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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of the plane through point (2, 0, -1) and perpendicular to the line joining the two points (1, 2, 3) and (3, -1, 6)

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Solution :
Vector equation of the plane passing through point $\overrightarrow{a}$ and normal to $\overrightarrow{n}$ is
$(r-\overrightarrow{a})=0$
$\overrightarrow{r} - \overrightarrow{n} = \overrightarrow{a} . \overrightarrow{n} $
Here $\overrightarrow{a}= 2 \hat i - \hat k$
Hence the vector equation of the plane is
$\overrightarrow{r}. (2 \hat i - 3\hat j +3 \hat k) =(2 \hat i - \hat k) .(2 \hat i - 3 \hat j +3 \hat k)$
((ie) $\overrightarrow{r} .(2 \hat i - 3 \hat j +3 \hat k) =4-3=1$
Hence $\overrightarrow{r}( 2\hat i - 3\hat j +3 \hat k) = 4-3 =1$
Hence $\overrightarrow{r} (2 \hat i -3 \hat j +3 \hat)=1$ is the required vector equation of the plane .
The Cartesian equation is
$(a \hat i + y \hat j +z \hat k) (2 \hat i - 3\hat j +3 \hat k) =1$
(ie) $2x-3y+3z =1$ is the required equation of the plane .
answered Sep 3, 2015 by meena.p
 

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