Solução_Calculo_Stewart_6e

F.
132 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ETTX.10
CHAPTER 12
(b) The intersection of this sphere with the yz-plane is the set of points on the sphere whose x-coordinate is 0. Putting x =0
into the equation, we have (y − 2) 2 +(z − 1) 2 =68,x=0which represents a circle in the yz-plane with center (0, 2, 1)
and radius √ 68.
(c) Completing squares gives (x − 4) 2 +(y +1) 2 +(z +3) 2 = −1+16+1+9=25. Thus the sphere is centered at
(4, −1, −3) and has radius 5.
3. u · v = |u||v| cos 45 ◦ =(2)(3) √ 2
2 =3√ 2. |u × v| = |u||v| sin 45 ◦ =(2)(3) √ 2
2 =3√ 2.
By the right-hand rule, u × v is directed out of the page.
5. For the two vectors to be orthogonal, we need h3, 2,xi·h2x, 4,xi =0 ⇔ (3)(2x) + (2)(4) + (x)(x) =0 ⇔
x 2 +6x +8=0 ⇔ (x +2)(x +4)=0 ⇔ x = −2 or x = −4.
7. (a) (u × v) · w = u · (v × w) =2
(b) u · (w × v) =u · [− (v × w)] = −u · (v × w) =−2
(c) v · (u × w) =(v × u) · w = − (u × v) · w = −2
(d) (u × v) · v = u · (v × v) =u · 0 =0
9. For simplicity, consider a unit cube positioned with its back left corner at the origin. Vector representations of the diagonals
11.
joining the points (0, 0, 0) to (1, 1, 1) and (1, 0, 0) to (0, 1, 1) are h1, 1, 1i and h−1, 1, 1i. Letθ be the angle between these
two vectors. h1, 1, 1i·h−1, 1, 1i = −1+1+1=1=|h1, 1, 1i| |h−1, 1, 1i| cos θ =3cosθ ⇒ cos θ = 1 3
⇒
θ =cos −1 1
3
≈ 71 ◦ .
−→
AB = h1, 0, −1i, −→
AC = h0, 4, 3i,so
(a) a vector perpendicular to the plane is −→
AB × −→
AC = h0+4, −(3 + 0), 4 − 0i = h4, −3, 4i.
(b) 1 2
−→
AB × −→
√
AC = 1 √
2 16 + 9 + 16 = 41
. 2
13. Let F 1 be the magnitude of the force directed 20 ◦ away from the direction of shore, and let F 2 be the magnitude of the other
force. Separating these forces into components parallel to the direction of the resultant force and perpendicular to it gives
F 1 cos 20 ◦ + F 2 cos 30 ◦ =255 (1),andF 1 sin 20 ◦ − F 2 sin 30 ◦ sin 30 ◦
=0 ⇒ F 1 = F 2 (2). Substituting (2)
sin 20◦ into (1) gives F 2 (sin 30 ◦ cot 20 ◦ +cos30 ◦ )=255 ⇒ F 2 ≈ 114 N. Substituting this into (2) gives F 1 ≈ 166 N.
15. The line has direction v = h−3, 2, 3i. LettingP 0 =(4, −1, 2), parametric equations are
x =4− 3t, y = −1+2t, z =2+3t.
17. A direction vector for the line is a normal vector for the plane, n = h2, −1, 5i, and parametric equations for the line are
x = −2+2t, y =2− t, z =4+5t.