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Q.

  Let a=2i^+j^2k^ and b=i^+j^. If c is a vector such  that ac=|c|,|ca|=22 and the angle between 

a×b and c is 30, then |(a×b)×c| is equal to 

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a

2

b

3/2

c

d

3

answer is B.

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Detailed Solution

|(a×b)×c|=|a×b||c|sin30=12|a×b||c|----1

  We have, a=2i^+j^2k^ and b=i^+j^

  a×b=2i^2j^+k^ or  |a×b|=9=3

 Also given |ca|=22

 or  |ca|2=8

 or  |c|2+|a|22ac=8

 Given |a|=3 and ac=|c|, using these we get 

|c|22|c|+1=0

or  (|c|1)2=0or  |c|=1

 Substituting values of |a×b| and |c| in (i), we get 

|(a×b)×c|=12×3×1=32

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