Question

In the following case, find the coordinates of the foot of the perpendicular drawn from the origin:
3y +4z -6 =0

Answer 1

Given plane is 0x + 3y +4z -6 =0 ...(ii)

$\therefore$ Dr's of any line perpendicular to plane (ii) are 0,. 3,4.

Hence, equation of the line through origin and at right angles to plane (ii) are $\frac{x-0}{0}=\frac{y-0}{3}=\frac{z-0}{4}$

Any point on this line is (0,3t,4t). This point lies in the plane (ii)

If $0+3\times \left(3t\right)+4\times \left(4t\right)-6=0i.e.,t=\frac{6}{25}$

Hence, the required foot of the perpendicular is

$(0+3\times \frac{6}{25},4\times \frac{6}{25})$ or $(0,\frac{18}{25},\frac{24}{25})$