Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x-3y+4z-6=0.$

Answer 1

Given equation of plane is

$2x-3y+4z-6=0$

$\Rightarrow 2x-3y+4z=6$

Let point $P(x_1,y_1,z_1)$ be foot of perpendicular from origin.


Direction ratios of the normal vector to the plane are

$2,-3,4$

Direction ratios of vector $\overrightarrow{OP}$ is

$x_1,y_1,z_1$

As vector $\overrightarrow{OP}$ is parallel to normal vector $\overrightarrow{n}$, so the direction ratios are proportional,

$\frac{x_1}{2}=\frac{y_1}{-3}=\frac{z_1}{4}=k$

$\Rightarrow x_1=2k,y_1=-3k,z_1=4k$

As this point lies on the plane, so

$2x_1-3y_1+4z_1-6=0$

$\Rightarrow 2\left(2k\right)-3(-3k)+4\left(4k\right)-6=0$

$\Rightarrow 4k+9k+16k=6$

$\Rightarrow k=\frac{6}{29}$

Therefore,

$x_1=2k=\frac{12}{29}$

$y_1=-3k=\frac{-18}{29}$

$z_1=4k=\frac{24}{29}$

Hence, the required coordinate of the foot of perpendicular are $(\frac{12}{29},\frac{-18}{29},\frac{24}{29})$