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

116 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
19. a + b = (5 + (-3) ,-12 + (-6)) = (2, - 18)
2a + 3b = (10, -24) + (-9, -18) = (1, -42)
lal = y'5 2 + ( - 12) 2 = .jl69 = 13
Ia- b l = 1(5- ( -3), - 12- ( -6))1 = 1(8, -6)1 = y'8 2 + ( - 6)2 = v'100 = 10
21. a+ b = (i'+ 2j - 3 k) + ( -2 i - j + 5 k) = - i + j + 2k
2a + 3b = 2 (i + 2j - 3 k) + 3 ( -2 i - j + 5 k) = 2 i + 4j - 6 k - 6 i - 3j + 15 k = - 4 i + j + 9k
lal = y'12 + 22 + ( -3)2 = v'i4
Ia - b l = l(i+ 2j - 3 k) - (-2 i -j +5 k)l = l3 i + 3 j - 8k l = y'3 2 +3 2 + (-8)2 = v'82
23. The vector - 3 i + 7 j has length l-3 i + ~ j I = y' ( -3)2 + 7 2 = v'58, so by Equation 4 the unit vector with the same
d . . . 1 ( 3. 7 ' ) 3 . 7 .
1rect1on IS r,:n - 1 + J =- r.;o 1 + r.;oJ·
v58 v58 v58
25. The vector 8 i - j + 4 k has length 18 i - j + 4 k l = y'8 2 + ( - 1) 2 + 4 2 = J8I = 9, so by Equation 4 the unit vector with
the same direction is ~ (8 i - j + 4 k ) = ~ i - ~ j + ~ k.
27.
y
. .j3
From the figure, we sec that tan 0 = - 1
- = J3 => B = 60°.
X
29. From the figure, we see that the x-component ofv is
y
v1 = I vi cos(1r / 3) = 4 ·. ~ = 2 and they-component is
V2 = I vi sin( 1r / 3) = 4 · .,;; = 2 v'3. Thus
V = (th, V2) = (2, 2 .j3 ).
v
Vz
v,
X
31. The velocity vector v makes an angle of 40° with the horizontal and
has magnitude equal to the speed at which the footba ll was thrown.
From the figure, we see that the horizontal component ofv is
I vi cos 40° = 60 cos 40° ~ 45.96 ft/s and the vertical component
is I vi sin 40° = 60 sin 40° ~ 38.57 fils.
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