Question

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin,
$5y+8=0$

Answer 1

Given, $5y+8=0$
$\Rightarrow$ $0x-5y+0z=\mathrm{8.....}\left(1\right)$
The direction ratios of normal are $0,-5$ and $0$.
Now $\sqrt{0+{(-5)}^2+0}=5$
Therefore, the direction cosines of the normal to the plane are $0,-1$ and $0$

and the distance of plane from the origin is $=\left|\frac{5\left(0\right)+8}{\sqrt{0^2+5^2+0^2}}\right|=\frac{8}{5}$