Reduce the equation 2x – 3y – 6z = 14 to the normal form and hence find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Question

Philoid · Accepted Answer

The given equation of the plane is

2x – 3y – 6z = 14 ……(i)

Now,

Dividing (i) by 7, we get

……(ii)

The Cartesian Equation of the normal form of a plane is

lx + my + nz = p ……(iii)

where l, m and n are direction cosines of normal to the plane and p is the length of the perpendicular from the origin to the plane.

Comparing (ii) and (iii), we get

Direction cosine: l = , m = , n = and

Length of the perpendicular from the origin to the plane: p = 2.