Question

If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation

Answer 1

Direction ratios of the normal to the plane are 2, −1, 2 

∴ a = 2, b = –1 and c = 2

∴ Direction cosines are,

l = `2/sqrt(2^2 + (-1)^2 + 2^2)`

m = `(-1)/sqrt(2^2 + (-1)^2 + 2^2)`

n = `2/sqrt(2^2 + (-1)^2 + 2^2)`

∴ l = `2/3, "m" = (-1)/3, "n" = 2/3`

Cartesian equation of a plane which is at the distance of 6 units from the origin is

lx + my + nz = 6

∴ `2/3x - 1/3y + 2/3z` = 6

∴ 2x – y + 2z = 18