Question

Let $P$ be a point in the first octant, whose image $Q$ in the plane $x+y=3$ (that is, the line segment $PQ$ is perpendicular to the plane $x+y=3$ and the mid-point of $PQ$ lies in the plane $x+y=3$) lies on the $z-$axis. Let the distance of $P$ from the $x-$axis be $5.$ If $R$ is the image of $P$ in the $xy-$plane, then the length of $PR$ is

Answer 1

Let Point $P(\alpha ,\beta ,\gamma )$ and $R(\alpha ,\beta ,-\gamma )$
$Q(x,y,z)$
$\therefore \frac{x-\alpha }{1}=\frac{y-\beta }{1}=\frac{z-\gamma }{0}=\frac{-2(\alpha +\beta -3)}{2}$
$\Rightarrow x=3-\beta ,y=3-\alpha ,z=\gamma$
As $Q(3-\beta ,3-\alpha ,\gamma )$ lies on $z-$axis
$\therefore \alpha =3,\beta =3$
$\therefore P(3,3,\gamma )$ where distance of P from $x-$axis is $5$.
$\therefore \sqrt{{\beta }^2+{\gamma }^2}=5\Rightarrow \gamma =4$
$\therefore P(3,3,4),Q(0,0,4),R(3,3,-4)$
Hence, $PR=2\gamma =2\times 4=8.$