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# Equation of A Plane In The Normal Form

In Geometry, the term “normal” defines an object such as a vector or line that should perpendicular to the given object. Here, you are going to have a look at the equation of a plane in the normal form. This can be determined if two things are known. One is normal to the plane and the other one is the distance of the plane from the origin. In this section, you will learn the equation of a plane in the vector as well as Cartesian form.

## Equation of a Plane in the Normal and Cartesian Form

The vector form of the equation of a plane in normal form is given by:

$$\begin{array}{l}\vec{r}.\hat{n} = d\end{array}$$

Where

$$\begin{array}{l}\vec{r}\end{array}$$
is the position vector of a point in the plane, n is the unit normal vector along the normal joining the origin to the plane and d is the perpendicular distance of the plane from the origin.

Let P (x, y, z) be any point on the plane and O is the origin. Then, we have,

$$\begin{array}{l}\vec{OP}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\end{array}$$

Now the direction cosines of

$$\begin{array}{l}\hat{n}\end{array}$$
as l, m and n are given by:

$$\begin{array}{l}\hat{n}= l\hat{i}+m\hat{j}+n\hat{k}\end{array}$$

From the equation

$$\begin{array}{l}\vec{r}.\hat{n}\end{array}$$
= d we get

$$\begin{array}{l}(x\hat{i}+y\hat{j}+z\hat{k}).(l\hat{i}+m\hat{j}+n\hat{k})= d\end{array}$$

Thus, the Cartesian form of the equation of a plane in normal form is given by:

lx + my + nz = d

### Equation of Plane in Normal Form Examples

An example is given here to understand the equation of a plane in the normal form.

Example 1:

A plane is at a distance of

$$\begin{array}{l}\frac{9}{\sqrt{38}}\end{array}$$
from the origin O. From the origin, its normal vector is given by 5
$$\begin{array}{l}\hat{i}\end{array}$$
+ 3
$$\begin{array}{l}\hat{j}\end{array}$$
– 2
$$\begin{array}{l}\hat{k}\end{array}$$
.

What is the vector equation for the plane?

Solution:

Let the normal vector be:

$$\begin{array}{l}\vec{n} = 5\hat{i}+3\hat{j}-2\hat{k}\end{array}$$

We now find the unit vector for the normal vector. It can be given by:

$$\begin{array}{l}\hat{n}=\frac{\vec{n}}{|\vec{n}|}\end{array}$$

$$\begin{array}{l}\hat{n}=\frac{5\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{25+9+4}}\end{array}$$

$$\begin{array}{l}\hat{n}=\frac{5\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{38}}\end{array}$$

So, the required equation of the plane can be given by substituting it in the vector equation is:

$$\begin{array}{l}\vec{r}.(\frac{5}{\sqrt{38}}\hat{i}+\frac{3}{\sqrt{38}}\hat{j} + \frac{-2}{\sqrt{38}}\hat{k}) = \frac{9}{\sqrt{38}}\end{array}$$

Example 2:

Find the cartesian equations for the following planes.

(a)

$$\begin{array}{l}\vec{r}.(\hat{i}+\hat{j}-\hat{k}) = 2\end{array}$$

(b)

$$\begin{array}{l}\vec{r}.(2\hat{i}+3\hat{j}-4\hat{k}) = 1\end{array}$$

Solution:

(a)

$$\begin{array}{l}\vec{r}.(\hat{i}+\hat{j}-\hat{k}) = 2\end{array}$$
—-(1)

We know that for any arbitrary point, P(x, y, z)on the plane, the position vector is given as:

$$\begin{array}{l}\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}\end{array}$$

Now, substitute the value of

$$\begin{array}{l}\vec{r}\end{array}$$
in equation (1), we get

$$\begin{array}{l} ( x\hat{i}+y\hat{j}+z\hat{k}) .(\hat{i}+\hat{j}-\hat{k}) = 2\end{array}$$

⇒ x + y – z = 2

Thus the cartesian equation of the plane is x + y – z = 2.

(b)

$$\begin{array}{l}\vec{r}.(2\hat{i}+3\hat{j}-4\hat{k}) = 1\end{array}$$
—-(2)

We know that for any arbitrary point, P(x, y, z)on the plane, the position vector is given as:

$$\begin{array}{l}\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}\end{array}$$

Now, substitute the value of

$$\begin{array}{l}\vec{r}\end{array}$$
in equation (2), we get

$$\begin{array}{l} ( x\hat{i}+y\hat{j}+z\hat{k}) .(2\hat{i}+3\hat{j}- 4\hat{k}) = 1\end{array}$$

⇒ 2x + 3y – 4z = 1

Thus the cartesian equation of the plane is 2x + 3y – 4z = 1.

To solve more examples and learn the concept by watching vibrant video lessons on them, download BYJU’S – The Learning App.