If →a,→b,→c are three non coplanar, non zero vectors and →r is any vector in space, then (→a×→b)×(→r×→c)+(→b×→c)×(→r×→a)+(→c×→a)×(→r×→b) is equal to λ[→a→b→c]. Then the value of λ is
A
1
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B
−1
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C
2
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D
−2
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Solution
The correct option is C2 Let →T1=(→a×→b)×(→r×→c) →T1=[→a→b→c]→r−[→a→b→r]→c
Let →T2=(→b×→c)×(→r×→a) →T2=[→b→c→a]→r−[→b→c→r]→a=[→a→b→c]→r−[→b→c→r]→a
Let →T3=(→c×→a)×(→r×→b) →T3=[→c→a→b]→r−[→c→a→r]→b=[→a→b→c]→r−[→c→a→r]→b
Adding →T1,→T2,→T3 ⇒3[→a→b→c]→r−{[→a→b→r]→c+[→b→c→r]→a+[→c→a→r]→b}
As →r is any vector in space so, ⇒3[→a→b→c]→r−[→a→b→c]→r ⇒2[→a→b→c]→r=λ[→a→b→c]→r
Hence, λ=2