# In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

Question:

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) $z=2$

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

(c) $2 x+3 y-z=5$

(d) $5 y+8=0$

Solution:

(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)

The direction ratios of normal are 0, 0, and 1.

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

$0 . x+0 . y+1 . z=2$

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

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x+y+z=1$          ..(1)

The direction ratios of 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

$\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}$                $\ldots(2)$

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.

Therefore, the direction cosines of the normal are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$, and $\frac{1}{\sqrt{3}}$ and the distance of normal from the origin is $\frac{1}{\sqrt{3}}$ units.

(c) 2x + 3y ­− = 5 … (1)

The direction ratios of normal are 2, 3, and −1.

$\therefore \sqrt{(2)^{2}+(3)^{2}+(-1)^{2}}=\sqrt{14}$

Dividing both sides of equation (1) by $\sqrt{14}$, we obtain

$\frac{2}{\sqrt{14}} x+\frac{3}{\sqrt{14}} y-\frac{1}{\sqrt{14}} z=\frac{5}{\sqrt{14}}$

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.

Therefore, the direction cosines of the normal to the plane are $\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$, and $\frac{-1}{\sqrt{14}}$ and the distance of normal from the origin is $\frac{5}{\sqrt{14}}$ units.

(d) $5 y+8=0$

⇒ 0x − 5y + 0z = 8 … (1)

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

$\therefore \sqrt{0+(-5)^{2}+0}=5$

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

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

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.

Therefore, the direction cosines of the normal to the plane are $0,-1$, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.