Thomas Calculus 13th [Solutions]

880 Chapter 12 Vectors and the Geometry of Space
58.
2 2 2 2 2 2 2 2 2 2 2 1 2
3x 3y 3z 2y 2z 9 x y y z z 3 x y y z 2 z 1 3 2
3 3 3 9 3 9 9
2 2 2
2 1 1 29
( x 0) y z the center is at 1 1
29
0, , and the radius is
3 3 3
3 3
3
59. (a) the distance between ( x, y, z ) and ( x , 0, 0) is
(b) the distance between ( x, y, z ) and (0, y , 0) is
(c) the distance between ( x, y, z ) and (0, 0, z ) is
2 2
y z
2 2
x z
2 2
x y
60. (a) the distance between ( x, y, z ) and ( x, y , 0) is z
(b) the distance between ( x, y, z ) and (0, y, z ) is x
(c) the distance between ( x, y, z ) and ( x, 0, z ) is y
61.
AB
BC
2 2 2
1 ( 1) ( 1 2) (3 1) 4 9 4 17
2
2 2
(3 1) 4 ( 1) (5 3) 4 25 4 33
CA
2 2 2
( 1 3) (2 4) (1 5) 16 4 16 36 6
Thus the perimeter of triangle ABC is 17 33 6.
62.
PA
2 2 2
(2 3) ( 1 1) (3 2) 1 4 1 6
2 2 2
PB (4 3) (3 1) (1 2) 1 4 1 6
Thus P is equidistant from A and B.
63.
( x
2
x) y ( 1)
2
( z
2
z) ( x
2
x) ( y
2
3) ( z
2
z) ( y
2
1) ( y
2
3) 2y 1 6y
9
y 1
64.
2 2 2 2 2 2 2 2 2 2
( x 0) ( y 0) ( z 2) ( x x) ( y y) ( z 0) x y ( z 2) z
2 2
2 y
x y 4z 4 0 z x 1
4 4
2
65. (a) Since the entire sphere is below the xy -plane, the point on the sphere closest to the xy -plane is the point
at the top of the sphere, which occurs when x 0 and
2 2 2
y 3 0 (3 3) ( z 5) 4 z 5 2
z 3 (0, 3, 3).
(b) Both the center (0, 3, 5) and the point (0, 7, 5) lie in the plane z 5, so the point on the sphere
closest to (0, 7, 5) should also be in the same plane. In fact it should lie on the line segment between
2 2 2
(0, 3, 5) and (0, 7, 5), thus the point occurs when x 0 and z 5 0 ( y 3) ( 5 5) 4
y 3 2 y 5 (0, 5, 5).
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