Question

Determine the direction cosines of the normal to plane and the distance from the origin:

2x +3y -z = 5

Answer 1

Given, equation of plane is 2x +3y -z = 5

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

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

Dividing both sides of Eq. (i) by $\sqrt{14}$, we obtain

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

Which 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

$\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}and-\frac{1}{\sqrt{14}}$ and the distance of normal to the origin is $\frac{5}{\sqrt{14}}$ units.