Let →a=−^i−^k,^b=−^i+^j and →c=^i+2^j+3^k be three given vectors. If →r is a vector such that →r×→b=→c×→b and →r⋅→a=0, then the value of →r⋅→b=
A
9
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B
8
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C
7
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D
6
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Solution
The correct option is A9 →r×→b=→c×→b
Taking cross product with →a, we get →a×(→r×→b)=→a×(→c×→b) ⇒(→a⋅→b)→r−(→a⋅→r)→b=→a×(→c×→b) ⇒→r=(→a⋅→b)→c−(→a⋅→c)→b(∵→a⋅→b=1,→a⋅→r=0) →r=1(^i+2^j+3^k)+4(−^i+^j)→r=−3^i+6^j+3^k ⇒→r⋅→b=3+6=9