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Question

If a,b andcare three vectors such that [a b c]=1, then the value of [a+b b+c c+a]+[a×b b×c c×a]+[a×(b×c) b×(c×a) c×(a×b)] is

A
1
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B
2
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C
3
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D
0
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Solution

The correct option is C 3
Solving the equation by splitting in three parts:
Part - I
=[a+b b+c c+a]
=(a+b)[(b+c)×(c+a)]
=(a+b)[b×c+b×a+c×c+c×a]
=(a+b)[b×c+b×a+0+c×a]
=[a b c]+[a b a]+[a c a]+[b b c]+[b b a]+[b c a]
=[a b c]+0+0+0+0+[b c a]
=2[a b c]=2(1)=2
Part - II
=[a×b b×c c×a]
=(a×b)[(b×c)×(c×a)]
=(a×b){[b c a]c[b c c]a}
=(a×b){[b c a]c0}
=(a×b){c[a b c]}
=[a b c][a b c]
=[a b c]2=12=1
Part - III
=[a×(b×c) b×(c×a) c×(a×b)]
={a×(b×c)}{(b×(c×a))×(c×(a×b))}
=0
Adding Parts I,II,III
2+1+0=3

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