# 4  In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.    (d)      5y + 8 = 0

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 $5y+8=0$.

or written as $0x-5y+0z=8$

The direction ratios of the normal of the plane are $0,\ -5,\and\ 0$.

Therefore $\sqrt{(0)^2+(-5)^2+(0)^2} = 5$

So, now dividing both sides of the equation by $5$ we will obtain,

$-y=\frac{8}{5}$

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;

$\left ( 0,-1(\frac{8}{5}),0 \right )$  or  $\left ( 0,\frac{-8}{5},0 \right )$.

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