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

5. (a)
un
(d)
(b)
(c)
u
uNw
SECTION 12.2 VECTORS 0
w~
u
v+w
(e)
(f)
u
u
u -:Gv
~ v Dw
v+u+w u-w-v
115
--->
7. Because the taif of dis the midpoint of QR we have QR = 2d, and by .the Triangle Law,
a + 2d = b =? 2d=b-a =? d=!{b-a) = th - ~a. AgainbytheTriangleLawwehavec + d = bso
c = b - d = b - ( ~ b - ~a) = ~ ~ + ~b .
9. a= (3 - (- 1), 2 - 1}=(4, 1} 11.a=(2 - (-1), 2 - 3} = (3,-1}
y
B(3, 2) A(-1, 3)
y
0 X
X
13. a = ·(2 - 0, 3 - .3,-1 - 1} = (2, 0,-2} 15. (-1, 4} + (6, - 2} = (-1 + 6, 4 + ( - 2)} = (5, 2)
A(0, 3, 1)
y
X
X
17. (:~, 0, 1) + (0, 8, 0) = (3 + 0, 0 + 8, 1 + 0)
= (3, 8, 1)
(3, 0, 1)
X
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