Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 13 REVIEW ET CHAPTER 12 ¤ 131
17. (a) See Exercise 13.4.43 [ ET 12.4.43].
(b) See Example 8 in Section 13.5 [ ET 12.5].
(c) See Example 10 in Section 13.5 [ ET 12.5].
18. The traces of a surface are the curves of intersection of the surface with planes parallel to the coordinate planes. We can find
the trace in the plane x = k (parallel to the yz-plane) by setting x = k and determining the curve represented by the resulting
equation. Traces in the planes y = k (parallel to the xz-plane) and z = k (parallel to the xy-plane) are found similarly.
19. SeeTable1inSection13.6[ET12.6].
1. True, by Theorem 13.3.2 [ ET 12.3.2], property 2.
3. True. If θ istheanglebetweenu and v, then by Theorem 13.4.6 [ ET 12.4.6],
|u × v| = |u||v| sin θ = |v||u| sin θ = |v × u|.
(Or, by Theorem 13.4.8 [ ET 12.4.8], |u × v| = |−v × u| = |−1||v × u| = |v × u|.)
5. Theorem 13.4.8 [ ET 12.4.8], property 2 tells us that this is true.
7. This is true by Theorem 13.4.8 [ ET 12.4.8], property 5.
9. This is true because u × v is orthogonal to u (see Theorem 13.4.5 [ ET 12.4.5]), and the dot product of two orthogonal vectors
is 0.
11. If |u| =1, |v| =1and θ is the angle between these two vectors (so 0 ≤ θ ≤ π), then by Theorem 13.4.6 [ ET 12.4.6],
|u × v| = |u||v| sin θ =sinθ, which is equal to 1 ifandonlyifθ = π (that is, if and only if the two vectors are orthogonal).
2
Therefore, the assertion that the cross product of two unit vectors is a unit vector is false.
13. This is false. In R 2 , x 2 + y 2 =1represents a circle, but (x, y, z) | x 2 + y 2 =1 represents a three-dimensional surface,
namely, a circular cylinder with axis the z-axis.
15. False. For example, i · j = 0 but i 6=0 and j 6=0.
17. This is true. If u and v are both nonzero, then by (7) in Section 13.3 [ET 12.3], u · v =0implies that u and v are orthogonal.
But u × v = 0 implies that u and v are parallel (see Corollary 13.4.7 [ET 12.4.7]). Two nonzero vectors can’t be both parallel
and orthogonal, so at least one of u, v must be 0.
1. (a) The radius of the sphere is the distance between the points (−1, 2, 1) and (6, −2, 3), namely,
[6 − (−1)]2 +(−2 − 2) 2 +(3− 1) 2 = √ 69. By the formula for an equation of a sphere (see page 804 [ET 768]),
an equation of the sphere with center (−1, 2, 1) and radius √ 69 is (x +1) 2 +(y − 2) 2 +(z − 1) 2 =69.