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Q.

If a,b and c are unit vectors such that a+b+c=0 and a,b=π3, then a×b+b×c+c×a=
 

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a

0

b

3

c

32

d

332

answer is C.

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Detailed Solution

|a|=|b|=|c|=1     |a×b|1×1×32 a+b+c=0    0+a×b+a×c=0   a×b=c×a b×a+0+b×c=0  b×c=a×b |a×b|+|b×c|+|c×a|=3|a×b|                                             =332         

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