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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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If $P(3,2,6)$ is a point and $Q$ is a point on the line $\overrightarrow r=(\hat i-\hat j+2\hat k)+\lambda(-3\hat i+\hat j+5\hat k),$ then the value of $\lambda$ for which $\overrightarrow {PQ}$ is parallel to the plane $x-4y+3z=1$ is ?

$\begin{array}{1 1} (a)\:\:\large\frac{1}{4}\:\:\:\qquad\:\:(b)\:\:-\frac{1}{4}\:\:\:\qquad\:\:(c)\:\:\frac{1}{8}\:\:\:\qquad\:\:(d)\:\:-\frac{1}{8} \end{array} $

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Any point $Q$ on the given line can be expressed in terms of $\lambda$
$i.e., \:\: Q(-3\lambda+1,\lambda-1,5\lambda+2)$
Given: $P(3,2,6)$
$\overrightarrow {PQ}=(-3\lambda-2,\lambda-3,5\lambda-4)$
It is given that $\overrightarrow {PQ}$ is parallel to the plane $x-4y+3z-1=0$
$\therefore\:\overrightarrow {PQ}.\overrightarrow n=0$ where $\overrightarrow n$ is normal to the plane.
$\Rightarrow\:(-3\lambda-2,\lambda-3,5\lambda-4).(1,-4,3)=0$
$\Rightarrow\:-3\lambda-2-4\lambda+12+15\lambda-12=0$
$\Rightarrow\:8\lambda=2\:\:or\:\:\lambda=\large\frac{1}{4}$
answered Dec 27, 2013 by rvidyagovindarajan_1
 

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