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Question

Let r be a unit vector satisfying r×a=b, where |a|=3 and |b|=2. Then

A
r=23(a+a×b)
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B
r=13(±a+a×b)
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C
r=14(±a+a×b)
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D
r=23(±a+a×b)
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Solution

The correct option is B r=13(±a+a×b)
r×a=b
a×(r×a)=a×b
3r(ar)a=a×b (1)
We know that,
(ra)2+|(r×a)|2=|r|2|a|2
(ra)2+|b|2=3
ra=±1
From (1),r=13(a×b±a)

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