If vector →a is collinear with vector →b=3^i+6^j+6^k and →a⋅→b=27. Then →a is equal to
A
^i+2^j+2^k
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B
2^i+^j+^k
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C
−^i−2^j−2^k
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D
2^i−2^j+2^k
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Solution
The correct option is A^i+2^j+2^k Given,→a and →b are collinear.
Hence, →a=k→b ⇒→a⋅→b=27⇒k(→b⋅→b)=27,(∵→b=3^i+6^j+6^k)⇒(9+36+36)k=27⇒k=13
So, →a=^i+2^j+2^k