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Question

Let a=2^i+^j2^k and b=^i+^j. If c is a vector such that a.c=|c|, |ca|=22 and the angle between (a×b) and c is 30 ,then |(a×b)×c| is equal to

A
23
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B
32
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C
2
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D
3
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Solution

The correct option is B 32
In this question, vector c is not given, therefore, we cannot apply the formulae of (a×b×c) (vector triple product).
Now, |(a×b)×c|=|a×b||c|sin 30 ...(i)Again,|a×b|=∣ ∣ ∣^i^j^k212110∣ ∣ ∣=2^i2^j+^k|a×b|=22+(2)2+1=4+4+1=9=3Since,|ca|=22 [given]|ca|2=8(ca).(ca)=8c.cc.aa.c+a.a=8|c|2+|a|22|c|=8|c|2+92|c|=8|c|22|c|+1=0(|c|1)2=0|c|=1From, Eq.(i),|(a×b)×c|=(3)(1).(12)=32


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