CALCULUS III FALL 1999 HOMEWORK 12 – ANSWERS

§11.4 Questions 6,24,28,30; §11.5 Questions 8,16,34,38,54,64

6.

a × b = | \begin{matrix}i & j & k \\ 1 & 2 & −1 \\ 3 & −1 & 7 \end{matrix} | = 13i − 10j − 7k.

24.

(a) The plane is parallel to the vectors 4i + 9j + 3k and 7i + 5j + 6k. So the required orthogonal vector is

(4i + 9j + 3k) × (7i + 5j + 6k) = | \begin{matrix}i & j & k \\ 4 & 9 & 3 \\ 7 & 5 & 6 \end{matrix} | = 39i − 3j − 43k.

(b) The area of the triangle is \frac{1}{2}|39i − 3j − 43k| = \frac{1}{2} \sqrt{3379}.

28. The parallelepiped is generated by the vectors 2i + 3j + 3k, −i − j − k and 6i − 2j + 2k. So the volume is

|| \begin{matrix}2 & 3 & 3 \\ −1 & −1 & −1 \\ 6 & −2 & 2 \end{matrix} || = 4.

CALCULUS III  HOMEWORK 12 – ANSWERS  FALL 1999 2.

30. These four points can be used to form a parallelepiped, as in the previous question. The points are coplanar (lie in a single plane) precisely when this parallelepiped has zero volume. The parallelepiped is generated by the vectors i + 4j + 5k, 2i − j + k and 5i + 2j + 7k. The volume of the parallelepiped is therefore

|| \begin{matrix}1 & 4 & 5 \\ 2 & −1 & 1 \\ 5 & 2 & 7 \end{matrix} || = 0, as required.

8. The vector v = 4i−3j−(1/2)k is parallel to the line. So the vector equation of the line can be written as

r = −i + 4j + k + t(4i − 3j − (1/2)k).

It follows that the parametric eautions are

x = −1 + 4t,  y = 4 − 3t,  z = 1 − \frac{t}{2},

and the symmetric equations are

\frac{x = 1}{4} = −\frac{y − 4}{3} = −(2z − 2).

16. The parametric equations of the lines are

x = 1 + 2t, y = t, z = 1 + 4t  and  x = s, y = −2 + 2s, z = −2 + 3s.

Note that putting t = 0 and s = 1 yields the point of intersection (1, 0, 1).

34. Note that v = 5i + j − k is parallel to the line. Also (0, 1, 0) is a point on the line. Hence the plane is also parallel to the vector w = −i − j + k. Consequently v × w = −4j − 4k is orthogonal to the plane. It follows that the equation of the plane is y + z = 1.

38. The appropriate parameter value is given by the equation

1 + t = 1 − 2 + 2t + t = −1 + 3t  and so  t = 1.

Consequently the point of intersection is (0, 1, 2).

54. The normals to the given planes are

a = i − k  b = j + 2k  c = i + j − 2k.

The required plane contains a × b = i − 2j + k and c. So the normal to the plane is (i − 2j + k) × c = 3i + 3j + 3k. Finally note that (1, 3, 0) is a point of the plane and so the equation of the plane is x + y + z = 4.

64. The required distance is

\frac{|12 + 12 + 7 − 5|}{\sqrt{16 + 36 + 1}} = \frac{26}{\sqrt{53}}.