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

SECTION 12.4 THE CROSS PRODUCT , 0 123
59. Let a = (a1, a2, a3) and= (b1, b2, b3).
Property 2:
a· b = (a1, a2, a3 ) · (b1, b2, bs) = a1b1 + a2b2 + asbs
= b1a1 + ~a2 + b3a3 = (b1, ~. bs ) · (a1 , a2, as) = b ·a
Property 4:
(ca) · b = (ca1, ca2, ca3) · (b1, b2, bs) = (ca1)b1 + (ca2)b2 + (ca3)b3
= c(a1b1 + a2b2 + asbs) = c (a· b) = a1(cb1) + a2(~) + a 3(cbs)
= (a1 , a2, as)· (cbt, c~, cbs) = a· (c b)
Property 5: 0 ·a= (0, 0, 0) · (a1, a2, a3) = {O)(al) + (O)(a2) + (O)(a3) = 0
61 . Ia · b l = l lal lbl cos8l = lallbl lcos 81. Since lcos81 ~ 1, Ia · bl = lal lbl lcos 81 ~ lal lbl.
Note: We have equality in the case ofcos8 = · ± 1, so 8 = 0 or 8 = 1r, thus equality when a and bare parallel.
63. (a) t he Parallelogram Law states that the sum of the squares of the
lengths of the diagonals of a parallelogram equals the sum of the
squares of its (four) sides.
(b) Ia + b l 2 =(a + b) · (a + b)= lal 2 + 2(a ·b)+ lbl 2 and Ia - b l 2 = (a- b)· (a- b)= lal 2 -
2(a · b)+ lb l 2 •
Adding these two equations gives Ia + b l 2 + Ia- b l 2 = 2 lal 2 + 2 lbl 2 .
12.4 The Cross Product
1. a x b = 6 0
j
0 8
k
-2 = I~
0
-2 . 6 - 2 . 6 o I
1 - J + k
0 0 0 0 8
1
= [0 - (- 16)]i - (0 - O)j + (48 - O) k = 16 i + 48k
1
1
1
Now (a x b)· a = (16, 0, 48) · (6, 0, -2) = 96 + 0 - 96 = 0 and (a x b)· b = (16, 0, 48) · (0, 8, 0) = 0 + 0 + 0 = 0, so
a x b is orthogonal to both a and b .
j
k
3. a x b = 1 3 - 2 = I ~ ~~I i - ~-~ - ~ lj + l ~~ ~I k
-1 0 5
·= (15 - O)i - (5 - 2) j + [0 - (- 3))k = 15 i -3j +3k
Since (a x b) · a= {15i - 3j + 3k) · (i + 3j - 2k) = 15 - 9-6 = O, a x bisorthogonal to a.
Since (ax b)· b = (15 i - 3j + 3 k) · ( - i + 5 k) = - 15 + 0 + 15 = 0, a x b is orthogonal to b .
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