Question

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

5y + 8 = 0

Answer 1

Given equation is 5y + 8 = 0 , which can be written as

$0x+(-1)y+0z=\frac{8}{5}$

Which 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 distance of normal from the origin is 8/5 units.