# Equation of Plane in Normal Form

Here you will learn equation of plane in normal form with example.

Let’s begin –

## Equation of Plane in Normal Form

### (a) Vector Form

The vector equation of a plane normal to unit vector $$\hat{n}$$ and at a distance d from the origin is

$$\vec{r}.\hat{n}$$ = d

Remark 1 : The vector equation of ON is $$\vec{r}$$ = $$\vec{0}$$ + $$\lambda$$ $$\hat{n}$$ and the position vector of N is d$$\hat{n}$$ as it is at a distance d from the origin from the origin O.

### (b) Cartesian Form

If l, m, n are direction cosines of the normal to a given plane which is at a distance p from the origin, then the equation of the plane is

lx + my + nz = p

Note : The equation $$\vec{r}.\vec{n}$$ = d is in normal form if $$\vec{n}$$ is a unit vector and in such a case d on the right hand side denotes the distance of the plane from the origin. If $$\vec{n}$$ is not a unit vector, then to reduce the equation $$\vec{r}.\vec{n}$$ = d to normal form divide both sides by | $$\vec{n}$$ | to obtain

$$\vec{r}$$.$$\vec{n}\over |\vec{n}|$$ = $$d\over |\vec{n}|$$ $$\implies$$ $$\vec{r}.\hat{n}$$ = $$d\over |\vec{n}|$$

Example : Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector $$2\hat{i} + \hat{j} + 2\hat{k}$$.

Solution : Here, d = 8 and $$\vec{n}$$ = $$2\hat{i} + \hat{j} + 2\hat{k}$$

$$\therefore$$ $$\hat{n}$$ = $$\vec{n}\over |\vec{n}|$$ = $$2\hat{i} + \hat{j} + 2\hat{k}\over \sqrt{4 + 1 + 4}$$

= $${2\over 3}\hat{i} + {1\over 3}\hat{j} + {2\over 3}\hat{k}$$

Hence, the required equation of the plane is

$$\vec{r}$$.($${2\over 3}\hat{i} + {1\over 3}\hat{j} + {2\over 3}\hat{k}$$) = 8

[ By using $$\vec{r}.\hat{n}$$ = d ]

or, $$\vec{r}$$.($$2\hat{i} + \hat{j} + 2\hat{k}$$) = 24