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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 30o, 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
a×b=∣ ∣ijk212110∣ ∣=2i2j+k
|a×b|=4+4+1=3
|ca|=22
|ca|2
=(ca)2=8
|c|22a.c+|a|2=8
|c|22|c|+9=8
|c|22|c|+1|c|=1
|(a×b)×c|=|(a×b)c|sin30o=32

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