Question

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin,
$2x+3y-z=5$

Answer 1

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