Question

(1) Find the coordinates of the foot of the perpendicular drawn from the origin to :
(i) $2x+3y+4z-12=0$

(2) Find the coordinates of the foot of the perpendicular drawn from the origin to :
(ii) $3y+4z-6=0$

(3) Find the coordinates of the foot of the perpendicular drawn from the origin to :
(iii) $x+y+z=1$

(4) Find the coordinates of the foot of the perpendicular drawn from the origin to :
(iv) $5y+8=0$

Answer 1

(1) Given equation of plane is

$2x+3y+4z-12=0$
Let the foot of perpendicular from origin be $P(x_1,y_1,z_1)$

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

Direction ratios are : $x_1,y_1,z_1$

Normal vector, $\overrightarrow{n}=2\hat{i}+3\hat{j}+4\hat{k}$

Direction ratios are : $2,3,4$

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

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

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

As the point $P(x_1,y_1,z_1)$ lies on the plane, so

$2\left(2k\right)+3\left(3k\right)+4\left(4k\right)-12=0$

$\Rightarrow 4k+9k+16k=12$
$\Rightarrow 29k=12$
$\Rightarrow k=\frac{12}{29}$

$\therefore x_1=2k=\frac{24}{29}$

$y_1=3k=\frac{36}{29}$

$z_1=4k=\frac{48}{29}$

Hence, required coordinates of foot of perpendicular is : $(\frac{24}{29},\frac{36}{29},\frac{48}{29})$

(2) Given equation of plane is

$3y+4z-6=0$
Let the foot of perpendicular from origin be $P(x_1,y_1,z_1)$

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

Direction ratios are : $x_1,y_1,z_1$

Normal vector, $\overrightarrow{n}=0\hat{i}+3\hat{j}+4\hat{k}$

Direction ratios are : $0,3,4$

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

$\therefore \frac{x_1}{0}=\frac{y_1}{3}=\frac{z_1}{4}=k$

$\Rightarrow x_1=0,y_1=3k,z_1=4k$

As the point $P(x_1,y_1,z_1)$ lies on the plane, so

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

$\Rightarrow 9k+16k=6$
$\Rightarrow k=\frac{6}{25}$
$\therefore x_1=0$

$y_1=3k=\frac{18}{25}$

$z_1=4k=\frac{24}{25}$

Hence, required coordinates of foot of perpendicular is : $(0,\frac{18}{25},\frac{24}{25})$

(3)Given equation of plane is

$x+y+z=1$
Let the foot of perpendicular from origin be $P(x_1,y_1,z_1)$

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

Direction ratios are : $x_1,y_1,z_1$

Normal vector, $\overrightarrow{n}=\hat{i}+\hat{j}+\hat{k}$

Direction ratios are : $1,1,1$

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

$\therefore \frac{x_1}{1}=\frac{y_1}{1}=\frac{z_1}{1}=k$

$\Rightarrow x_1=k,y_1=k,z_1=k$

As the point $P(x_1,y_1,z_1)$ lies on the plane, so

$k+k+k=1$

$\Rightarrow k=\frac{1}{3}$
$\therefore x_1=k=\frac{1}{3}$

$y_1=k=\frac{1}{3}$

$z_1=k=\frac{1}{3}$

Hence, required coordinates of foot of perpendicular is : $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$

(4)Given equation of plane is : $5y+8=0$
$0x+5y+0z=-8$

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

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

Direction ratios are : $x_1,y_1,z_1$

Normal vector, $\overrightarrow{n}=0\hat{i}+5\hat{j}+0\hat{k}$

Direction ratios are : $0,5,0$

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

$\therefore \frac{x_1}{0}=\frac{y_1}{5}=\frac{z_1}{0}=k$

$\Rightarrow x_1=0,y_1=5k,z_1=0$

As the point $P(x_1,y_1,z_1)$ lies on the plane, so

$0+5\left(5k\right)+0+8=0$

$\Rightarrow k=\frac{-8}{25}$
$\therefore x_1=0$

$y_1=5k=-\frac{8}{5}$

$z_1=0$

Hence, required coordinates of foot of perpendicular is : $(0,-\frac{8}{5},0)$