Question

Find the equations of the planes parallel to the plane x – 2y + 2z – 4 = 0 which is a unit distance from the point (1, 2, 3).

Answer 1

Let the equation of plane parallel to the given plane be x – 2y + 2z + λ = 0  ....(1)

Here `barn = hati - 2hatj + 2hatk`,

∴ `hatn = (hati - 2hatj + 2hatk)/3`

Equation of plane in normal form is

`barr*hatn` = p,

i.e. `barr*((hati - 2hatj + 2hatk))/3 = -λ/3`

P.v. of given point = `bara = hati + 2hatj + 3hatk`

∴ `bara*hatn = (1 - 4 + 6)/3` = 1

∵ Distance of the plane from the given point is 1  ...(Given)

∴ `|p - (bara*hatn)|` = 1

∴ `|-λ/3 - 1|` = 1

∴ –λ – 3 = ± 3,

∴ λ = – 6 or 0

∴ From (1), equations of planes are

x – 2y + 2z = 0

or x – 2y + 2z – 6 = 0