Question

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x-3y+4z=6$ is?

Answer 1

Assume a point $P(x_1,y_1,z_1)$ on the given plane

Since perpendicular to plane is parallel to normal vector

Vector $\overrightarrow{OP}$ is parallel to normal vector $\overrightarrow{n}$ to the plane.

Given equation of plane is

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

Since $\overrightarrow{OP}$ and $\overrightarrow{n}$ are parallel, their direction ratios are proportional.

$\overrightarrow{OP}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

Direction rations $=x_1,y_1,z_1$

$\therefore a_1=x_1,b_1=y_1,c_1=z_1$

$\overrightarrow{n}=2\hat{i}-3\hat{j}+4\hat{k}$

Direction rations $=2,-3,4$

$\therefore a_2=2,b_2=-3,c_2=4$

Direction ratios are proportional

So, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$

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

$x_1=2k,y_1=3k,z_1=4k$

Also , point $P(x_1,y_1,z_1)$ lies in the plane

Putting $P(2k,-3k,4k)$

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

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

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

$29k=6$

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

So, $x_1=2k=2\times \frac{6}{29}=\frac{12}{29}$

$y_1=-3k=-3\times \frac{6}{29}=\frac{-18}{29}$

$z_1=4k=4\times \frac{6}{29}=\frac{24}{29}$

Therefore, coordinates of foot of perpendicular are $(\frac{12}{29},\frac{-18}{29},\frac{24}{29})$