Thomas Calculus 13th [Solutions]

Section 12.5 Lines and Planes in Space 903
41. S(0, 1, 1), 4y 3z 12 and P (0, 3, 0) is on the plane PS 4j k and n 4j 3k
d PS n 16 3 19
| n| 16 9 5
42. S(2, 2, 3), 2x y 2z 4 and P (2, 0, 0) is on the plane PS 2j 3k and n 2i j 2k
d PS n 2 6 8
| n| 4 1 4 3
43. S(0, 1, 0), 2x y 2z 4 and P (2, 0, 0) is on the plane PS 2i j and n 2i j 2k
d PS n 4 1 0 5
| n| 4 1 4 3
44. S(1, 0, 1), 4x y z 4 and P ( 1, 0, 0) is on the plane PS 2i k and n 4i j k
8 1 9 3 2
d PS n | n| 16 1 1 18 2
45. The point P (1, 0, 0) is on the first plane and S (10, 0, 0) is a point on the second plane PS 9 i,
and n i 2j 6k is normal to the first plane the distance from S to the first plane is
d PS n 9 9
| | , which is also the distance between the planes.
n 1 4 36 41
46. The line is parallel to the plane since v n i j 1 k i 2j 6k 1 2 3 0. Also the point S(1, 0, 0)
2
when t 1 lies on the line, and the point P (10, 0, 0) lies on the plane PS 9 i . The distance from S to
the plane is d PS n 9 9
n
,
1 4 36 41
which is also the distance from the line to the plane.
1 1 2 1 1
47. n1
i j and 2 1 1
2 2 2 cos n n
n i j k
n n
cos cos
2 9 2 4
1 2
1 1 2 1 1
48. n1 5i j k and 5 2 3
2 2 3 cos n n
n i j k
n n
cos cos (0)
27 14
2
1 2
1 1 2 1 1
49. n1 2i 2j 2k and 4 4 2 1
2 2 2 cos n n
n i j k
n n
cos cos 1.76 rad
12 9 3 3
1 2
1 1 2 1
50. n1
i j k and 1
2 cos n n
n k
n n
cos 0.96 rad
3 1
1 2
1 1 2 1 1
51. n1 2i 2j k and 2 4 1
5
2 2 cos n n
n i j k
n n
cos cos 0.82 rad
9 6 3 6
1 2
1 1 2 1 1
52. n1 4j 3k and 8 18 26
2 3 2 6 cos n n
n i j k
n n
cos cos 25 49
35
0.73 rad
1 2
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