. Let each quantity equal 0:
P 3, 0, 3 other answers possible
cos T
u ? v
u v
8 1
17 9
7
3 17
7 17
51
5, 8, 6
any nonzero multiple of v is correct
32. By equating like variables, you have
(i) 3t 1 3s 1, (ii) 4t 1 2s 4, and
27. L1: v
L2 : v
L3: v
3, 2, 4
6, 4, 8
6, 4, 8
6, 2, 5 on line
6, 2, 5 on line
6, 2, 5 not online
(iii) 2t 4 s 1.
From (i) you have s
t and from (iii), t
t , and consequently from (ii),
3. The lines do not intersect.
L4 : v
6, 4, 6
not parallel to L1 , L2 , nor L3
33. Writing the equations of the lines in parametric form you
L1 and L2 are identical. L1

28. L1: v 2, 6, 2
L2 and is parallel to L3 .

3, 0, 1 on line
have
x
x
3t
1 4s
2 t
2 s
1 t
3 3s.
L2 : v
2, 1, 3
1, 1, 0 on line
For the coordinates to be equal, 3t
1 4s and
L3: v
2, 10, 4
1, 3, 1 on line
2 t
2 s. Solving this system yields t
17
7
and
L4 : v
2, 1, 3
5, 1, 8 on line
11. When
using these values for s and t, the z
L2 and L4 are parallel, not identical, because 1, 1, 0 is
not on L4 .
coordinates are not equal. The lines do not intersect.
29. L1: v
L2 : v
L3: v
L4 : v
4, 2, 3
2, 1, 5
8, 4, 6
2, 1, 1.5
8, 5, 9 on line

8, 5, 9 on line
L1 and L3 are identical.

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