If →a,→b and→care 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 C3 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]→c−0} =(→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