Thomas Calculus 13th [Solutions]

Chapter 12 Practice Exercises 915
27. The desired vector is n v or v n since n v is perpendicular to both n and v and, therefore, also parallel to
the plane.
28. If a 0 and b 0, then the line by c and i are parallel. If a 0 and b 0, then the line ax c and j are
parallel. If a and b are both 0, then ax by c contains the points c , 0
a
and 0, c b
the vector
ab c i c j
a b
c b i a j and the line are parallel. Therefore, the vector bi
aj is parallel to the line
ax by c in every case.
29. The line L passes through the point P (0, 0, 1) parallel to v i j k . With PS 2i 2j k and
PS
i j k
v 2 2 1 (2 1) i (2 1) j (2 2) k i 3j 4 k , we find the distance d
26 78
.
3 3
1 1 1
| PS v| 1 9 16
| v| 1 1 1
30. The line L passes through the point P (2, 2, 0) parallel to v i j k . With PS 2i 2j k and
PS
i j k
v 2 2 1 (2 1) i ( 2 1) j ( 2 2) k i 3j 4 k , we find the distance d
26 78
.
3 3
1 1 1
31. Parametric equations for the line are x 1 3 t, y 2, z 3 7 t.
| PS v| 1 9 16
| v| 1 1 1
32. The line is parallel to PQ 0i j k and contains the point P (1, 2, 0) parametric equations are
x 1, y 2 t,
z t for 0 t 1.
33. The point P (4, 0, 0) lies on the plane x y 4, and PS (6 4) i 0 j ( 6 0) k 2i 6k with n i j
d
| n PS| 2 0 0 2
| n| 1 1 0 2
2.
34. The point P (0, 0, 2) lies on the plane 2x 3y z 2, and PS (3 0) i (0 0) j (10 2) k 3i 8k with
| n PS| n 2i 3j k d
6 0 8 14 14.
| n| 4 9 1 14
35. P (3, 2, 1) and n 2 i j k (2)( x 3) (1) y ( 2) (1)( z 1) 0 2x y z 5
36. P ( 1, 6, 0) and n i 2j 3 k (1) x ( 1) ( 2)( y 6) (3)( z 0) 0 x 2y 3z
13
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