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Question

# If →a,→b,→c are non coplanar non zero vectors, then the value of (→a×→b)×(→a×→c)+(→b×→c)×(→b×→a)+(→c×→a)×(→c×→b) is

A
[abc]2(a+b+c)
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B
[abc](a+b+c)
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C
0
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D
None of these
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Solution

## The correct option is D [→a→b→c](→a+→b+→c)(→a×→b)×(→a×→c)+(→b×→c)×(→b×→a)+(→c×→a)×(→c×→b) =→b(→a⋅(→a×→c))−→a(→b⋅(→a×→c))+→c⋅(→b⋅(→b×→a))−→b(→c⋅(→b×→a))+ →a⋅(→c⋅(→c×→b))−→c(→a⋅(→c×→b))=→b[→a→a→c]−→a[→b→a→c]+→c[→b→b→a]−→b[→c→b→a]+ →a[→c→c→b]−→c[→a→c→b]⟶∗ we khow that ,(1)[→a→a→b]=0 and (2)[→a→b→c]=[→c→a→b]=[→b→c→a]=−[→b→a→c]=− [→c→b→a]=−[→a→c→b] Hence by using the concepts of (1)& (2)eqn.⟶∗ can be written as, =→b⋅0+→a[→a→b→c]+→c⋅0+→b[→a→b→c]+→a⋅0+→c[→a→b→c] =[→a→b→c](→a+→b+→c) Hence, (B) is the correct answer.

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