C H A P T E R 1 1
Analytic Geometry in Three Dimensions
Section 11.1 The Three-Dimensional Coordinate System
. . . . . . 1029
Section 11.2 Vectors in Space . . . . . . . . . . . . . . . . . . . . . 1037
Section 11.3 The Cross Product of Two Vectors
. . . . . . . . . . . 1042
Section 11.4 Lines and Planes in Space . . . . . . . . . . . . . . . . 1048
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1055
Problem Solving
Practice Test
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1061
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1068
C H A P T E R 1 1
Analytic Geometry in Three Dimensions
Section 11.1
■
■
The Three-Dimensional Coordinate System
You should be able to plot points in the three-dimensional coordinate system.
The distance between the points x1, y1, z1 and x2, y2, z2 is
d x2 x12 y2 y12 z2 z12.
The midpoint of the line segment joining the points x1, y1, z1 and x2, y2, z2 is
x1 x2 y1 y2 z1 z2
,
,
.
2
2
2
The equation of the sphere with center h, k, j and radius r is
x h2 y k2 z j2 r2.
You should be able to find the trace of a surface in space.
■
■
■
Vocabulary Check
1. three-dimensional
2. xy-plane, xz-plane, yz-plane
x x2 y1 y2 z1 z 2
5. 1
,
,
2
2
2
8. trace
4. Distance Formula
7. surface, space
1. A1, 4, 3, B1, 3, 2, C3, 0, 2
z
3.
3. octants
6. sphere
2. A6, 2, 3, B2, 1, 2 C2, 3, 0
4.
z
3
5
(−3, − 2, −1)
4
3 (2, 1, 3)
2
(− 1, 2, 1)
(3, 0, 0)
y
−2
1
3
4
5
x
2
2
3
4
y
−5 −4 −3 −2
1
2
4
5
1
−4
−3
−5
6.
3
−4
2 −3
1
−5 −4 −3 −2
(3, −1, 0)
4
x
y
2
1
−2
1
2
3
−5
3
z
5
4
(4, 0, 4) 3
2
1
−4 −3
−3
−4
2
−2
−2
z (− 4, 2, 2)
1
−3
x
5.
−4
6
x
5
−4
y
1 2 3
−3
−4
−5
5 6
(0, 4, −3)
1029
1030
Chapter 11
Analytic Geometry in Three Dimensions
8. x 6, y 1, z 1 ⇒ 6, 1, 1
7. x 3, y 3, z 4: 3, 3, 4
9. y z 0, x 10: 10, 0, 0
10. x 0, y 2, z 8 ⇒ 0, 2, 8
11. Octant IV
12. Octant VI
13. Octants I, II, III, IV
(above the xy-plane)
14. Octants III, IV, VII, or VIII
15. Octants II, IV, VI, VIII
16. Octants I, II, VII, or VIII
17. d 5 02 2 02 6 02
18. d 7 12 0 02 4 02
25 4 36
36 16
65 units
52
213
19. d 7 32 4 22 8 52
20. d 4 22 1 12 9 62
42 22 32
4 9
16 4 9
13
29
5.385
21. d 6 12 0 42 9 22
22. d 1 22 1 32 7 72
7 2 42 7 2
9 16
49 16 49
25
114
5
10.677
23. d 1 02 0 32 10 02
24. d 2 02 4 62 0 32
1 9 100
4 100 9
110 10.488
113
25. d1 2 02 5 02 2 22 4 25 29
d2 0 02 4 02 0 22 16 4 20 25
d3 0 22 4 52 0 22 4 1 4 9 3
d12 d22 d32 29
26. d1 4 22 4 12 1 22 36 25 1 62
d2 4 22 4 52 1 02 4 1 1 6
d3 2 22 1 52 2 02 16 36 4 56 214
d12 d22 d 32 62
Section 11.1
The Three-Dimensional Coordinate System
1031
27. d1 2 02 2 02 1 02 4 4 1 9 3
d2 2 02 4 02 4 02 4 16 16 36 6
d3 2 22 4 22 4 12 36 9 45 35
d12 d22 9 36 45 d32
28. d1 1 12 3 02 1 12 3
d2 1 12 0 02 3 12 2
d3 1 12 0 32 3 12 13
d32 13 d12 d22
29. d1 5 12 1 32 2 22 16 4 16 36 6
d2 5 12 1 12 2 22 36 4 40 210
d3 1 12 1 32 2 22 4 16 16 36 6
d1 d3 Isosceles triangle
30. d1 7 52 1 32 3 42 4 4 1 9 3
d2 3 72 5 12 3 32 16 16 32 42
d3 3 52 5 32 3 42 4 4 1 9 3
d1 d3 3. Isosceles triangle
31.
3 2 0, 22 0, 4 2 0 32, 1, 2
32. Midpoint:
1 2 2, 5 2 2, 12 2 32 , 27, 12
33. Midpoint:
3 2 3, 62 4, 10 2 4 0, 1, 7
34. Midpoint:
12 3, 5 2 7, 32 1 1, 6, 2
35. Midpoint:
6 2 4, 22 2, 5 2 6 1, 0, 112
36. Midpoint:
32 6, 5 2 4, 5 2 8 92, 92, 132
37. Midpoint:
22 7, 8 2 4, 10 2 2 52, 2, 6
38. Midpoint:
9 2 9, 52 2, 1 2 4 9, 27, 23
39. x 32 y 22 z 42 16
40. x 32 y 42 z 32 4
41. x 02 y 42 z 32 32
42. x 22 y 12 z 82 36
x2 y 42 z 32 9
43. Radius Diameter
5
2
x 32 y 72 z 52 52 25
44. Radius Diameter
4: x 02 y 52 z 92 42 16
2
1032
Chapter 11
45. Center:
Analytic Geometry in Three Dimensions
30 00 06
3
,
,
, 0, 3
2
2
2
2
Radius:
3 32
Sphere:
x 2 46. Center:
3
2
2
0 02 0 32 y 02 z 32 2 1 2 4 2 6
1
,
,
, 1, 4
2
2
2
2
Radius:
2 21
Sphere:
x 2 1
2
2
94 9 454
45
4
2 12 2 42 y 12 z 42 94 9 4 614
61
4
25
25
47. x 2 5x 4 y 2 z 2 4
x 52 2 y 2 z 2 254
5
Center: 2, 0, 0
Radius:
5
2
48. x 2 y 2 8y 16 z 2 16
x 2 y 42 z2 16
Center: 0, 4, 0
Radius: 4
49. x 2 4x 4 y 2 2y 1 z 2 6z 9 10 4 1 9
x 22 y 12 z 32 4
Center: 2, 1, 3
Radius: 2
50. x2 6x 9 y2 4y 4 z2 9 9 4
x 32 y 22 z2 4
Center: 3, 2, 0
Radius: 2
51. x2 4x 4 y2 z2 8z 16 19 4 16
x 22 y2 z 42 1
Center: 2, 0, 4
Radius: 1
52. x2 y2 8y 16 z2 6z 9 13 16 9
x2 y 42 z 32 12
Center: 0, 4, 3
Radius: 12 23
Section 11.1
The Three-Dimensional Coordinate System
x 2 y 2 z 2 2x 23 y 8z 73
9
53.
x 2 2x 1 y 2 23 y z 2 8z 16 739 1 19 16
2
x 12 y 13 z 42 9
1
9
Center: 1, 13, 4
Radius: 3
5
54. x2 y2 z2 x 3y 2z 2
x2 x 14 y2 3y 94 z2 2z 1 52 14 94 1
x 12 2 y 32 2 z 12 1
1 3
Center: 2, 2, 1
Radius: 1
9x 2 6x 9y 2 18y 9z 2 1
55.
x 2 23 x 19 y 2 2y 1 z 2 19 19 1
Center:
x 13 2 y 12 z 2 1
13, 1, 0
Radius: 1
1
33 1
y2 8y 16 z2 2z 1 16 1
4
4
4
56. x2 x x 21
Center:
2
y 42 z 12 9
12, 4, 1
Radius: 3
57.
58. yz trace x 0: y 32 z2 25 Circle
z
(x − 1) 2 + z 2 = 36
z
(0, −3, 0)
−8
4
−6
−2
2
4
(1, 0, 0)
x
6
− 10
−2
4
2
−4
6
10
8
2
−2
y
2
y
6
x
8
−6
(y + 3) 2 + z 2 = 25
59.
z
(y − 3) 2 + z 2 = 5
2
(− 2, 3, 0)
x
2
2
y
−6
−8
1033
1034
Chapter 11
Analytic Geometry in Three Dimensions
60. xy trace z 0: x 2 y 12 3 Circle
z
61.
6
z
5
4
−4
2
−4
(0, 1, − 1)
−2
2
4
x
6
−4
−6
y
x2
+ (y −
1) 2
5
x
4
3
2
2
3
4
5
y
=3
62. x2 y2 6y z2 8z 16 21 16
z
x2 y2 6y z 42 5
z1 4 5 x2 y2 6y
−6
z2 4 5 x2 y2 6y
−4
2
−2
2
x
63. The length of each side is 3.
Thus, x, y, z 3, 3, 3.
(b) Assume the north and south poles are on the
z-axis. Lines of longitude that run north-south
are traces of planes containing the z-axis. These
shapes are circles of radius 3963 miles.
y
4
64. x 4, y 4, z 8
66. (a) x 2 y 2 z 2 39632.
−4
−2
4, 4, 8
65. d 165 ⇒ r 165
2 82.5
x 2 y 2 z 2 165
2 2
(d) The prime meridian is a trace of a plane containing
the z-axis. It is a semi-circular arc running from
pole to pole.
(e) The equator is the trace of the plane containing the
x- and y-axes.
(c) Latitudes are traces of planes perpendicular to
the z-axis. These shapes are circles.
67. False. x is the directed distance from the yz-plane
to P.
68. False. The trace could be a single point, or empty.
69. In the xy-plane, the z-coordinate is 0.
In the xz-plane, the y-coordinate is 0.
In the yz-plane, the x-coordinate is 0.
70. It is a plane.
71. The trace is a circle, or a single point.
72. The trace will be a line in the xy-plane (unless the
plane is the xy-plane).
73. xm x2 x1
⇒ x2 2xm x1
2
Similarly for y2 and z2,
x2, y2, z2 2xm x1, 2ym y1, 2z m z1.
74. x2 2xm x1 25 3 7
y2 2ym y1 28 0 16
z2 2zm z1 27 2 12
7, 16, 12
Section 11.1
9
9
2
4
4
75. v 2 3v v
3
2
v
2
78. x 2 3x x 32
2
z 72
17
3
±
2
2
v
3 17
±
2
2
9
9
1
4
4
49
49
19 4
4
76. z2 7z 17
4
79.
2
7
55
±
2
2
z
7 5
± 5
2 2
13
3
±
2
2
x
3 13
±
2
2
x
y
v 32
v 3
18
tan 32
x
5
4
2
89
5
±
4
4
10
1
±
2
2
x
5 89
±
4
4
1 10
±
2
2
83. v 4i 5j, Quadrant I
Quadrant II
22
5 5
±
2
2
89
16
x
12
5
4
5
5
±
2
2
10
4
82. v 1, 2
2
5
25
25
4
80. x 2 x 2
16
16
y
81. v 3i 3j, Quadrant IV
2
x
1 9 1
4 4 4
2
1035
25
25
5 4
4
x 52
4y 2 4y 9
y 12
x
77. x 2 5x 125
4
z
y2 y 13
4
The Three-Dimensional Coordinate System
5
v 16 25 41
2
⇒ 116.6
1
tan 5
⇒ 51.34
4
3
tan 1 ⇒
3
45 or 315
84. v 10, 7
85. u v 4, 1
Quadrant IV
v 100 49 149
tan 7
⇒ 325.0
10
3, 5
86. u v 1, 0
43 15
20
7
2
87. a0 1, an an1 n2
a1 1 12 2
a2 2 22 6
a3 6 32 15
a4 15 42 31
1
First differences:
Second differences:
Neither model
2
1
6
4
3
15
9
5
31
16
7
2, 6
1036
Chapter 11
Analytic Geometry in Three Dimensions
88. a0 0, an an1 1
a1 0 1 1
a2 1 1 2
a3 3
a4 4
1
0
1
First difference
2
1
Second difference
0
3
1
0
4
1
0
Linear model
89. a1 1, an an1 3
a2 1 3 2
a3 2 3 5
a4 5 3 8
a5 8 3 11
1
First differences:
Second differences:
2
3
5
8
3
11
3
0
3
0
0
14
24
Linear model
90. a1 4, an an1 2n
a2 4 22 0
a3 0 23 6
a4 6 24 14
a5 14 25 24
4
First difference
6
0
4
Second difference
6
2
8
2
10
2
Quadratic model
91. x 52 y 12 49
92. x 32 y 62 81
93. y 12 4px 4, p 3
94. x h2 4py k p 5,h, k 2, 5
y 12 43x 4
x 22 45 y 5
y 12 12x 4
x 22 20 y 5
95. a 3, b 2, center: 3, 3, horizontal major axis
x 32 y 32
1
9
4
Section 11.2
96. Center: 0, 3 Vertical major axis length 9 ⇒ a c 3 ⇒ b2 a2 c2 Vectors in Space
1037
9
2
81
45
9
4
4
x 02 y 32
1
454
814
97. Center: 6, 0, horizontal transverse axis
98. Center: 3, 5
Vertical transverse axis
a 2, c 6, b2 c2 a2 36 4 32
a 4, c 5, b2 c2 a2 25 16 9
x 62 y 2
1
4
32
y 52 x 32
1
16
9
Section 11.2
■
■
■
■
Vectors in Space
Vectors in space v v1, v2, v3 have many of the same properties as vectors in the plane.
The dot product of two vectors u u1, u2, u3 and v v1, v2, v3 in space is u v u1v1 u2v2 u3v3.
Two nonzero vectors u and v are said to be parallel if there is some scalar c such that u cv.
You should be able to use vectors to solve real life problems.
Vocabulary Check
1. zero
2. v v1i v2 j v3 k
4. orthogonal
5. parallel
1. v 0 2, 3 0, 2 1 2, 3, 1
3. component form
2. (a) v 1 1, 4 4, 0 4 0, 0, 4
(b)
z
z
3
3
2
2
−3
1
−2
−2
− 3 (− 2, 3, 1)
1
2
2
−4
−3
−2
−1
1
−3
−2
−1
1
3
1
4
2
3
3
y
x
−2
−3
−4
x
3. (a) v 1 6, 1 4, 3 2
7, 5, 5
(b) v 7 2 52 5 2
49 25 25
99
311
(c)
−4
11
v
1
7, 5, 5 7, 5, 5
33
v 311
2
3
4
y
(0, 0, −4)
4. (a) v 0 7, 0 3, 2 5 7, 3, 3
(b) v 49 9 9 67
(c) Unit vector:
67
1
7, 3, 3 7, 3, 3
67
67
1038
Chapter 11
Analytic Geometry in Three Dimensions
z
5. (a)
(b)
z
〈2, 2, 6〉
6
4
5
3
−4
2 −3
4
3
2
2
3
1
1
3
2
3
4
〈−1, −1, − 3〉
4
4
−3
x
−2
4
1
1
2
3
y
− 4 − 3 −2
y
−4 − 3 − 2
1
−4
x
(c)
(d)
z
z
4
3 3 9
, ,
2 2 2
5
4
3
−4
2 −3
3
〈0, 0, 0〉
2
1
y
−4 −3 −2
2
2
1
3
−4 − 3 − 2
4
2
3
4
−4
−3
x
3
−3
x
−2
4
2
−2
4
3
y
1
1
6. v 1, 2, 2
(b) 2v 2, 4, 4
(a) v 1, 2, 2
z
z
4
4
3
3
−4
2 −3
−4
2 −3
1
1
y
− 6 − 5 −4 − 3
1
〈1, − 2, − 2〉
y
−4 − 3 − 2
2
2
3
−2
3
4
−4
5
5
(d) 2 v 2 , 5, 5
1
1
(c) 2 v 2, 1, 1
z
z
− 5 , 5, 5
2
6
2
1
2
−3
x
−4
1
1
−2
4
−3
x
〈− 2, 4, 4〉
− 1 , 1, 1
5
2
−2
4
3
−2
−1
2
−4
2 −3
y
1
1
2
1
−1
y
−3 −2
x
2
3
−2
4
1
1
−2
x
7. z u 2v 1, 3, 2 21, 2, 2 3, 7, 6
8. z 71, 3, 2 1, 2, 2 15 5, 0, 5 7, 19, 13
9. 2z 4u w ⇒ z 124u w 1241, 3, 2 5, 0, 5 10. z u v 1, 3, 2 1, 2, 2 0, 1, 0
12, 6, 32
2
3
4
5
Section 11.2
11. v 7, 8, 7
Vectors in Space
1039
12. v 22 02 52 4 25 29
49 64 49 162 92
13. v 42 32 72
14. v 22 12 62 41
16 9 49 74
15.
v 1 1, 0 3, 1 4 0, 3, 5
v 0 32 52 34
17. (a)
8, 3, 1
u
u
74
(b) 74
1
8i 3j k 8, 3, 1
74
74
16.
v 1 0, 2 1, 2 0 1, 3, 2
v 1 9 4 14
18. (a)
(b)
u
3, 5, 10
1
3i 5j 10k
134
134
u
1
3i 5j 10k
134
74
1
8i 3j k 8, 3, 1
74
74
19. u v 4, 4, 1
2, 5, 8
20. u v 34 110 61 28
8 20 8 4
21. u v 2, 5, 3
9, 3, 1
22. u v 06 34 62 0
18 15 3 0
23. cos uv
8
⇒ 124.45
u v 825
24. cos 25. cos uv
120
⇒ 109.92
u v 170073
26. cos 27. 32 8, 4, 10 12, 6, 15 ⇒ parallel
uv
5
⇒ 49.80
u v 106
uv
100
⇒ 65.47
u v 464125
28. u v 2 3 5 10 0 and
u cv ⇒ neither
29. u v 3 5 2 0 ⇒ orthogonal
30. 8u 8 1, 12, 1 8, 4, 8 v ⇒ parallel
31. v 7 5, 3 4, 1 1 2, 1, 2
32. v 4 2, 8 7, 1 4 2, 1, 3
u 4 7, 5 3, 3 1 3, 2, 4
u 0 4, 6 8, 7 1 4, 2, 6
Since u and v are not parallel, the points are not
collinear.
Since u 2v, the points are collinear.
33. v 1 1, 2 3, 5 2 2, 1, 3
34. v 1 0, 5 4, 6 4 1, 1, 2
u 3 1, 4 2, 1 5 4, 2, 6
u 2 1, 6 5, 7 6 1, 1, 1
Since u 2v, the points are collinear.
Since u and v are not parallel, the points are
not collinear.
1040
Chapter 11
Analytic Geometry in Three Dimensions
35. v 2, 4, 7 q1 1, q2 5, q3 0 ⇒
2 q1 1
4 q2 5
7 q3
q1 3
⇒ q2 1
q3 7
⇒
Terminal point is 3, 1, 7.
36. 4, 1, 1 x 6, y 4, z 3 ⇒ x, y, z 10, 5, 2
37. v 4, 32, 14 q1 2, q2 1, q3 32
4 q1 2 ⇒ q1 6
3
2
q2 1 ⇒ q2 52
14 q3 32 ⇒ q3 74
Terminal point: 6, 52, 74 38.
52, 12, 4 x 3, y 2, z 12
3 7
⇒ x, y, z 11
2 , 2, 2 cu c2 4c2 9c2 c 14 3 ⇒
c±
3
314
±
14
14
41. v q1, q2, q3
⇒ c 12
6
6 ⇒ c ± 6
24
6
42. v lies in xz-plane ⇒ y 0.
Since v lies in the yz-plane, q1 0. Since v makes
an angle of 45, q2 q3. Finally, v 4 implies
that q22 q32 16. Thus, q2 q3 22
and v 0, 22, 22 , or q2 22 and
q3 22 and v 0, 22, 22 .
v 10sin 60, 0, cos 60 53, 0, 5, or
v 10sin 60, 0, cos 60 53, 0, 5
→
43. AB 0, 70, 115. F1 C10, 70, 115
→
AC 60, 0, 115. F2 C260, 0, 115
→
AD 45, 65, 115. F3 C3 45, 65, 115
F1 F2 F3 0, 0, 500. Thus
60C2 45C3 0
70C1
40. c u c u c 4 4 16 c 24 12
39. cu ci 2cj 3ck
65C3 0
115C1 115C2 115C3 500
Solving this system yields C1 Thus,
F1 202.919
N
F2 157.909
N
F3 226.521
N
104
28
112
, C2 , C3 .
69
23
69
Section 11.2
44. (a) sin 18
L
sin1
T
θ
θ
18
L
8
cos Vectors in Space
L
8
cos
sin1
18
L
8
L2 182
8L
L 2 182
1041
T
18 in.
L
Domain: L > 18
(b) L
T
(c)
30
20
18.4
25
11.5
30
10
35
9.3
40
9.0
45
8.7
50
8.6
L = 18
T=8
0
100
0
Vertical asymptote: L 18
Horizontal asymptote: T 8
The minimum tension in each cable is 8 pounds and the minimum cable length is 18 inches.
(d) 10 8
cos sin1
sin1
18
L
⇒ cos sin1
18L 108 45
18L cos 45
1
53
18
4
sin cos1
L
5
L
90
30 inches
3
45. True. cos 0 ⇒ 90
47. If u v < 0, then cos < 0
and the angle between u and v
is obtuse, 180 > > 90.
46. True
48. Let v v1, v2, v3 and u u1, u2, u3.
su
Then tv tv1, tv2, tv3 su 1 tv
u tv u1 tv1, u2 tv2, u3 tv3
u
and su tv su1 tv1, su2 tv2, su3 tv3 u 1 tv
v
The endpoints of these three vectors are collinear, as indicated in the figure.
So, the figure is a line.
49. (a) x t , y 3t 2
(b) x t 1, y 3t 1 2 3t 1
50. (a) x t, y 2
t
(b) x t 1, y 2
t1
tv
1042
Chapter 11
Analytic Geometry in Three Dimensions
52. (a) x t, y 4t 3
51. (a) x t , y t 2 8
(b) x t 1, y 4t 13
(b) x t 1, y t 12 8 t 2 2t 7
Section 11.3
■
The Cross Product of Two Vectors
The cross product of two vectors u u1i u2 j u3k and v v1i v2 j v3k is given by
u v u2v3 u3v2i u1v3 u3v1j u1v2 u2v1k
i
u1
v1
j
u2
v2
k
u3 .
v3
■
The cross product satisfies the following algebraic properties.
(a) u v v u
(b) u v w u v u w
(c) cu v cu v u cv
(d) u 0 0 u 0
(e) u u 0
(f) u v w u v w
■
The following geometric properties of the cross product are valid, where is the angle between the vectors
u and v:
(a) u v is orthogonal to both u and v.
(b) u v u v sin (c) u v 0 if and only if u and v are scalar multiples.
(d) u v is the area of the parallelogram having u and v as sides.
■
The absolute value of the triple scalar product is the volume of the parallelepiped having u, v, and w
as sides.
u
u1
v w v1
w1
u2
v2
w2
u3
v3
w3
Vocabulary Check
1. cross product
2. 0
3. u v sin 4. triple scalar product
i
1. j i 0
1
j
1
0
k
0 k
0
i
2. k j 0
0
2
2
2
−2
(−1, 0, 0)
−1
1
(0, 0, − 1)
−2
1
−2
−1
−−11
x
−2
1
−1
k
1 i
0
z
z
−2
j
0
1
−1
2
y
x
−2
1
2
y
Section 11.3
i
3. i k 1
0
j
0
0
k
0 j
1
The Cross Product of Two Vectors
i
4. k i 0
1
2
2
−2
1
(0, 1, 0)
1
−1
2
−2
x
j
2
1
2
y
2
y
2
i
5. u v 3
0
−1
−1
1
−1
−2
1
−2
−1
(0, −1, 0)
x
k
1 j
0
z
z
−2
j
0
0
k
5 3, 3, 3
1
−2
i
6. u v 6
4
u v u 3, 3, 3 3, 2, 5 0
j
8
1
k
3 29, 36, 38
4
u v v 3, 3, 3 0, 1, 1 0
i
7. u v 10
7
j
0
0
k
6 0, 42, 0
0
i
8. u v 5
2
u v u 0, 42, 0 10, 0, 6 0
j
5
2
k
11 7, 37, 20
3
u v v 0, 42, 0 7, 0, 0 0
i
9. u v 6
1
j
2
3
k
1 7, 13, 16
2
7i 13j 16k
i
10. u v 1
1
2
11. u v j
i
0
1
1
4
j
0
3
i
13. u v 1
0
i
14. u v 1
0
k
52 32, 32, 32 32 i 32 j 32 k
3
2
3
4
j
0
1
j
0
1
k
6 18, 6, 0
1
18i 6j
i
12. u v 2
3
j
0
0
1
3
k
1 1, 2, 1
2
i 2j k
k
2 0 2i 1 0j 1 0k 2i j k
1
k
0 2j 29 k
3
1043
1044
15.
Chapter 11
Analytic Geometry in Three Dimensions
i
uv 3
0
j
1
1
k
0 i 3j 3k
1
i
16. u v 1
1
u
u
v
1
i 3j 3k
v
19
19
i
17. u v 3
j
2
34
1
2
19
Unit vector 18.
1
71, 44, 25
7602
7602
7602
i
j
k
7 14
5 70i 175j 392k
14 28 15
189,189 21429
w 712 442 252 7602
uv
u v 702 1752 3922
Consider the parallel vector 71, 44, 25 w.
Unit vector Unit vector 71, 44, 25
19.
i
uv 1
1
j
1
1
k
1 2i 2j
1
u
u
1
v
1
2i 2j
v
22
2
2
2
i
i
1
2
2
2
j
0
0
1
10, 25, 56
3429
429
1287
10, 25, 56
i
uv 1
2
Unit vector j
2
1
k
2 6i 6j 3k
2
uv
1
6i 6j 3k
u v 9
2
2
1
i j k
3
3
3
j
j
i
21. u v 0
1
20.
uv
1
70, 175, 392
u v 21429
u v 36 36 9 9
u v 22
Unit vector 6
3
2
uv
i j k
u v
7
7
7
1, 3, 3
k
71 11 5
5 , ,
20
5 4
1
10
k
0 6i 3j 2k.
3
u v 36 9 4 7
u v 19
Unit vector j
2
0
k
1 j
1
Area u v j 1 square unit
i
22. u v 1
1
j
2
0
k
2 2i j 2k
1
Area u v 2i j 2k
4 1 4 3 square units
i
23. u v 3
2
j
4
1
k
6 26i 3j 11k
5
Area u v 262 32 112
806 square units
i
24. u v 2
1
j
3
2
Area u v 82
k
2 8, 10, 7
4
102 72
213 square units
Section 11.3
j
2
2
1045
i
25. u v 2
0
The Cross Product of Two Vectors
i
26. u v 4
5
k
3 12, 6, 4
3
j
3
0
k
2 3, 6, 15
1
Area u v 32 62 152
Area u v 122 62 42
270 330 square units
14 square units
\
27. (a) AB 3 2, 1 1, 2 4 1, 2, 2 is parallel to
\
DC 0 1, 5 3, 6 8 1, 2, 2.
\
\
AD 3, 4, 4 is parallel to BC 3, 4, 4.
\
(b) AB
\
AD \
Area AB
\
(c) AB
i
1
3
j
2
4
\
AD k
2 16, 2, 10
4
162
22 102 360 610 square units
AD 1, 2, 2 3, 4, 4 0
\
⇒ not a rectangle
→
28. (a) AB 1, 2, 3
→
CD 1, 2, 3
Opposites are parallel and same length. Thus ABCD form a parallelogram.
i
j
k
→ →
2
3 10, 14, 6
(b) AB AC 1
5
4
1
→ →
Area AB AC 102 142 62 283 square units
→ →
(c) AB AC 5 8 3 16 0 ⇒ not a rectangle.
29. u 1, 2, 3, v 3, 0, 0
uv
i
1
3
j
2
0
30. u 2 1, 0 4, 2 3 1, 4, 1
k
3 0, 9, 6
0
Area 12 u v 1281 36 3213 square units
v 2 1, 2 4, 0 3 3, 6, 3
i
uv
1
3
Area j
4
6
k
1 6, 6, 18
3
1
1
2
2 u v 2 6 1
2 396 311 square
62 182
units
31. u 2 2, 2 3, 0 5 4, 5, 5
32. u 2 2, 4 4, 0 0 4, 8, 0
v 3 2, 0 3, 6 5 1, 3, 11
v 0 2, 0 4, 4 0 2, 4, 4
i
u v 4
1
i
u v 4
2
Area 1
2 u j
5
3
v k
5 40, 49, 17
11
1
2
2 40
124290 square units
492 172
Area j
8
4
1
2 u v
1
2 1280
k
0 32, 16, 0
4
1
2
2 32
162
85 square units
1046
Chapter 11
Analytic Geometry in Three Dimensions
2
33. u v w 4
0
3
4
0
3
0
4
2
34. u v w 0
0
0
3
0
1
36. u v w 2
0
4
0
3
216 316 30 16
2
35. u v w 1
4
3
1
3
1
0
1
1
w 0
1
Volume u
39. u v
1
1
0
0
1 112
1
1
38. u v w 0
3
v w 2 cubic units
2
0
0
Volume u v
2
2
2
Volume u v
0
2
5
v w 12 cubic units
4
w 0
0
w 9 9 cubic units
2
2
0
1
2
1
12 21 4 1(0 4 16
41. u 4, 0, 0, v 0, 2, 3, w 0, 5, 3
u v
3
3
3
1
40. u v w 1
2
0 26 20 12
Volume u
1
3
0
19 19 39 9
0
w 0
3
7
4
6
10 12 412 0 76 6
21 31 17 2
37. u v
1
0 6
1
w 16 cubic units
→
→
→
42. AB 1, 1, 0, AC 1, 0, 2, AD 0, 1, 1
0
3 421 84
3
u v
1
w 1
0
1
0
1
0
2 12 11 3
1
Volume 3 cubic units
Volume 84 84 cubic units
1
43. V cos 40 j sin 40 k
2
F pk
i
(a) V F 0
0
j
1
cos 40
2
0
1
p cos 40 i
2
p
T V F cos 40
2
k
1
sin 40
2
p
k
j
(b)
p
T
15
5.75
20
7.66
25
9.58
30
11.49
35
13.41
40
15.32
45
17.24
V
40°
F
Section 11.3
44. V 0.16cos 30 j sin 30 k
j
VF 0
k
0.16 cos 30 0.16 sin 30
0
2000
0
1047
k
F 2000k
i
The Cross Product of Two Vectors
20000.16 cos 30 i
1603 i
j
60°
V
F
30°
T V F 1603 ft-lb
46. False. u v v
45. True. The cross product is not
defined for two-dimensional
vectors.
u
47. If the magnitudes of two vectors
are doubled, the magnitude of
the cross product will be four
times as large.
i
j
48. u v cos sin cos sin k
0 cos sin sin cos k
0
y
u
α2β
Area of triangle formed by the unit vectors u and v is
1
1
2 baseheight 2 1 sin .
The area is also given by
1
2 u v 1
2
α
v
cos sin sin cos β
x
Notice that cos sin sin cos is negative.
Thus, sin sin cos cos sin .
1
52. cos 930 cos 210 55. tan
1
51. sin 690 sin 330 2
50. tan 300 3
49. cos 480 cos 120 2
3
2
53. sin
7
1
19
sin
6
6
2
15
7
tan
1
4
4
56. tan
54. cos
10
4
tan
3
3
3
0 7 8 56
3, 4: z 6 3 7 4 46
15, 0: z 6 15 7 0 90
58. 0, 8: z 6
57. z 6x 4y
At 0, 5: z 60 45 20
At 0, 0: z 60 40 0
20
At 20
3 , 0: z 6 3 40 40
Minimum: 46 at 3, 4.
At 6, 4: z 66 44 52
Maximum: Unbounded
The maximum value of z, z 52, is found at 6, 4.
The minimum value of z, z 0, is found at 0, 0.
y
y
20
18
16
14
10
8
8
6
4
2
(0, 5)
(6, 4)
4
2
2
4
6
8
(0, 8)
(3, 4)
(15, 0)
2 4 6
( 203 , 0)
(0, 0)
x
10
3
17
5
cos
6
6
2
10 12 14 16
x
1048
Chapter 11
Section 11.4
■
■
■
■
■
Analytic Geometry in Three Dimensions
Lines and Planes in Space
The parametric equations of the line in space parallel to the vector a, b, c and passing through the point
x1, y1, z1 are
x x1 at, y y1 bt, z z1 ct.
The standard equation of the plane in space containing the point x1, y1, z1 and having normal vector
a, b, c is
ax x1 b y y1 c z z1 0.
You should be able to find the angle between two planes by calculating the angle between their
normal vectors.
You should be able to sketch a plane in space.
The distance between a point Q and a plane having normal n is
PQ n
D projnPQ \
\
n
where P is a point in the plane.
Vocabulary Check
\
PQ
1. direction,
t
2. parametric equations
3. symmetric equations
4. normal
5. ax x1 b y y1 cz z1 0
2. (a) x x1 at 3 3t
1. x x1 at 0 t
y y1 bt 0 2t
y y1 bt 5 7t
z z1 ct 0 3t
z z1 ct 1 10t
(a) Parametric equations: x t, y 2t, z 3t
Parametric equations:
x 3 3t, y 5 7t, z 1 10t
(b) Symmetric equations:
x
y
z
1 2 3
(b) Symmetric equations:
1
4
3. x x1 at 4 t, y y1 bt 1 t, z z1 ct 0 t
2
3
1
4
(a) Parametric equations: x 4 t, y 1 t, z t
2
3
Equivalently: x 4 3t, y 1 8t, z 6t
(b) Symmetric equations:
x4 y1
z
3
8
6
x3 y5 z1
3
7
10
Section 11.4
Lines and Planes in Space
1049
4. x x1 at 5 4t, y y1 bt 0 0t, z z1 ct 10 3t
(a) Parametric equations: x 5 4t, y 0, z 10 3t
(b) Symmetric equations:
x 5 z 10
,y0
4
3
5. x x1 at 2 2t, y y1 bt 3 3t, z z1 ct 5 t
(a) Parametric equations: x 2 2t, y 3 3t, z 5 t
(b) Symmetric equations:
x2 y3
z5
2
3
7. (a) v 1 2, 4 0, 3 2 1, 4, 5
6. (a) v 3, 2, 1
Point: 2, 0, 2
x 1 3t, y 2t, z 1 t
(b) Symmetric equations:
x 2 t, y 4t, z 2 5t
x1
y
z1
3
2
1
(b)
8. (a) v 8, 5, 12
Point: 2, 3, 0
Parametric equations: x 2 8t, y 3 5t, z 12t
(b) Symmetric equations:
x2 y3
z
8
5
12
9. (a) v 1 3, 2 8, 16 15 4, 10, 1
Point: 3, 8, 15
x 3 4t, y 8 10t, z 15 t
(b)
x 3 y 8 z 15
4
10
1
10. 2, 3, 1, 1, 5, 3
(a) Let P 2, 3, 1, Q 1, 5, 3
\
\
V PQ 1 2, 5 3, 3 1 1, 8, 4
Direction numbers: a 1, b 8, c 4
Choose P as the initial point:
x 2 t, y 3 8t, z 1 4t
(b)
x2 y3 z1
1
8
4
x2
y
z2
1
4
5
1050
Chapter 11
Analytic Geometry in Three Dimensions
11. 3, 1, 2, 1, 1, 5
(a) v 1 3, 1 1, 5 2 4, 0, 3
Parametric: x 3 4t, y 1, z 2 3t
(b) Since b 0, there are no symmetric equations.
12. 2, 1, 5, 2, 1, 3
(a) Let P 2, 1, 5, Q 2, 1, 3
\
\
V PQ 2 2, 1 1, 3 5 0, 2, 8
Direction numbers: a 0, b 2, c 8
Choose P as the initial point:
x 2, y 1 2t, z 5 8t
(b) Since the direction number a 0, no set of symmetric equations are possible.
1
1
1
13. , 2, , 1, , 0
2
2
2
12, 12 2, 0 21
32, 52, 21
(a) v 1 Direction numbers: 3, 5, 1
1
1
Parametric: x 3t, y 2 5t, z t
2
2
(b) Symmetric:
2x 1 y 2 2z 1
6
5
2
32, 5 32, 4 2
92, 132, 6
, or 9, 13, 12
14. (a) v 3 Point: 3, 5, 4
Parametric equations: x 3 9t, y 5 13t, z 4 12t
(b) Symmetric equations:
x3 y5 z4
9
13
12
z
15.
16.
z
3
6
−3
2
−3
−2
−1
1
2
3
1
−6
(0, 2, 1)
2
−4
2
1
2
−6
4
(5, 1, 5)
−2
4
3
6
y
x
2
4
6
y
x
17. ax x1 b y y1 cz z1 0
18. ax x0 b y y0 cz z0 0
1x 2 0 y 1 0z 2 0
0x 1 0 y 0 1z 3 0
x20
z30
19. 2x 5 1 y 6 2z 3 0
20. 0x 0 3 y 0 5z 0 0
2x y 2z 10 0
3y 5z 0
Section 11.4
Lines and Planes in Space
1051
21. n 1, 2, 1 ⇒ 1x 2 2 y 0 1z 0 0
x 2y z 2 0
23. u 1 0, 2 0, 3 0 1, 2, 3
22. n 1, 1,2
1x 0 1 y 0 2z 6 0
v 2 0, 3 0, 3 0 2, 3, 3
x y 2z 12 0
i
1
nuv
2
j
2
3
k
3 3, 9, 7
3
3x 0 9 y 0 7z 0 0
3x 9y 7z 0
3x 9y 7z 0
24. u 2, 6, 2, v 3, 3, 0
uv
i
2
3
j
6
3
k
2 6, 6, 24
0
n 1, 1, 4
25. u 3 2, 4 3, 2 2 1, 1, 4
v 1 2, 1 3, 0 2 1, 4, 2
nuv
i
1
1
j
1
4
k
4 18, 6, 3
2
Plane: 1x 4 1 y 1 4z 3 0
18x 2 6 y 3 3z 2 0
x y 4z 7 0
18x 6y 3z 24 0
6x 2y z 8 0
26. u 4, 0, 2, v 1, 2, 5
i
uv 4
1
j
0
2
27. n j: 0x 2 1 y 5 0z 3 0
y50
k
2 4, 22, 8
5
n 2, 11, 4
Plane: 2x 1 11 y 1 4z 2 0
2x 11y 4z 5 0
28. 1 2, 1 2, 1 1 3, 1, 2 and 2, 3, 1 are parallel to plane.
i
n 3
2
j
1
3
k
2 7, 1, 11
1
7x 2 1 y 2 11z 1 0
7x y 11z 5 0
29. n1 5, 3, 1, n2 1, 4, 7
n1
n2 5 12 7 0; orthogonal
31. n1 2, 0, 1, n2 4, 1, 8
n1
n2 8 8 0;
orthogonal
30. n1 3, 1, 4, n2 9, 3, 12
3n1 9, 3, 12 n2 ⇒ parallel planes
32. n1 1, 5, 1
n2 5, 25, 5 5n1 ⇒ parallel
1052
Chapter 11
Analytic Geometry in Three Dimensions
33. (a) n1 3, 4, 5, n2 1, 1, 1; normal vectors to planes
cos n1 n2 6
n1 n2
503
6
150
(b) 3x 4y 5z 6
Equation 1
xyz2
Equation 2
⇒ 60.67
3 times Equation 2 added to Equation 1 gives
7y 8z 0
8
y z.
7
8
1
Substituting back into Equation 2, x 2 y z 2 z z 2 z.
7
7
Letting t z7, we obtain x 2 t, y 8t, z 7t.
34. (a) n1 1, 3, 1, n2 2, 0, 5
cos n1 n2 n1 n2
7
1129
7
⇒ 66.93
319
1
(b) 2x 5z 3 0 ⇒ x 5z 3
2
1
3
1
1
1
⇒ y z
Then 3y x z 2 5z 3 z 2 z 2
2
2
2
6
5
3
1
1
Let z t. Parametric equations: x t , y t , z t
2
2
2
6
1
3
or equivalently, let z 2t and you obtain x 5t , y t , z 2t.
2
6
35. (a) n1 1, 1, 1, n2 2, 5, 1; normal vectors to planes
cos (b)
n1 n2 2
n1 n2
330
xyz0
Equation 1
2x 5y z 1
Equation 2
2
90
⇒ 77.83
2 times Equation 1 added to Equation 2 gives
7y z 1
y
z1
.
7
Substituting back into Equation 1, x z y z Letting z t, x 6t 1
t1
,y
.
7
7
z 1 6z 1 1
6z 1.
7
7
7 7
Equivalently, let y t, z 7t 1 and x 6t 1.
36. The planes are parallel because n1 2, 4, 2 is a multiple of n2 3, 6, 3.
The planes do not intersect.
Section 11.4
38. 2x y 4z 4
37. x 2y 3z 6
6
5
4
3
(0, 0, 2)
−5
z
6
5
4
3
(0, − 4, 0)
6
(0, 3, 0)
2
2
4 5
−4
−2
(2, 0, 0)
y
5
6
−2
−1
6
(2, 0, 0)
(0, 0, 5)
3
5
2
2
5
4
3
z
−1
−2
2
−2
2
(0, 3, 0)
4
5
6
−6
−7
3
−2
6
y
x
(0, 5, 0) 6
4
6
x
−2
−1
y
42. x 3z 6
z
z
−1
2
3
3
4
5
4 (0, 2, 0)
6
5
x
41. 3x 2y z 6
3
2
−2
−1
−1
(4, 0, 0)
y
3
6
x
40. y z 5
6
−2
4
(6, 0, 0)
−2
2
2
3
6
4
3
2
(0, 0, 1)
−2
3
1053
39. x 2y 4
z
z
x
Lines and Planes in Space
(6, 0, 0)
−4
(0, 0, −6)
(0, 0, − 2)
y
−6
y
x
PQ n
D
\
43.
44. P 4, 0, 0 on plane, Q 3, 2, 1, n 1, 1, 2
n
\
PQ 1, 2, 1
P 1, 0, 0 on plane, Q 0, 0, 0,
PQ n 1
D
\
\
n 8, 4, 1, PQ 1, 0, 0
n
1, 0, 0 8, 4, 1 8 8
D
64 16 1
81
6
1
6
6
6
9
PQ n
D
\
45.
n
P 2, 0, 0 on plane, Q 4, 2, 2,
\
n 2, 1, 1, PQ 2, 2, 2
D
2, 2, 2 2, 1, 1 6
4
26
3
6
47. (a) z 0.81x 0.36y 0.2
46. P 6, 0, 0 on plane, Q 1, 2, 5,
→
PQ 7, 2, 5 , n 2, 3, 1
→
PQ n
3
3
314
D
n
14
14
14
(b) The approximations are very similar to the actual
values of z.
Year
x
y
z (Actual)
z (Model)
1999
6.2
7.3
7.8
7.85
2000
6.1
7.1
7.7
7.70
2001
5.9
7.0
7.4
7.50
2002
5.7
7.0
7.3
7.34
2003
5.6
6.9
7.2
7.22
(c) If the consumption of the two types of milk
increases (or decreases), so does the consumption
of the third type of milk.
1054
Chapter 11
Analytic Geometry in Three Dimensions
48. The plane containing P6, 0, 0, S0, 0, 0, T1, 1, 8 has normal vector
6, 0, 0
1, 1, 8 or n1 0, 8, 1.
i
6
1
j
0
1
k
0 0, 48, 6
8
The plane containing P6, 0, 0, Q6, 6, 0, and R7, 7, 8 has normal vector
0, 6, 0
i
1, 1, 8 0
1
or n2 8, 0, 1.
j
6
1
k
0 48, 0, 6,
8
The angle between two adjacent sides is given by
cos n1 n2 n1 n2
1
6565
1
⇒ 89.12.
65
49. False. They might be skew lines, such as:
50. True
L1: x t, y 0, z 0 (x-axis)
and L2: x 0, y t, z 1
51. The lines are parallel:
32 10, 18, 20 15, 27, 30
52. (a) Sphere: x 42 y 12 z 12 4
(b) Two planes parallel to given plane. Let Q x, y, z be a point on one of these planes, and pick P 0, 0, 10
on the given plane. By the distance formula,
→
PQ n
x, y, z 10 4, 3, 1
2
26
n
± 226 4x 3y z 10
4x 3y z 10 ± 226 (Two planes parallel to given plane)
53. x2 y 2 102 100
54. 3
y
⇒ tan 1 ⇒ y x (line)
4
x
r 3 cos 55.
r2 3r cos x2 y2 3x
56. r 1
⇒ 2r r cos 1 ⇒ 2x2 y2 x 1
2 cos ⇒ 2x2 y2 x 1 ⇒ 4x2 y2 x2 2x 1 ⇒ 3x2 4y2 2x 1
57. r2 49
r7
Review Exercises for Chapter 11
58. x2 y2 4x 0
1055
y5
59.
r sin 5
r2 4r cos 0
r 5 csc r 4 cos 0 ⇒ r 4 cos 2x y 1 0
60.
2r cos r sin 1
r2 cos sin 1
r
1
sin 2 cos Review Exercises for Chapter 11
2.
1. (a) and (b)
5 (0, 0, 5)
4
z
3
3
(−3, 3, 0)
2 −3
−4
2 −3
1
−5 −4
(5, −1, 2)
1
y
−2
2
3
4
x
1
3. 5, 4, 0
z
1
2
y
−4 −3 −2
3
2
3
−2
1
2
3
−2
4
−3
1
−3
x
−4
(2, 4, −3)
−5
5. d 5 42 2 02 1 72
4. y-axis ⇒ x z 0
0, 7, 0
1 4 36
41
6. d 2 12 3 32 4 02
9 36 16
61
7. d1 3 02 2 32 0 22 9 25 4 38
d2 0 02 5 32 3 22 4 25 29
d3 0 32 5 22 3 02 9 49 9 67
d12 d22 38 29 67 d32
8. d1 4 02 3 02 2 42 16 9 4 29
d2 4 42 5 32 5 22 4 9 13
d3 4 02 5 02 5 42 16 25 1 42
d 12 d 22 d 32 42
9. Midpoint:
8 5 2 6 3 7
13
,
,
, 2, 5
2
2
2
2
10. Midpoint:
7 2 1, 1 2 1, 42 2 4, 0, 1
1056
Chapter 11
Analytic Geometry in Three Dimensions
10 2 8, 6 2 2, 122 6 1, 2, 9
11. Midpoint:
12. Midpoint:
52 7, 32 9, 1 2 5 6, 6, 2
13. x 22 y 32 z 52 1
14. x 32 y 22 z 42 16
15. Radius: 6
16. Radius 15
2
x 12 y 52 z 22 36
x 2 y 42 z 12 225
4
17. x2 4x 4 y2 6y 9 z2 4 4 9
x 22 y 32 z2 9
Center: 2, 3, 0
Radius: 3
18. x2 10x 25 y2 6y 9 z2 4z 4 34 25 9 4
x 52 y 32 z 22 4
Center: 5, 3, 2
Radius: 2
19. (a) xz-trace y 0: x2 z2 7, circle
(b) yz-trace x 0: y 32 z2 16, circle
z
z
(y − 3) 2 + z 2 = 16
x2 + z2 = 7
4
4
2
−2
−2
(0, 3, 0)
2
4
x
2
−4
−2
2
4
4
x
6
2
4
6
y
y
(0, 3, 0)
20. (a) xy-trace z 0: x 22 y 12 9 circle
z
(b) yz-trace x 0: 4 y 12 z2 9
y 12 z2 5 circle
(x + 2) 2 + (y − 1) 2 = 9
z
4
(y − 1) 2 + z 2 = 5
4
2
(− 2, 1, 0)
x
4
2
6
y
x
6
y
(−2, 1, 0)
21. Initial point: 2, 1, 4
\
22. (a) PQ 3 2, 2 1, 3 2
5, 3, 1
Terminal point: 3, 3, 0
(b) PQ 35
(a) v 3 2, 3 1, 0 4
1, 4, 4
35
1
(c) Unit vector: 5, 3, 1
35 5, 3, 1
35
\
(b) v 12 42 42 33
(c) u v
1
4
4
,
,
v
33 33
33
Review Exercises for Chapter 11
1057
23. Initial point: 7, 4, 3
Terminal point: 3, 2, 10
(a) v 3 7, 2 4, 10 3
10, 6, 7
(b) v 102 62 72 185
10
6
7
v
(c) u v 185, 185, 185
\
24. (a) PQ 5 0, 8 3, 6 1
5, 11, 7
\
(b) PQ 195
195
1
5, 11, 7
5, 11, 7
(c) Unit vector: 195
195
25. u v 10 46 35 9
26. u v 82 45 22 0
27. u v 21 10 11 1
28. u v 21 13 22 5
29. Since u v 0, the angle is 90.
30. cos uv
12 5 2
1145
u v
15
1145
⇒ 47.61
2, 4, 12
31. Since 23 39, 12, 21
26, 8, 14
, the
vectors are parallel.
32. u v 8, 5, 8
33. First two points: u 3, 4, 1
34. First two points: 1, 5, 4
16 20 4 0
Orthogonal
Last two points: v 0, 2, 6
Last two points: 2, 10, 8
Since u cv, the points are not collinear.
Since, 2, 10, 8
21, 5, 4
, the 3 points
are collinear.
35. Let a, b, and c be the three force vectors determined by A0, 10, 10, B4, 6, 10 and C4, 6, 10.
a a
0, 10, 10
1
1
a 0,
,
10 2
2
2
b b
c c
4, 6, 10
2 3
5
b
,
,
152
38 38 38
4, 6, 10
2
3
5
c
,
,
152
38 38 38
Must have a b c 300k. Thus:
2
38
1
2
1
2
a a 3
38
5
38
—CONTINUED—
b b b 2
38
3
38
5
38
c 0
c 0
c 300.
1058
Chapter 11
Analytic Geometry in Three Dimensions
35. —CONTINUED—
From the first equation b c. From the second equation,
From the third equation,
1
10
a 300 b. Thus,
2
38
6
10
b 300 b
38
38
Finally, a 2
1
6
a b.
2
38
16
7538
b 300 and b c 115.58.
4
38
⇒
63875 4 38 2252
2
159.10.
36. Let a, b, c be the three force vectors determined by A0, 10, 10, B4, 6, 10 and C4, 6, 10.
1
a a 0, 10, 10
102 a 0,
,
1
2 2
238 , 338 , 538
b b 4, 6, 10
152 b
238 , 338, 538
c c 4, 6, 10
152 c
We must have a b c 200k. Thus,
2
38
1
2
1
2
a a 3
38
5
38
b b b 2
38
3
38
5
38
c 0
c 0
c 200
Solving this system, a 106.1, b c 77.1.
Thus, the tensions are 106.1, 77.1 and 77.1 pounds.
i
37. u v 2
1
39.
j
8
1
i
j
2
u v 3
10 15
1
7602
71, 44, 25
i
40. u v 0
1
j
0
0
i
38. u v 10
5
k
5 71, 44, 25
2
u v 7602
Unit vector:
k
2 10, 0, 10
1
k
4 4j ⇒ unit vector: j 0, 1, 0
12
j
15
3
k
5 15, 25, 105
0
Review Exercises for Chapter 11
41. First two points: 3, 2, 3
42. u 1, 0, 1
, v 1, 0, 1
opposite sides parallel
and equal length.
Last two points: 3, 2, 3
First and third points: 2, 2, 0
i
3
2
j
2
2
k
3 6, 6, 10
0
1059
Area 6, 6, 10
36 36 100
Adjacent sides: u 1, 0, 1
, w 0, 2, 0
i
uw 1
0
j
0
2
k
1 2, 0, 2
0
Area u w 4 4 22 square units
172 243 square units
43. The parallelogram is determined by the three
vectors with initial point 0, 0, 0.
u 3, 0, 0
, v 2, 0, 5
, w 0, 5, 1
3
u v w 2
0
0
0
5
0
5 75
1
44. u 2, 0. 0
, v 0, 4, 0
, w 0, 0, 6
2
u v w 0
0
Volume u v
0
4
0
0
0 48
6
w 48 cubic units
Volume 75 75 cubic units
45. v 3 1, 6 3, 1 5
4, 3, 6
, point: 1, 3, 5
(a) Parametric equations: x 1 4t, y 3 3t, z 5 6t
(b) Symmetric equations:
x1 y3 z5
4
3
6
47. Use 2v 4, 5, 2
, point: 0, 0, 0.
46. (a) v 5, 20, 3
(b)
x 5t, y 10 20t, z 3 3t
(a) Parametric equations: x 4t, y 5t, z 2t
y 10 z 3
x
5
20
3
(b) Symmetric equations:
49. u 5, 0, 2
, v 2, 3, 8
48. (a) v 1, 1, 1
x 3 t, y 2 t, z 1 t
(b)
x
y
z
4 5 2
x3 y2 z1
or
1
1
1
x3y2z1
i
uv 5
2
j
0
3
k
2 6, 36, 15
8
n 2, 12, 5
ax x0 b y y0 cz z0 0
2x 0 12 y 0 5z 0 0
2x 12y 5z 0
50. u 5, 5, 2
, v 3, 5, 2
i
nuv 5
3
j
5
5
k
2 0, 16, 40
2
Plane: 0x 1 16 y 3 40z 4 0
2 y 3 5z 4 0
2y 5z 14 0
51. n k, normal vector
Plane: 0x 5 0 y 3 1 z 2 0
z20
1060
Chapter 11
Analytic Geometry in Three Dimensions
52. n 1, 1, 2
, point: 0, 0, 6
53. 3x 2y 3z 6
1x 0 1 y 0 2z 6 0
z
x y 2z 12 0
(0, 0, 2)
(0, − 3, 0)
1
x y 2z 12 0
1
(2, 0, 0)
2
3
1
y
−2
x
55. 2x 3z 6
54. 5x y 5z 5
56. 4y 3z 12
z
z
z
−2
2
(0, − 5, 0)
4
−4
2
1
−4
−2
−2
(1, 0, 0)
3
2
4
4
1
(0, 0, − 1)
2
(0, 0, −2) 3
x
2
y
−2
3
x
4
y
(0, 3, 0)
−1
1
−1
1
−1
2
1
(3, 0, 0)
1
−1
2
4
y
−2
−3
(0, 0, −4)
x
\
57. n 2, 20, 6
, P 0, 0, 1 in plane, Q 2, 3, 10, PQ 2, 3, 9
PQ n 2
D
\
n
440
110
1
0.0953
110
110
→
58. D PQ n
n
→
Q 1, 2, 3, P 2, 0, 0 on plane. PQ 1, 2, 3
, n 2, 1, 1
D
1, 2, 3
2, 1, 1
6
1
6
6
6
\
59. n 1, 10, 3
, P 2, 0, 0 in plane, Q 0, 0, 0, PQ 2, 0, 0
PQ n
D
\
n
2
1 100 9
2
2110 110
0.191
110
55
110
→
60. D PQ n
n
→
Q 0, 0, 0, P 0, 0, 12 on plane. PQ 0, 0, 12
, n 2, 3, 1
D
0, 0, 12
2, 3, 1
14
61. False. a b b a
12
614
7
14
62. True. See page 831.
63. u u 3, 2, 1
941
14
u2
3, 2, 1
Problem Solving for Chapter 11
i
64. u v 3
2
j
2
4
k
1 10, 11, 8
3
i
vu 2
3
j
4
2
k
3 10, 11, 8
1
Thus, u v v
66. u v w u u.
uw
i
3
1
j
2
2
1, 2, 1 6
u v u w 11 5 6
i
1, 2, 1 3
1
u v 10, 11, 8 (Exercise 64)
65. u v w 3, 2, 1
j
2
2
k
1 4, 4, 4
1
k
1 6, 7, 4
2
u v u w 10, 11, 8 6, 7, 4 4, 4, 4
u v w
Problem Solving for Chapter 11
1. (a)
(c) w 1, 2, 1 a1, 1, 0 b0, 1, 1
z
1a
3
2
−3
1
−2
−1
2
−2
2ab
−3
1b
v
u 1
2
3
3
y
Hence, a b 1.
(d) w 1, 2, 3 a1, 1, 0 b0, 1, 1
x
(b) w au bv a1, 1, 0 b0, 1, 1
1a
0 a, a b, b ⇒ a b 0
2ab
3b
Impossible
2. This set is a sphere:
x x12 y y12 z z12 16
4. (a) u v 4, 7.5, 2
(b) u v 8.7321
(c) u 26 5.0990
(d) v 9.0139
3. Programs will vary. See online website.
1061
1062
Chapter 11
Analytic Geometry in Three Dimensions
5. The largest angle in a triangle is always opposite the longest side of the triangle. First, determine the lengths of
uv
the three sides. Then, once the largest angle has been identified, use the fact that cos , where u and v
u v
are defined to be the vectors that form . If u v 0, the angle is a right angle. If u v > 0, the angle is acute.
If u v < 0, the angle is obtuse.
(a) A: 1, 2, 0
B: 0, 0, 0
C: 2, 1, 0
dAB 5, dAC 10, dBC 5
Angle B is largest.
\
\
BA 1, 2, 0, BC 2, 1, 0
\
BA
BC 0
\
⇒ The triangle is a right triangle.
(b) A: 3, 0, 0
B: 0, 0, 0
C: 1, 2, 3
dAB 3, dAC 29, dBC 14
Angle B is largest.
\
\
BA 3, 0, 0, BC 1, 2, 3
\
BA
BC 3
\
< 0 ⇒ The triangle is an obtuse triangle.
(c) A: 2, 3, 4
B: 0, 1, 2
C: 1, 2, 0
dAB 24, dAC 50, dBC 6
Angle B is largest.
\
\
BA 2, 4, 2, BC 1, 1, 2
\
BA
BC 10
\
< 0 ⇒ The triangle is an obtuse triangle.
(d) A: 2, 7, 3
B: 1, 5, 8
C: 4, 6, 1
dAB 178, dAC 189, dBC 107
Angle B is largest.
\
\
BA 3, 12, 5, BC 5, 1, 9
\
BA
BC 48
\
⇒ The triangle is an acute triangle.
Problem Solving for Chapter 11
\
6. PQ 1 0 0, 1 0, 0 4 0, 1, 4
\
PQ 2 \
23 0, 21 0, 0 4
23, 12, 4
PQ 3 3
2
0,
3 1
1
0, 0 4 , , 4
2
2 2
\
\
\
Note that PQ 1 PQ 2 PQ 3 17.
z
Unit vectors:
P(0, 0, 4)
1
u1 0, 1, 4
17
u2 1
17
(
Q3 −
3, 1,0
2 2
)
23, 12 , 4
Q 1(0, − 1, 0)
3 1
1
u3 , , 4
2 2
17
y
Q2
(
3, 1,0
2 2
)
x
The unit vectors u1, u2, u3 give the directions of the force in each leg. Since the
legs are the same length, the total weight is distributed equally among the legs. So,
F1 120
40
u1 0, 1, 4
3
17
F2 120
40 3 1
u , , 4
3 2 17 2 2
F3 3 1
120
40
u , , 4
3 3 17
2 2
7. Let A lie on the y-axis and the wall on the x-axis. Then, A 0, 10, 0, B 8, 0, 6, C 10, 0, 6 and
\
\
AB 8, 10, 6, AC 10,10, 6.
\
AB 82 102 6 2 102
\
AC 102 102 62 259
\
AB
420
84
F1 420
8, 10, 6 4, 5, 3
2
AB 102
\
\
F2 650
AC
650
650
10, 10, 6 5, 5, 3
59
AC 259
\
F F1 F2 4842 559650, 5284 559650, 3842 3650
59
185.526, 720.099, 432.059
F 860.0 lb
8. Note that cos uv
. Therefore,
u v
u v sin u v 1 cos2 u v
1
u v2
u2v2 u v2
u2v2
u12 u22 u32v12 v22 v32 u1v1 u2v2 u3v32
u2v3 u3v22 u1v3 u3v12 u1v2 u2v12
u v sin u v
Since u and v are orthogonal, 90 and sin 1. So, u v u v, u, v orthogonal.
1063
1064
Chapter 11
Analytic Geometry in Three Dimensions
9. u a1, b1, c1, v a2, b2, c2, w a3, b3, c3 v
i
w a2
a3
j
b2
b3
k
c2 b2c3 b3c2 i a2c3 a3c2 j a2b3 a3b2 k
c3
i
u v w a1
b2c3 b3c2 u v
j
b1
a3c2 a2c3 k
c1
a2b3 a3b2 w b1a2b3 a3b2 c1a3c2 a2c3 i a1a2b3 a3b2 c1b2c3 b3c2 j
a1a3c2 a2c3 b1b2c3 b3c2 k
a2a1a3 b1b3 c1c3 a3a1a2 b1b2 c1c2 i
b2a1a3 b1b3 c1c3 b3a1a2 b1b2 c1c2 j
c2a1a3 b1b3 c1c3 c3a1a2 b1b2 c1c2 k
10.
u wv u vw
u1
v1
w1
u2
v2
w2
u3
v
v3 u1 2
w2
w3
u1
v2
w2
u1i v3
v
u2 1
w3
w1
v3
v
i i u2 1
w3
w1
v2
w2
11.
j
v2
w2
v2
w2
v2
w2
v
v2
k k
w2
v3
j u3k w3
v3
v
i 1
w3
w1
k
v3 u v
w3
v3w
v3
v
j j u3 1
w3
w1
v3
v
i u2 j 1
w3
w1
u1i u2j u3k i
u v1
w1
v3
v
u3 1
w3
w1
v1
w1
v2
k
w2
v3
v
j 1
w3
w1
v2
k
w2
w
w area of base and projv wu height of parallelepiped
Therefore, the volume is
V heightarea of base projv wu v
w
u v w
v
v w
w
u
u v
w
w .
v
|| proj v 3 w u ||
(a) O 0, 0, 0
12.
A 18 cos 30, 18 sin 30, 0 93, 9, 0
A
OA 93, 9, 0
\
18 in.
θ
30°
O
F
\
M OA
\
F OA F sin 1440
932 92 02 60 sin 1080 sin 0
180
0
—CONTINUED—
Problem Solving for Chapter 11
1065
12. — CONTINUED—
22 540
\
(b) M 1080 sin45 1080
2 763.7 in-lb
\
(c) M 1080 sin has its maximum value at 90. In order to generate the maximum torque,
the force should be applied in a direction perpendicular to the wrench handle.
\
13. (a) In inches: AB 15j 12k
400
5
In feet: AB j k
4
\
0
F 200cos j sin k
\
(b) AB
i
j
F 0
AB
5
4
1
200 cos 0
\
k
200 sin −300
250 sin 200 cos i
F 250 sin 200 cos 25 10 sin 8 cos \
(c) When 30: AB
F 25 10
180
12 8 23 255 43 298.2
(d) From the graph we see that the maximum value occurs when 51.34.
\
(e) From the graph we see that the zero occurs when 141.34, the angle making AB parallel to F.
14. The area of the triangle is one-half of the area of any of the 3 parallelograms having the following adjacent sides:
b and c, b and a, c and a
B
So,
b c a c a
Area 2
2
a
c
b 2
A
C
b c a c a b b
b c sin A a c sin B a b sin C
Divide by a b c:
sin A sin B sin C
a
b
c
15. First insect: x 6 t, y 8 t, z 3 t
Second insect: x 1 t, y 2 t, z 2t
(a) When t 0 the first insect is located at 6, 8, 3 and the second insect is
located at 1, 2, 0.
d 1 62 2 82 0 32 70 inches
20
(b) d 1 t 6 t2 2 t 8 t2 2t 3 t2
52 2t 62 t 32
−1
5t 2 30t 70
11
0
t
0
1
2
3
4
5
6
7
8
9
10
d
70
45
30
5
30
45
70
105
150
205
270
—CONTINUED—
1066
Chapter 11
Analytic Geometry in Three Dimensions
15. —CONTINUED—
(c) The distance between the two insects appears to lessen in the first 3 seconds, but then begins to
increase with time.
(d) When t 3, the insects get within 5 inches of each other.
16. (a) x 2 4t, y 3, z 1 t ⇒ direction vector u 4, 0, 1
Let P be the point on the line with t 0:
P 2 4
\
0, 3, 1 0 2, 3, 1, Q 1, 5, 2
PQ 1 2, 5 3, 2 1 3, 2, 3
\
PQ
i
u 3
4
j
2
0
k
2
3 0
1
3
3
i
1
4
3
3
j
1
4
2
k
0
2i 9j 8k 2, 9, 8
\
PQ
u 22 92 82 149
u 42 02 12 17
\
PQ u 149 2533
D
2.9605
u
17
17
(b) x 2t, y 3 t, z 2 2t ⇒ direction vector u 2, 1, 2
Let P be the point on the line with t 0:
P 2
\
0, 3 0, 2 2 0 0, 3, 2, Q 1, 2, 4
PQ 1 0, 2 3, 4 2 1, 1, 2
\
PQ
i
u 1
2
j
1
1
k
1
2 1
2
2
1
i
2
2
2
1
j
2
2
1
k
1
0i 2j k 0, 2, 1
\
PQ
u 02 22 12 5
u 22 12 22 9 3
\
PQ u 5
0.7454
D
u
3
17. (a) u 0, 1, 1 direction vector of line determined by P1 and P2
z
6
5
4
3 P
2
\
D
P1Q u 2, 0, 1 0, 1, 1
u
2
P1
1, 2, 2
3
32
2
2
2
(b) The shortest distance to the line segment is P1Q 2, 0, 1 5.
4
x
3
2
1
Q
2 3
4
y
Problem Solving for Chapter 11
1
1
18. (a) x t 3, y t 1, z 2t 1 ⇒ direction vector u 1, , 2
2
2
Let Q 4, 3, s.
Let P be the point on the line corresponding to t 0: P 3, 1, 1
\
PQ 4 3, 3 1, s 1 1, 2, s 1
\
PQ
u
i
j
1
2
1
2
1
\
u s1
2
s1 1
2
2
2
i
1
1
7s
5
7s
5
i s 3j k , s 3,
2
2
2
2
PQ
k
u 7s
2
1 12
2
2
2
s 32 22 5
2
2
1
s1
j
2
1
5s2 10s 110
2
21
2
\
D
PQ u 5s2 10s 110 105s2 210s 2310
u
21
21
(b)
10
215
15
210
Minimum distance is D 2.23607 at s 1.
(c) D 105s2 210s 2310
21
As s approaches very large (very positive) or very small (very negative) values,
the expression under the radical is dominated by the term 105s 2:
105s 2 210s 2310 105s2, s “large”
So, at large values of s ,
D
105s2
21
±
105
Asymptotes: D ±
21
s.
105
21
s 1
2
1 k
2
1067
1068
Chapter 11
Chapter 11
Analytic Geometry in Three Dimensions
Practice Test
1. Find the lengths of the sides of the triangle with vertices 0, 0, 0, 1, 2, 4, and 0, 2, 1.
Show that the triangle is a right triangle.
2. Find the standard form of the equation of a sphere having center 0, 4, 1 and radius 5.
3. Find the center and radius of the sphere x2 y2 z2 2x 4z 11 0.
4. Find the vector u 3v given u 1, 0, 1 and v 4, 3, 6.
5. Find the length of 12v if v 2, 4, 6.
6. Find the dot product of u 2, 1, 3 and v 1, 1, 2.
7. Determine whether u 1, 1, 1 and v 3, 3, 3 are orthogonal, parallel, or neither.
8. Find the cross product of u 1, 0, 2 and v 1, 1, 3. What is v
u?
9. Use the triple scalar product to find the volume of the parallelepiped having adjacent edges u 1, 1, 1,
v 0, 1, 1, and w 1, 0, 4.
10. Find a set of parametric equations for the line through the points 0, 3, 3 and 2, 3, 4.
11. Find an equation of the plane passing through 1, 2, 3 and perpendicular to the vector n 1, 1, 0.
12. Find an equation of the plane passing through the three points A 0, 0, 0, B 1, 1, 1, and C 1, 2, 3.
13. Determine whether the planes x y z 12 and 3x 4y z 9 are parallel, orthogonal or neither.
14. Find the distance between the point 1, 1, 1 and the plane x 2y z 6.