Question

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

Answer 1

The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)

The direction ratios of normal are 0, 0, and 1.

`:.sqrt(0^2 + 0^2 + 1^2)  = 1`

Dividing both sides of equation (1) by 1, we obtain

0.x + 0.y + 1.z = 2

This 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 the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.