Question

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

2x + 3y – z = 5

Answer 1

2x + 3y ­− z = 5 … (1)

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

`:.sqrt((2)^2 + (3)^3 + (-1)^2) = sqrt(14)`

Dividing both sides of equation (1) by `sqrt14`, we obtain

This equation 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 `2/sqrt14, 3/sqrt14  and (-1)/sqrt(14)` the distance of normal from the origin is `5/sqrt14` units.