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Q.
The image of the point with position vector $i +3 k$ in the plane $r \cdot( i + j + k )=1$ is
ManipalManipal 2011
Solution:
Let $Q$ be the image of the point $P( i +3 k )$ in the plane $r \cdot( i + j + k )=1 .$ Then, $P Q$ is normal to the plane. Since, $P Q$ passes through $P$ and in normal to the given plane, therefore equation of $P Q$ is $r =( i +3 k )+\lambda( i + j + k )$
Since, $Q$ lies on the line $P Q$, so let the position vector of
$Q$ be $( i +3 k )+\lambda( i + j + k )$
i.e. $(1+\lambda) i +\lambda j +(3+\lambda) k$
Since, $R$ is the mid-point of $P Q$, therefore position vector of $R$ is
$\frac{(1+\lambda) i +\lambda\, j +(3+\lambda) k + i +3 k }{2}$
$=\left(\frac{\lambda+2}{2}\right) i +\frac{\lambda}{2} j +\left(\frac{6+\lambda}{2}\right) k$
$=\left(\frac{\lambda}{2}+1\right) i +\frac{\lambda}{2} j +\left(3+\frac{\lambda}{2}\right) k$
Since, $R$ lies on the plane
$r \cdot( i + j + k )=1$
$\therefore \left[\left(\frac{\lambda}{2}+1\right) i +\frac{\lambda}{2} j +\left(3+\frac{\lambda}{2}\right) k \right]$
$\cdot( i + j + k )=1$
$\Rightarrow \left[\frac{\lambda}{2}+1+\frac{\lambda}{2}+3+\frac{\lambda}{2}\right]=1$
$\Rightarrow \lambda=-2$
Hence, the position vector of $Q$ is $- i -2\, j + k$.
