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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane passing through the points (1,—-1,2) and (2,—-2,2) and perpendicular to the plane 6x-—2y+2z = 9.

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Solution :
Vector eqution of the line joining the points $(1,-1,2) $ and $(2,-2,2)$ is
$\overrightarrow{a} =(2-1) \hat i +(-2-1)\hat j +(2-2)\hat k$
$\qquad= \hat i - \hat j$
Given equation of the plane is $6x-2y+2z=0$
$\overrightarrow{n} = 6 \hat i - 2 \hat j+ 2 \hat k$
Equation of the plane passing through $\overrightarrow{a} $ and normal to $\overrightarrow{n}$ is $(\overrightarrow{r}-\overrightarrow{a}). \overrightarrow{n} =0$ (ie) $\overrightarrow{r}.\overrightarrow{n}= \overrightarrow{a} . \overrightarrow{n} $
Hence the vector equation of the plane is $\overrightarrow{r} .( 6 \hat i - 2 \hat j +2 \hat k) =6+2=8$
=> $\overrightarrow{r}.(6 \hat i -2 \hat j +2 \hat k)=8$
The Cartesian form of equation is
$(x \hat i + y \hat j + z \hat k) .(6 \hat i - 2\hat j +2 \hat k)=8$
((ie) $6x-2y+2z=8$ is the required equation of the plane .
answered Sep 3, 2015 by meena.p

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