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 ⇒3→r−(→a⋅→r)→a=→a×→b⋯(1)
We know that, (→r⋅→a)2+|(→r×→a)|2=|→r|2|→a|2 ⇒(→r⋅→a)2+|→b|2=3 ⇒→r⋅→a=±1 ∴ From (1),→r=13(→a×→b±→a)