#### notes

Consider a plane whose perpendicular distance from the origin is d (d ≠ 0). in following fig.

If `vec (ON)` is the normal from the origin to the plane, and `hat n ` is the unit normal vector along `vec (ON)`. Then `vec (ON)` = d . `hat n` . Let P be any point on the plane. Therefore , `vec (NP)` is perpendicular to `vec (ON)`.

Therefore, `vec (NP) . vec (ON) = 0` ...(1)

Let `vec r` be the position vector of the point P, then `vec (NP) = vec r - d . hat n` (as `vec (ON) + vec (NP) = vec (OP)`)

Therefore, (1) becomes

`(vec r - d . hat n) . d hat n = 0`

or `(vec r - d.hat n). hat n =0` (d ≠ 0)

or `vec r . hat n - d hat n . hat n = 0`

i.e., `vec r . hat n = d` (as `hat n . hat n = 1`) ...(2)

This is the vector form of the equation of the plane.

**Cartesian form**

Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane. Then

`vec (OP) = vec r = x hat i + y hat j + z hat k`

Let l, m, n be the direction cosines of `hat n` . Then

`hat n = l hat i + m hat j + n hat k `

Therefore, (2) gives

`(x hat i + y hat j + z hat k) . (l hat i + m hat j + n hat k) = d`

i.e. lx + my + nz = d ... (3)