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# 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

$$\begin{array}{l}\vec{a} \end{array}$$
and perpendicular to the vector
$$\begin{array}{l}\vec{N} \end{array}$$
. Let us consider a point P(x, y, z) lying on this plane and its position vector is given by
$$\begin{array}{l}\vec{r} \end{array}$$
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:

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

From the figure given above it can be seen that,

$$\begin{array}{l}\vec{AP} \end{array}$$
= (
$$\begin{array}{l}\vec{r} \end{array}$$
–
$$\begin{array}{l}\vec{a} \end{array}$$
)

Substituting this value in

$$\begin{array}{l}\vec{AP} \end{array}$$
$$\begin{array}{l}\vec{N} \end{array}$$
=0, we have (
$$\begin{array}{l}\vec{r} \end{array}$$
–
$$\begin{array}{l}\vec{a} \end{array}$$
).
$$\begin{array}{l}\vec{N} \end{array}$$
=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,

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

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

$$\begin{array}{l}\vec{N} = A\hat{i}+B\hat{j}+C\hat{k}\end{array}$$

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

$$\begin{array}{l}(\vec{r}-\vec{a}). \vec{N} = 0\end{array}$$

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

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

$$\begin{array}{l}A(x-x_{1})+ B(y-y_{1})+ C(z-z_{1})= 0\end{array}$$

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.