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Question

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


We have,
a=2^i+^j2^k|a|=4+1+4=3and b=^i+^j|b|=1+1=2Now, |ca|=3|ca|2=9(ca).(ca)=9|c|2+|a|22c.a=9Again, |(a×b)×c|=3|a×b||c| sin 30o=3|c|=6|a×b|But a×b=∣ ∣ ∣^i^j^k212110∣ ∣ ∣=2^i2^j+^k |c|=64+4+1=2
FromEqs. (i) and (ii), we get
(2)2+(3)22c.a=9 c.a=2


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