Question

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

5y + 8 = 0

Answer 1

5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of normal are 0, −5, and 0.

`:. sqrt(0+(-5)^2 + 0)` = 5

Dividing both sides of equation (1) by 5, we obtain

`-y = 8/5`

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 0, −1, and 0 and the distance of normal from the origin is `8/5` units.