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Question

Let a=2^i+^j2^k and b=^i+^j. Let c be a vector such that |ca|=3, |(a×b)×c|=3 and the angle between c and a×b be 30. Then ac is equal to:

A
258
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B
2
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C
5
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D
18
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Solution

The correct option is B 2
|a|2=4+1+4=9
and a×b=2^i2^j+^k
|a×b|=9
Also
|ca|=3
|c|2+|a|22ca=9
|c|22ca=0 (1)
Also, |(a×b)×c|=3
|a×b||c|sin30=3
3|c|×12=3
|c|=2
From (1),ca=ac=2

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