If a,b,c are non-coplanar vectors and λ is a real number, then [λ(→a+→b)λ2→bλ→c]=[→a→b+→c→b] for
A
exactly two values of λ
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B
exactly three values of λ
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C
no value of λ
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D
exactly one value of λ
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Solution
The correct option is D no value of λ (λ(→a+→b)×λ2→b).λ→c=λ4((→a+→b)×→b).→c=λ4[abc]R.H.S=(→a×(→b+→c)).→b=[→a→c→b]⇒λ4[abc]=−[abc]⇒λ4=−1 which is not possible.