Question

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

(i) $z=2$

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

(iii) $2x+3y-5=0$

(iv) $5y+8=0$

Answer 1

(i) The equation of the plane is $z=2$ or $0x+0y+z=\mathrm{2....}\left(1\right)$

The direction ratios of normal are $0,0$ and $1$
$\therefore$ $\sqrt{0^2+0^2+1^2}=1$
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 $l,m,n$ are the direction cosines of normal to the plane and $d$ is the distance of the perpendicular drawn from the origin.
Therefore, the direction cosines are $0,0,1$ and the distance of the from the origin is $2$ units.

(ii) $x+y+z=\mathrm{1......}\left(1\right)$

The direction ratios of normal are $1,1$ and $1$.
$\therefore$ $\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{3}$
Dividing both sides of equation (1) by $\sqrt{3}$, we obtain
$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}.....\left(2\right)$
This equation is of the form $lx+my+nz=d$, where $l,m,n$ 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.

(iii) $2x+3y-z=\mathrm{5.........}\left(1\right)$

The direction ratios of normal are $2,3$ and $-1$
$\therefore$ $\sqrt{{\left(2\right)}^2+{\left(3\right)}^2+{(-1)}^2}=\sqrt{14}$
Dividing both sides 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 of the form $lx+my+nz=d$, where $l,m,n$ 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

(iv) $5y+8=0$

$\Rightarrow$ $0x-5y+0z=\mathrm{8.....}\left(1\right)$
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 $l,m,n$ 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.