If →a,→b,→c are three non coplanar vectors and a vector →α is such that →α=p(→b×→c)+q(→c×→a)+r(→a×→b) and →α⋅(→a+→b+→c)=1, then [→a→b→c] is equal to
A
p+q+r
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B
1p+q+r
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C
2(p+q+r)
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D
2p+q+r
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Solution
The correct option is B1p+q+r Given: →α=p(→b×→c)+q(→c×→a)+r(→a×→b) and →α⋅(→a+→b+→c)=1
Now, →α⋅→a=p[→a→b→c] →α⋅→b=q[→a→b→c] →α⋅→c=r[→a→b→c]
Adding them, we have: →α⋅(→a+→b+→c)=(p+q+r)[→a→b→c] ⇒1=(p+q+r)[→a→b→c]⇒[→a→b→c]=1p+q+r