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Question

Let a=2i+j2k 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 300 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
We have, a×b=2i2j+k
|(a×b)×c|=|a×b||c|sin300
=22+(2)2+12|c|12=32|c|
Given |ca|=22
(ca)2=9|c|2+|a|22c.a=8
|c|22|c|+1=0(|c|1)2=0|c|=1
|(a×b)×c|=32

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