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