Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

120 D CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
15. ja j = vW + 3 2 = 5, jbj = }2 2 + (- 1)2 = v'5, and a· b = {4){2) + {3)(- 1) = 5. ~rom Corollary 6, we have
a· b 5 1 · . ( 1 ) ·
cos(}= jajjb/ = 5
. v'5 = vg· So the angle between a and b IS(}= cos- 1 v'5 ~ 63°.
17. jaj = .,j3 2 + ( -1)2 + 5 2 = .;35, jbj = .,j( - 2) 2 + 4 2 + 3 2 = v'29, and a · b ~ {3){ -2) + ( -1)(4) + {5)(3) = 5. Then
cos B = l:l · l~ l = .;35 ~ v'29 = v' 1 ~ 15
and the angle between a and b is. B = cos- 1 ( v'1~15 ) ~ 81°.
19. jaj = .J42 + ( -3)2 + 12 = .;26, jbj = .,j22 + 02 + ( -1)2 = v'5, and a. b = {4){2) + ( - 3)(0) + (1)( -1) = 7.
h (} a· b · 7 7 d 8
. _1 ( 7 )
T en cos = jajjbj = v'26 . v'5 = v'130 an = cos v'130 ~ .
52 o
21. Let p, q, and r be the angles at vertices P, Q, and R respectively.
--> -'-+
Then p is the angle between vectors PQ and P R , q is the angle
--> -->
between vectors QP and QR, and r is the angle between vectors
--> . ---+
RPaildRQ.
--> --> .
Thuscos = PQ·PR = (- 2 • 3 ) · ( 1 • 4 ) = - 2 + 12 =___!Q_and =cos- 1 (___!Q_) ~48° . Sim i larl,
p !?O!!nl .,j(- 2)2+32 .j12+ 42 y'l3ffi v'ffi p y'ffi y
---+ ---+
QP . QR . (2, -3) · (3, 1) 6 - 3 3 - 1 ( 3 )
75 o d
cosq = IQ?IIQRI = v'4 + 9 v'9 + 1 = v'13 v'f5 = v'13Q so q =cos v'l3Q ~ an
r ~ 180° - {48°+ 75°) = 57°.
. ,___.,2 ,___.,2 1--,-+12
,__., ,__.
. ~-~-n
Alternate solution: Apply the Law of Cosines three times as follows: cos p =
2 PQ PR
1
_ i?R( -I?QI 2 .- IQJil. 2 _ j?Q( -lnl 2 -I Q'RI 2
cos q - ,___.,, __., , and cosr - ---. .-.
· 2 PQ QR 2 PR QR
1 11 1
23. (a) a . b = ( -5){6) + (3)( - 8) + {7){2) = -40 -:/= 0, so a and bare not orthogonal. Also, since a is not a scalar multiple
ofb, a and bare not parallel.
(b) a · b = ( 4) ( -3) + (6)(2) = 0, so a and bare orthogonal (and not parallel).
(c) a· b = (-1){3) + (2){4) + {5)(-1) = 0, soaand bare orthogonal (and not parallel).
(d) Because a = - ~ b, a and b ~e parallel.
--+ ---+ ----+ --+ ----+ ----+
25. QP = (-1, - 3, 2), QR = (4, - 2, - 1), and QP · QR = -4 + 6-2 = 0. Thus QP and QRare orthogonal, so the angle of
the triangle at vertex Q is a right angle.
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