If →u,→v,→w are non-coplanar vectors and p,q are real numbers, then the equality [3→up→vp→w]−[p→vq→w→u]−[2→wq→vq→u]=0 holds for
A
exactly two values of (p,q)
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B
more than two but not all values of (p,q)
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C
all values of (p,q)
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D
exactly one value of (p,q)
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Solution
The correct option is D exactly one value of (p,q) Given: [3→up→vp→w]−[p→vq→w→u]−[2→wq→vq→u]=0 ⇒3p2[→u→v→w]−pq[→v→w→u]−2q2[→w→v→u]=0 ⇒(3p2−pq+2q2)[→u→v→w]=0
As [→u→v→w]≠0 ∴(3p2−pq+2q2)=0 ⇒2p2+p2−pq+q24+7q24=0 ⇒2p2+(p−q2)2+7q24=0 ⇒p=0,q=0,p=q2
only possible when p=q=0