Question

A plane passes through a fixed point (a, b, c). Show that the locus of the foot of the perpendicular to it from the origin is the sphere x² + y² + z² – ax – by – cz = 0.

Answer 1

Let the equation of the plane be

Ax + By + Cz + D = 0 ...(i)

which is passing through (a, b, c).

So, Aa + Bb + Cc + D = 0 ...(ii)

Let P(α, β, γ) be the foot of the perpendicular from the origin to the plane (i). The direction ratios of OP are (α, β, γ).

Therefore, (α, β, γ) be the direction ratios of the same line OP. So,

From Eq. (ii) and (iii), we get

Hence, the locus of the foot of the perpendicular P is

x² + y² + z² – (ax + by + cz) = 0.