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

j k
SECTION 12.4 THE CROSS PRODUCT . 0 125
17. axb = 2 - 1 3 = ~-~ :Ii - I~ : lj + I~
-1, k = ( -1-6) i - (2-12)j+ [4- (- 4)] k = - 7i+10j+8 k
2 .
4 2 1
j k
=1 - ~ ~ li -I:
21 .
b x a = 4 2 1 k = [6- (- 1)] i-{12 - 2)j + (-4 - 4) k = 7 i - lOj- 8 k
~ l j + 1: -1
2 -1 3
Notice a x b = - b x a here, as we know is always true by Property I of Theorem II .
19. By Theorem 8, the cross product of two vectors is orthogonal to both vectors. So we calculate
i j k 12 11 I 3 11 I 3 21
(3,2,1)x(-1, 1, 0) = 3 2 1 = i- · j + k= - i-j +5 k.
1 0 -1 0 - 1 1
- 1 1 0
. I b h ± (-1, -1,5) ±(-1, -1,5) h . ( 1 1 5 )
../3 , t at 1s, - M, -M ~ 37:3
3
So two umt vectors orthogona to ot are v' 1
+ 1
+ 25
=
21 . Let a= (a1, a2, as). Then
j
k
O x a= 0 0 0 0 lk= O,
= Ia: :31i - ~~ :sl j + 1:1 a2
a1 a2 as
j
k
axO = a1 a2 as
0 0 0
i _ I a1 as, . + I a1 a21 O
0 0 0 0 J 0
k = O.
= I a~ as I
25. ax (b+c) =a x (b1 +c1,b2 +c2,bs +c3)
= (a2(bs + cs)- a3(b2 + c2), as(b1 + c1)- a1(b3 + cs) , a1(b2 + c2)- a2(b1 + q))
= ((a2bs - asb2) + (a2cs - asc2), (asb1- a1bs) + (asc1 - a1c3), (a1b2 - a2b1) + (a1c2 - a2c1))
= (a2bs- asb2 , aab1 - a1bs, a1b2- a2b1 ) + (a2cs - asc2, asc1- a1c3, a1c2 - a2c1)
= (a x b) + (a X c)
@) 20 12 _Cengoge Learning. All Rights Rcsc:n·cd. Moy no! be scanned. copied, orduplicnlcd, or posted lo o publicly ot:ccssiblc wcbsi!c, in whole or in p:trt.