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Question

Let a=2^i+^j2^k and b=^i+^k. 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=

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
ca=22
squaring both sides, we get
c22c+9=8
c22c+1=0
(c1)2=0
c=1
a×b=∣ ∣ ∣^i^j^k212110∣ ∣ ∣
=2^i2^j+^k on simplification
a×b=4+4+1=9=3
Now , (a×b)×c=a×bcsin300
=3×1×12=32

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