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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The equation of the plane through the point $(1,2,-1)$ and $\perp$ to the line of intersection of the planes $\overrightarrow r.(3\hat i-\hat j+\hat k)=1$ and $\overrightarrow r.(\hat i+4\hat j-2\hat k)=2$ is?

$\begin{array}{1 1} (a)\:\overrightarrow r.(2\hat i+7\hat j-13\hat k)=1\:\qquad\:(b)\:\overrightarrow r.(2\hat i-7\hat j-13\hat k)=1\:\qquad\:(c)\:\overrightarrow r.(2\hat i+7\hat j+13\hat k)=0\:\qquad\:(d)\overrightarrow r.(-2\hat i+7\hat j-13\hat k)=1 \end{array} $

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  • Vector $\perp$ to $\overrightarrow a\:and\:\overrightarrow b\:is \:\overrightarrow a\times\overrightarrow b$
Given line of intersection of planes $\overrightarrow r.(3\hat i-\hat j+\hat k)=1.......(i)$
and $\overrightarrow r.(\hat i+4\hat j-2\hat k)=2.............(ii)$
Normal to $(i)\;and\:(ii) $ are $\overrightarrow n_1=(3,-1,1)\:\;and\:\:\overrightarrow n_2=(1,4,-2)$
The line of intersection of $(i)\:and\:(ii) $ is $\perp$ to normals to both the planes.
$\therefore$ The line is along $\overrightarrow n_1\times\overrightarrow n_2$
$\overrightarrow n_1\times\overrightarrow n_2=\left |\begin {array}{ccc} \hat i & \hat j & \hat k \\ 3 & -1 & 1 \\1 & 4 & -2\end {array}\right |=(-2,7,13)$
$\therefore$ $d.r.$ of the line of intersection is $(2,-7,-13)$
$\Rightarrow\:$The plane through $(1,2,-1) $ and $\perp$ to the line of intersection will have normal $\overrightarrow n$ =$(2,-7,-13)$
$\therefore$ The eqn. of the required plane is $[\overrightarrow r-(\hat i+2\hat j-\hat k)].\overrightarrow n=0$
$\Rightarrow\:\overrightarrow r.(2\hat i-7\hat j-13\hat k)=1$
answered Jan 4, 2014 by rvidyagovindarajan_1

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