Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y+8=0$.

Answer 1

Let the coordinates of the foot of perpendicular $P$ from the origin to the given plane be $(x_1,y_1,z_1)$.
We have, $5y+8=0$
$\Rightarrow$ $0x-5y+0z=\mathrm{8......}\left(1\right)$
The direction ratios of the normal are $0,-5$ and $0$
$\therefore$ $\sqrt{0+{\left(5\right)}^2+0}=5$
Dividing both sides of equation (1) by $5$, we obtain
$-y=\frac{8}{5}$
This equation is of the form $lx+my+nz=d$, where $l.m.n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by $(ld.md,nd)$
Therefore, the given coordinates of the foot of the perpendicular are
$(0,-1\left(\frac{8}{5}\right),0)$ i.e., $(0,-\frac{8}{5},0)$