Question

Distance of a Point from a Plane of three dimensional geometry.

Answer 1

Vector Form:

Let a point P whose position vector is \(\overrightarrow a\) and a plane π₁ whose equation is \(\overrightarrow r\) .n̂ = d.

Again, a plane π₂, through P parallel to the plane π₁

The unit vector normal to π₂ is n̂.

So, its equation is
\((\overrightarrow r - \overrightarrow a).\) n̂ = 0

i.e \(\overrightarrow r .\) n̂ = \(\overrightarrow a).\) n̂

Hence, distance of the plane PQ from origin.

ON' = |\(\overrightarrow a.\) n̂|

∴ Distance of the plane nx from P = |d - \(\overrightarrow a.\) n̂|

which is the length of the perpendicular from a point to the given line.

Remark:

(1) If the equation of the plane n₂ is in the form r .N = d, where N is normal to the plane, then

Cartesian Form :

Let equation of plane be ax + by + cz + d = 0 and P is a point outside the plane whose coordinates are (x₁ y₁ z₁). The coordinates of any point A lies on the plane are (x, y, z).

Perpendicular PN is drawn on the plane from the point P outside the plane.

But equation of the plane ax+ by + cz + d = 0

∴ ax + by + cz = - d

Putting these values in equation (2)

This is required distance.