If →a×(→b×→c)+→b×(→c×→a)=λ1→a+λ2→b+λ3→c then λ1+λ2+λ3=
A
−→b.→c+→a.→c
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B
→a.→c+→b.→c
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C
→a.→b+→b.→c+→c.→a
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D
none of these
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Solution
The correct option is A−→b.→c+→a.→c →a×(→b×→c)+→b×(→c×→a)=(→a.→c)→b−(→a.→b)→c+(→b.→a)→c−(→b.→c)→a=−(→b.→c)→a+(→a.→c)→b+(→b.→a−→a.→b)→c≡λ1→a+λ2→b+λ3→cλ1+λ2+λ3=→a.→c−→b.→cor−→b.→c+→a.→cHence,optionAiscorrectanswer.