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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\overrightarrow r=(2\hat i-\hat j+2\hat k)+\lambda(3\hat i+4\hat j+12\hat k)$ and the plane $\overrightarrow r.(\hat i-\hat j+\hat k)=5$ is ?

$\begin{array}{1 1} (a) \:\:\large\frac{6}{\sqrt 3}\:\:\qquad\:\:(b) \:\:13\:\:\qquad\:\:(c)\:\:\large\frac{16}{\sqrt 3}\:\:\qquad\:\:(d)\:\:\sqrt {149} \end{array} $

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1 Answer

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Given eqn. of the line is written as $\large\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$
Any point $P$ on this line can be expressed in terms of $\lambda$ as $P(3\lambda+2,4\lambda-1,12\lambda+2)$
Let this point $P$ be point of intersection of the line with
the given plane $x-y+z=5$.
$\therefore$ $P$ satisfies the equation of the plane.
$\Rightarrow\:3\lambda+2-4\lambda+1+12\lambda+2=5$
$\Rightarrow\:\lambda=0$
$\therefore P(2,-1,2)$ and the required distance from $A(-1,-5,-10) $ is $PA$
$=\sqrt {9+16+144}=\sqrt {169}=13$
answered Jan 2, 2014 by rvidyagovindarajan_1
 

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