If →a,→b,→c,→d are non-zero vectors, then [→a×→b→a×→c→d] can be simplified as:
A
(→a⋅→b)[→a→c→d]
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B
(→a⋅→d)[→a→b→c]
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C
(→b⋅→c)[→a→b→d]
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D
(→a⋅→c)[→b→c→d]
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Solution
The correct option is B(→a⋅→d)[→a→b→c] [→a×→b→a×→c→d]=(→a×→b)⋅{(→a×→c)×→d} =(→a×→b)⋅{(→a⋅→d)→c−(→c⋅→d)→a}=(→a⋅→d)[→a→b→c]−(→c⋅→d)[→a→b→a]=(→a⋅→d)[→a→b→c]