## Equation of Plane In Normal Form

In Geometry, the term “normal” is a vector or a line that is perpendicular to the given object.

Here, you will understand the equation of a plane in normal form, which can be determined if two things are known. The first is normal to the plane and the second is the distance of the plane from the origin. In this section, you will learn the way to derive the equation of plane in normal form and equation of a plane in the vector as well as Cartesian form.

Also, you will find a variety of concepts on the normal plane equation to help you in a quick understanding of the normal form of a plane along with the normal vector from a plane equation.

## Equation of Planes

Let us consider a known point on the following plane as $P_0 = (x_0, y_0, z_0)$ and a plane lies orthogonal to the plane ‘$\vec{n} = (a, b, c)$’, where this vector is the normal vector.

Assuming another point P, i.e., $P = (x, y, z)$ lying on the plane. Following this, $\vec{r_0}$ and $\vec{r}$ are the position vectors of $\vec{P_0}$ and $\vec{P}$, respectively.

Please note that the added vector $\vec{r} - \vec{r_0}$ will always lie in the plane. Also, we observe that by putting the normal vector on the plane, we don’t get the actual reason to expect this to be the case. We just kept it here to illustrate the point. However, it is completely possible that the normal vector remains in a way that does not touch the plane in any way.

Since $\vec{n}$ is orthogonal or perpendicular to the plane and is also perpendicular to the vector that lies in the above plane, in general, it is orthogonal to $\vec{r} - \vec{r_0}$. Now, from the dot product, we know that the product of two orthogonal vectors will be zero.

In expression form:

$\vec{n} \cdot (\vec{r} - \vec{r_0})$

$\Rightarrow \vec{n} \cdot \vec{r} = \vec{n} \cdot \vec{r_0}….(1)$

Therefore, the above equation (1) is the vector equation of the drawn plane. This is how we can derive the equation of plane in normal form.

Now, let’s say a point $P(x, y, z)$ lies on a plane and the vector $\vec{OP}$ can be written as:

$\vec{OP} = x \hat{i} + y \hat{j} + z \hat{k}$ (normal plane equation)

Similarly, we can write $\vec{n}$ in the following manner to obtain the normal vector of a plane equation:

$\vec{n} = l \hat{i} + m \hat{j} + n \hat{k}$

Here, l, m, and n are the direction cosines of the unit vector. Now, the dot product of the above two vectors will be:

$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot l \hat{i} + m \hat{j} + n \hat{k} = d$

So, we get the normal vector from the plane equation.

## Point Normal Form of Plane From Normal Form

If an equation of a line in a cartesian form is $ax + by + cz = D$, then the point normal form of a plane is given as;

$bc \times (x - a) + ac \times (y - 0) + ab \times (z - 0) = 0$

Or

simply, you can write in the following form:

$bc x + ac y + ab z = abc$

Or

$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$.

## Equation of Plane From Normal Vector

The equation of plane in normal form is written as:

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

Here, $\vec{r}$ is the position vector of a point lying on OP in the above plane; $\hat{n}$ is a unit normal vector that lies parallel to the normal and joins the origin to the plane.

Besides this, the magnitude of a unit vector is unity, while ‘d’ is the perpendicular distance of the plane from “O.”

## Normal Form of a Plane

The concept of the normal form of a plane is integral to three-dimensional geometry. One of the significance of this topic is to comprehend what it means to express or write the equation of a plane in normal form.

You must note that for writing the equation of a plane in normal form, two things are needed – you must know the normal to the plane along with the distance of the plane from the origin. So now we can look at the method in which an equation of a plane is expressed in normal form. Also, we will go through the Vector form and the Cartesian form in the following sections as well.

Now, let us go through the example of the plane equation from normal vector:

## Plane Equation From Normal Vector

Find the equation of the plane perpendicular to the vector $\vec{n} = (2, 3, 6)$, which passes through point A (1, 5, 3).

To solve the above problem, assume that any point lies on the plane, say $\vec{OR}$ and $\vec{AR}$ is a kind of line segment that also lies on the plane.

Now, we must note that the above-mentioned line segment must be orthogonal to (2, 3, 6); this means that the dot product of $\vec{AR}$ (i.e., $\vec{OR} - \vec{AR}$) with (2, 3, 6) comes out to be zero.

So, now write the mathematical expression of the same:

$(2, 3, 6) \cdot (\vec{OR} - \vec{OA}) = 0$

$\vec{OR} \cdot (2, 3, 6) = \vec{OA} \cdot (2, 3, 6)$

Also, we see that in the question, we have $\vec{OA} = (1, 5, 3)$, now, putting this value in the above equation:

We notice that $\vec{OR} \cdot (2, 3, 6)$ equals to $(1, 5, 3) \cdot (2, 3, 6)$

Now, solving for $(1, 5, 3) \cdot (2, 3, 6)$:

$(1, 5, 3) \cdot (2, 3, 6) = (1 \times 2) + (3 \times 5) + (3 \times 6) = 2 + 15 + 18 = 35$

Hence, we get the plane equation from the normal vector as: $2x + 3y + 6z = 35$.

## Derive the Equation of a Plane in Normal Form

Assume that the normal plane equation is $5\hat{i} + 3\hat{j} - 2\hat{k}$ and is at a distance of $\dfrac{9}{\sqrt{38}}$ from the origin, find the plane equation from the normal vector.

Here, $\vec{n} = 5\hat{i} + 3\hat{j} - 2\hat{k}$

The unit vector is:

$\hat{n} = \dfrac{\vec{n}}{|\vec{n}|} = \dfrac{5\hat{i} + 3\hat{j} - 2\hat{k}}{\sqrt{5^2 + 3^2 + 2^2}$

So, $\hat{n} = \dfrac{5\hat{i} + 3\hat{j} - 2\hat{k}}{\sqrt{38}}$

Now, $\vec{r} \cdot \hat{n} = d$

Here, $d = \dfrac{9}{\sqrt{38}}$

$\vec{r} \cdot \left(\dfrac{5\hat{i} + 3\hat{j} - 2\hat{k}}{\sqrt{38}}\right) = \dfrac{9}{\sqrt{38}}$

Hence, this is the required equation of a plane.

The equation of a plane in normal form is applied in many fields such as Physics and mathematics. Here, we have learnt about the equation of a plane in normal form in great detail by understanding normal vectors and the equation of planes.

## Conclusion:

A plane's vector equation is $n \cdot (r - r_0) = 0$, where n is a normal vector to the plane, r is any position vector in the plane, and r0 is a given position vector in the plane.

The cross product of any two non-parallel vectors in the plane produces the normal vector n.

Two planes are parallel if and only if their normal vectors are parallel; otherwise, their intersection is a line. The line's direction is a vector that is orthogonal to the normal vectors of the planes. A point on the line can be found by solving the system of equations consisting of the plane equations, which can be done by setting one of the coordinates to zero.

The absolute value of the dot product of the unit normal vector $\dfrac{n}{|n|}$ and the vector between p and $r_0$ is the distance between a point p and a plane $n \cdot (r - r_0) = 0$.

## FAQs on Equation of Plane In Normal Form

**1. Determine that if a line passes through the point $P(2, - 4, 3)$ and is perpendicular to the plane $x + 4y - 2z = 5$ is normal.**

When the given line is perpendicular to the plane, the direction vector of the line lies parallel to the normal to the plane. When the plane is $x + 4y - 2z = 5$, this means that the plane is normal.

**2. What is the equation of a plane in 3D?**

We can define the equation of a line in 3D as the space that comprises three points, which don’t lie on the same line or by a point, and a normal vector or a vector lying perpendicular to the plane.

Also, we call it the vector equation of a plane. Besides this, the general equation of a plane in 3D space is written as:

$Px + Qy + Ry = D$

$P \cdot 0 + Q \cdot 0 + R \cdot 0 + D = 0$

Hence, we get $S = 0$.

**3. State the general equation of a plane.**

The generation equation for the vector of a plane and a point passing through the plane is $(x – x_1) + b (y – y_1) + c (z – z_1) = 0.