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

SECTION 12.4 THE CROSS PRODUCT D 127
1 5 - 2
37. u · (v x w)= 3-1 0 =1 1- 1·
0 3 0 3
1- 5 1 1+(- 2) 1 -ll = 4 + 60 - 64 = 0,whichsaysthatthevolume
9 - 4 5 - 4 5 9 . .
5 9 - 4
of the parallelepiped determined by u, v and w is 0, and jhus these three vectors are coplanar.
39. The magnitude of the torque is Jr J = Jr x FJ = JrJJFJ sinO = (0.18 m)(60 N) sin(70 +lOt = 10.8 sin80° r::::: 10.6 N·m.
41 . Using the notation of the text, r = {0, 0.3, 0) and F has direction (0, 3, -4). The angle(} between them can be determined by
(0, 0.3, 0) . {0, 3, -4) . 9 0.9
cos 9 = => cos = =>
J(o, o.3, o)JI{o,3, -4)1 (0.3)(5)
cosO = 0.6 => (J r::::: 53.1°. Then Jr l = lri iFI sin(} =>
100 = 0.3IFisin53.1° => 1F i r:::::417N.
43. From Theorem 9 we have Ia x bl = la l lbl sin 9, where 0 is the angle between a and b, and from Th,eorem 12.3.3 we have
a· b = lal lb l cosO => Ja l lbl = a · be. Substituting the second equation into the first gives Ia x b l = a · b sin B, so
cos cos (J
Ia X b l = tan9. Here Ia X b l = 1(1, 2, 2)1 = V1 + 4 + 4 = 3, so tan() = Ia x bbl = ~ = J3 => () = 60° .
. a· b a · v3
45. (a)
The distance between a point and a line is the length of the perpendicular
from the point to th~ line, here In I =d. But referring to triangle PQS,
a
d = In I = IQP' sin (J = lbl sin 9. But 8 is the angle between QP = b
--+ Ia x bl
and QR = ~ ·Thus ~y Theorem 9, sinO = lai Jb l
d d = lb1 . 9 = Jbi Ja x bl = Ia x b J
an so 'I sm Jai Jbl laJ
--+ ---+
(b) a= QR = (- 1, - 2, - 1) and b = QP = (1, - 5, - 7}. Then
a x b = (( -2)( -7) - ( -1){ 7 5), ( - 1)(1)- ( - 1)( - 7) , ( -1)( -5)- ( - 2)(1)} = (9, - 8, 7).
Thus the distance is d = Ia [:I bl = ~ JBl + 64 + 49 = ji¥ = j¥.
47. From Theorem 9 we have Ia x b l = [al[b[ sin 9 so
Ja x b] 2 = [a[ 2 [b[ 2 sin 2 () = [a[ 2 lb[ 2 (1 - cos 2 B)
= fa i 2 Jb[ 2 - ([ai Jb [ cos 9) 2 = [a[ 2 lb[ 2 - (a · b) 2
by Theorem 12.3.3.
© 2012 Cengage l c'3ming. All Rights Reserved. May not be scanned, cor lcd, or duplicated, or j>ost