Question

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

x + y + z = 1

Answer 1

Given, plane is x +y +z=1

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

Also, $\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Dividing both sides of Eq. (i) by $\sqrt{3}$, we obtain

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$

Which is of the form lx +my +nz =d, where l,m,n are 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.