# Equation of A Plane

The equation of a plane in the three-dimensional space is defined with the normal vector and the known point on the plane. A Vector is a physical quantity that with its magnitude also has a direction attached to it. A normal vector means the line which is perpendicular to the plane. With reference to an origin, the position vector basically denotes the location or position (in a 3D Cartesian system) of a point. The Cartesian equation of a plane in 3 Dimensional space and vectors are explained in this article.

## Equation of a Plane in Three Dimensional Space

Generally, the plane can be specified using four different methods. They are:

• Two intersecting lines
• A line and point (not on a line)
• Three non-collinear points (Three points are not on the line)
• Two parallel and the non-coincident line
• The normal vector and the point

There are infinite planes that lie perpendicular to a specific vector. But only one unique plane exists to a specific point which remains perpendicular to the point while going through it

Let us consider a plane passing through a given point A having position vector   $\vec{a}$ and perpendicular to the vector  $\vec{N}$ . Let us consider a point P(x, y, z) lying on this plane and its position vector is given by  $\vec{r}$ as shown in the figure given below.

Position vector simply denotes the position or location of a point in the three-dimensional Cartesian system with respect to a reference origin.

For point P to lie on the given plane it must satisfy the following condition:

$\vec{AP}$  is perpendicular to $\vec{N}$  , i.e. $\vec{AP}$.$\vec{N}$ =0

From the figure given above it can be seen that,

$\vec{AP}$ = ( $\vec{r}$ – $\vec{a}$)

Substituting this value in  $\vec{AP}$$\vec{N}$ =0, we have ($\vec{r}$ – $\vec{a}$). $\vec{N}$ =0

This equation represents the vector equation of a plane.

We will assume that P, Q and R points are regarded as x1, y1, z1 and x2, y2, z2 in respectively to change the equation into the Cartesian system. A, B and C will be the assumed direction ratios. Thus,

$\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{a} = x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{N} = A\hat{i}+B\hat{j}+C\hat{k}$

Substituting these values in the vector equation of a plane, we have

$(\vec{r}-\vec{a}). \vec{N} = 0$

$(( x\hat{i}+y\hat{j}+z\hat{k})- (x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})). A\hat{i}+B\hat{j}+C\hat{k} = 0$

$[(x-x_{1})\hat{i}+ (y-y_{1})\hat{j}+ (z-z_{1})\hat{k}](A\hat{i}+B\hat{j}+C\hat{k})= 0$

$A(x-x_{1})+ B(y-y_{1})+ C(z-z_{1})= 0$

This gives us the Cartesian equation of a plane. To learn more about the equation of a plane in three dimensions and three-dimensional geometry download BYJU’S – The Learning App.