Monografias.com > Uncategorized
Descargar Imprimir Comentar Ver trabajos relacionados

Vectors and the geometry of space (página 2)




Enviado por Efrain Olivo



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.

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

24
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
© 2010 Brooks/Cole, Cengage Learning

Monografias.com

25
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

? E | 61.39q
(a) w1
projv u
§ u ? v ·
¨ v 2 ¸
6 1 7 4
12 42
1, 4
cos J |
14.1336
F
? J | 96.53q
(b) w 2
u w1
34
1, 4 2, 8
17
6, 7 2, 8
4, 1
40. F1: C1
300
F1
| 13.0931
44. u
9, 7 , v
1, 3
F2 : C2 | 6.3246
F2
F F1 F2
| 13.0931 20, 10, 5 6.3246 5, 15, 0
(a) w1
projv u
§ u ? v ·
¨ v 2 ¸
9 1 7 3
1 32
1, 3
230.239, 36.062, 65.4655
F | 242.067 lb

? 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

Monografias.com

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
© 2010 Brooks/Cole, Cengage Learning

Monografias.com

27
v
v
v
K
K
K

¨ 2

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
© 2010 Brooks/Cole, Cengage Learning

Monografias.com

28
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

Monografias.com

29
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

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

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

30
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
© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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
© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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

Monografias.com

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

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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.

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

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

Monografias.com

38
6
, z
x 7 y 2
4
y
y
2
x
2
y
k
j
x 3 y 5
§ 2 2 ·
© 3 3 ¹
2
11
y 4
Chapter 11

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

Monografias.com

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.

© 2010 Brooks/Cole, Cengage Learning

Monografias.com

Partes: 1, 2
 Página anterior Volver al principio del trabajoPágina siguiente 

Nota al lector: es posible que esta página no contenga todos los componentes del trabajo original (pies de página, avanzadas formulas matemáticas, esquemas o tablas complejas, etc.). Recuerde que para ver el trabajo en su versión original completa, puede descargarlo desde el menú superior.

Todos los documentos disponibles en este sitio expresan los puntos de vista de sus respectivos autores y no de Monografias.com. El objetivo de Monografias.com es poner el conocimiento a disposición de toda su comunidad. Queda bajo la responsabilidad de cada lector el eventual uso que se le de a esta información. Asimismo, es obligatoria la cita del autor del contenido y de Monografias.com como fuentes de información.

Categorias
Newsletter