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Question

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 C 2
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

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