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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The distance of the point $(1,1,1) $ from the plane passing through the points $(2,1,1),\:(1,2,1)\:and\:(1,1,2)$ is ?

$\large(a)\:\frac{1}{\sqrt 3}\:\:\:\qquad\:\:(b)\:\:1\:\:\:\qquad\:\:(c)\:\:\sqrt 3\:\:\:\qquad\:\:(d)\:\:2$
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  • Distance of a point $P$ from the plane $ABC$ is length of projection of $\overrightarrow {AP} $ on normal vector to the plane.
  • Projection of $\overrightarrow a$ on $\overrightarrow b$ is $\bigg|\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|}\bigg|$
Let $P(1,1,1),\:\:and\:\:A(2,1,1),\:B(1,2,1),\:C(1,1,2)$
Any two vectors on th the plane $ABC$ can be given by $\overrightarrow {AB},\:\overrightarrow {BC}$
$i.e.,$ $\overrightarrow {AB}=(-1,1,0)$ and $\overrightarrow {BC}=(0,-1,1)$
Normal to the plane is $\perp$ all to lines on the plane.
$\therefore $ Normal, $\overrightarrow n=\overrightarrow {AB}\times\overrightarrow {BC}=\left |\begin {array}{ccc} \hat i & \hat j & \hat k \\-1 & 1 & 0 \\ 0 & -1 & 1\end {array} \right |=(1,1,1)$
The distance of the point $P(1,1,1) $ on the plane is Projection of $\overrightarrow {AP}$ on $\overrightarrow n$
$\therefore $ The required distance of $P$ from the plane is $\bigg|\large\frac{\overrightarrow {AP}.\overrightarrow n}{|\overrightarrow n|}\bigg|$
$=\large\frac{(1,0,0).(1,1,0)}{\sqrt 3}=\frac{1}{\sqrt 3}$
answered Jan 5, 2014 by rvidyagovindarajan_1

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