Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

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.

equation of a plane

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.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

DOWNLOAD

App NOW