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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 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 A 32
(a×b)×c=a×b |c|sin30
Now,
We need |a×b| and |c|

a×b=a2b2ab2=9×29=3
We have
a.c=|c|
cosθ=1|a|=13
Now,
|ca|=a2+c22a.c=22
9+c22c=8
|c|=1

Hence combining all the data, we get
(a×b)×c=a×b |c|sin30=32

Hence, option B.

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