Solução_Calculo_Stewart_6e

F.
118 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ET TX.10 CHAPTER 12
35. a = −→
−→
−→
PQ = h2, 1, 1i, b = PR = h1, −1, 2i,andc = PS = h0, −2, 3i.
2 1 1
−1 2
a · (b × c) =
1 −1 2
=2
0 −2 3
−2 3 − 1 1 2
0 3 + 1 1 −1
=2− 3 − 2=−3,
0 −2
so the volume of the parallelepiped is 3 cubic units.
1 5 −2
−1 0
37. u · (v × w) =
3 −1 0
=1
5 9 −4
9 −4 − 5 3 0
5 −4 +(−2) 3 −1
=4+60− 64 = 0, which says that the volume
5 9 of the parallelepiped determined by u, v and w is 0, and thus these three vectors are coplanar.
39. The magnitude of the torque is |τ | = |r × F| = |r||F| sin θ =(0.18 m)(60 N)sin(70+10) ◦ =10.8sin80 ◦ ≈ 10.6 N·m.
41. Using the notation of the text, r = h0, 0.3, 0i and F has direction h0, 3, −4i. Theangleθ between them can be determined by
cos θ =
h0, 0.3, 0i·h0, 3, −4i
|h0, 0.3, 0i| |h0, 3, −4i|
⇒ cos θ = 0.9
(0.3)(5)
⇒ cos θ =0.6 ⇒ θ ≈ 53.1 ◦ .Then|τ | = |r||F| sin θ ⇒
100 = 0.3 |F| sin 53.1 ◦ ⇒ |F| ≈ 417 N.
43. (a) The distance between a point and a line is the length of the perpendicular
from the point to the line, here −→
PS = d. But referring to triangle PQS,
d = −→
PS = −→ −→
QP sin θ = |b| sin θ. Butθ is the angle between QP = b
and −→
|a × b|
QR = a. Thus by Theorem 6, sin θ =
|a||b|
and so d = |b| sin θ =
(b) a = −→
QR = h−1, −2, −1i and b = −→
QP = h1, −5, −7i. Then
|b||a × b|
|a||b|
=
|a × b|
.
|a|
a × b = h(−2)(−7) − (−1)(−5), (−1)(1) − (−1)(−7), (−1)(−5) − (−2)(1)i = h9, −8, 7i.
Thus the distance is d =
|a × b|
|a|
√
= √ 1
6 81 + 64 + 49 =
194 97
6
= . 3
45. (a − b) × (a + b) =(a − b) × a +(a − b) × b by Property 3 of Theorem 8
= a × a +(−b) × a + a × b +(−b) × b by Property 4 of Theorem 8
=(a × a) − (b × a)+(a × b) − (b × b) by Property 2 of Theorem 8 (with c = −1)
= 0 − (b × a)+(a × b) − 0 by Example 2
=(a × b)+(a × b) by Property 1 of Theorem 8
=2(a × b)
47. a × (b × c)+b × (c × a)+c × (a × b)
=[(a · c)b − (a · b)c]+[(b · a)c − (b · c)a]+[(c · b)a − (c · a)b] by Exercise 46
=(a · c)b − (a · b)c +(a · b)c − (b · c)a +(b · c)a − (a · c)b = 0