Question

Find the coordinates of the foot of the perpendicular drawn from the origin.
$x+y+z=1$

Answer 1

Given equation of the plane $x+y+z=1$

Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $\left(x_1,y_1,z_1\right)$

The direction cosines of the plane are $(1,1,1)$

Therefore,

$\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{1+1+1}=\sqrt{3}$

Divide equation $\left(1\right)$ on both sides by $\sqrt{3}$

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$

Therefore, the coordinates of the foot of the perpendicular are

$\begin{array}{c}\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}\right)\\ \Rightarrow \left(\frac{1}{3}\right),\left(\frac{1}{3}\right),\left(\frac{1}{3}\right)\end{array}$

Hence, this is the answer.