# 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.    (c)     x + y + z = 1

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given plane equation $x+y+z=1$.

The direction ratios of the normal of the plane are $1,\ 1,\and\ 1$.

Therefore $\sqrt{(1)^2+(1)^2+(1)^2} = \sqrt3$

So, now dividing both sides of the equation by $\sqrt3$ we will obtain,

$\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{1}{\sqrt3}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left ( \frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3} \right )$  or  $\left ( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right )$..

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