4  In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.     (b)     3y + 4z - 6 =  0

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

Given a plane equation $3y+4z-6=0$,

Or, $0x+3y+4z=6$

The direction ratios of the normal of the plane are $0,\ 3,\and\ 4$.

Therefore $\sqrt{(0)^2+(3)^2+(4)^2} = 5$

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

$0x+\frac{3}{5}y+\frac{4}{5}z = \frac{6}{5}$

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;

$\left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right )$  or  $\left ( 0, \frac{18}{25}, \frac{24}{25} \right )$.

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