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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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If $\overrightarrow u,\overrightarrow v,\overrightarrow w$ are non coplanar vectors and $p,q$ are real numbers so that $[3\overrightarrow u\:p\overrightarrow v\:p\overrightarrow w]-[p\overrightarrow v\:p\overrightarrow w\:q\overrightarrow u]-[2\overrightarrow w\:q\overrightarrow v\:q\overrightarrow u]=0$ then number of values of $p,q$ is

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  • $[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]=[\overrightarrow v\:\overrightarrow w\:\overrightarrow u]=[\overrightarrow w\:\overrightarrow u\:\overrightarrow u]$
$[3\overrightarrow u\:p\overrightarrow v\:p\overrightarrow w]-[p\overrightarrow v\:\overrightarrow w\:q\overrightarrow u]-[2\overrightarrow w\:q\overrightarrow v\:q\overrightarrow u]$
$=3p^2[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]-pq[\overrightarrow v\:\overrightarrow w\:\overrightarrow u]-2q^2[\overrightarrow w\:\overrightarrow v\:\overrightarrow u]$
$=[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]\:(3p^2-pq-2q^2)$
Given: $[3\overrightarrow u\:p\overrightarrow v\:p\overrightarrow w]-[p\overrightarrow v\:\overrightarrow w\:q\overrightarrow u]-[2\overrightarrow w\:q\overrightarrow v\:q\overrightarrow u]=0$
$\Rightarrow\:[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]\:(3p^2-pq-2q^2)=0$
But since $\overrightarrow u,\overrightarrow v,\overrightarrow w$ are non coplanar, $[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]\neq 0$
$\therefore\:3p^2-pq-2q^2=0$
$\Rightarrow\:3p^2-3pq+2pq-2q^2=0$
$\Rightarrow\:3p(p-q)+2q(p-q)=0$
$\Rightarrow\:p=q\:\:or\:\:p=\large\frac{2}{3}$$q$
answered Nov 23, 2013 by rvidyagovindarajan_1
 

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