Question

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
$x+y+z=1$.

Answer 1

Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be $(x_1,y_1,z_1)$
$x+y+z=\mathrm{1........}\left(1\right)$
The direction ratios of the normal are $1,1$ and $1$
$\therefore$ $\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{3}$
Dividing both sides of equation (1) by $\sqrt{3}$, we obtain
$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$
This equation is of the form $lx+my+mz=d$, 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 coordinates of the foot of the perpendicular are
$(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}})$ i.e., $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$