# In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

Question:

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) $2 x+3 y+4 z-12=0$

(b) $3 y+4 z-6=0$

(c) $x+y+z=1$

(d) $5 y+8=0$

Solution:

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1y1z1).

$2 x+3 y+4 z-12=0$

$\Rightarrow 2 x+3 y+4 z=12$

The direction ratios of normal are 2, 3, and 4.

$\therefore \sqrt{(2)^{2}+(3)^{2}+(4)^{2}}=\sqrt{29}$

Dividing both sides of equation (1) by $\sqrt{29}$, we obtain

$\frac{2}{\sqrt{29}} x+\frac{3}{\sqrt{29}} y+\frac{4}{\sqrt{29}} z=\frac{12}{\sqrt{29}}$

This equation is of the form lx + my + nz = d, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are $\left(\frac{2}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}}\right)$ i.e., $\left(\frac{24}{29}, \frac{36}{49}, \frac{48}{29}\right)$.

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1y1z1).

$3 y+4 z-6=0$

$\Rightarrow 0 x+3 y+4 z=6$

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

$\therefore \sqrt{0+3^{2}+4^{2}}=5$

Dividing both sides of equation (1) by 5, we obtain

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

This equation is of the form lx + my + nz = d, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are

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

(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1y1z1).

… (1)

The direction ratios of the normal are 1, 1, and 1.

$\therefore \sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

Dividing both sides of equation (1) by $\sqrt{3}$, we obtain

$\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}$

This equation is of the form lx + my + nz = d, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are

$\left(\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\right)$ i.e., $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) .$

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1y1z1).

$5 y+8=0$

$\Rightarrow 0 x-5 y+0 z=8 \ldots(1)$

The direction ratios of the normal are 0, −5, and 0.

$-y=\frac{8}{5}$

This equation is of the form lx + my + nz = d, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are

$\left(0,-1\left(\frac{8}{5}\right), 0\right)$ i.e., $\left(0,-\frac{8}{5}, 0\right)$