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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The distance of the point $(1,-5,9) $ from the plane $x-y+z=5$ measured along the line $x=y=z$ is ?

$\begin{array}{1 1} (a)\:\:\:10\sqrt 3\:\:\:\qquad(b)\:\:5\sqrt 3\:\:\:\qquad\:\:(c)\:\:3\sqrt {10}\:\:\:\qquad\:\:(d)\:\:3\sqrt 5

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Eqn. of the line through the point $P(1,-5,9) $ along (parallel) to the line $x=y=z$ is
$\large\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$...(i)
Let this line intersect the plane $x-y+z=5$ at the point $Q$
$\therefore Q$ is given by $Q(\lambda+1,\lambda-5,\lambda+9)$
$Q$ lies on the plane. $\therefore$ it satisfies the eqn. of the plane.
$\Rightarrow\:\lambda+1-\lambda+5+\lambda+9=5$
$\Rightarrow\:\lambda=-10$
and $Q(-9,-15,-1)$
Distance of the point $P$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is $\overline {PQ}$
$\therefore $ The required distance $\overline {PQ}=\sqrt {100+100+100}=10\sqrt 3$
answered Dec 27, 2013 by rvidyagovindarajan_1
 

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