#### notes

Consider a point P with position vector `vec a` and a plane `π_1` whose equation is `vec r . hat n = d` Fig.

Consider a plane `π_2` through P parallel to the plane `π_1`. The unit vector normal to `π_2` is

`hat n`.

Hence , its equation is `(vec r - vec a) . hat n = 0`

i.e., `(vec r . hat n = vec a . hat n)`

Thus, the distance ON′ of this plane from the origin is `|vec a . hat n|.` Therefore , the distance PQ from the plane `π_1` is above fig.

i.e. `ON - ON' = |d - vec a. hat n|`

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

**Cartesian form**

Let `P(x_1, y_1, z_1)` be the given point with position vector`vec a`

and Ax + By + Cz = D

be the Cartesian equation of the given plane. Then

`vec a = x_1 hat i + y_1 hat j + z_1 hat k`

`vec N = A hat i + B hat j + C hat k`

Hence, the perpendicular from P to the plane is

`|((x_1 hat i + y_1 hat j + z_1 hat k) . (A hat i + B hat j + C hat k) - D)/ sqrt (A^2 + B^2 +C^2)|`

`=|(Ax_1 + By_1 + Cz_1 - D)/sqrt(A^2 + B^2 +C^2)|`