Question

Find the equation of plane which is at a distance of 4 units from the origin and which is normal to the vector `2hati - 2hatj + hatk`.

Answer 1

The normal to the plane is `2hati - 2hatj + hatk`

The d.r.s. of the normal are 2, –2, 1

∴ `sqrt(4 + 4 + 1) = sqrt(9)` = 3

∴ d.c.s. of the normal are `2/3, (-2)/3, 1/3`

Now equation of plane in normal form is ℓx + my + nz = p where ℓ, m, n are the d.c.s. of the normal from the origin and p is the distance of the plane from the origin.

Here p = 4, 

ℓ = `2/3`,

m = `(-2)/3`,

n = `1/3`

∴ Equation of plane is

`(2x)/3 - (2y)/3 + z/3` = 4

i.e. 2x – 2y + z = 12