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  

a→
and perpendicular to the vector  
N→
. Let us consider a point P(x, y, z) lying on this plane and its position vector is given by  
r→
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:

AP→
 is perpendicular to
N→
 , i.e.
AP→
.
N→
=0

From the figure given above it can be seen that,

AP→
= (
r→
– 
a→
)

Substituting this value in  

AP→
N→
=0, we have (
r→
– 
a→
). 
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,

r→=xi^+yj^+zk^

 

a→=x1i^+y1j^+z1k^

 

N→=Ai^+Bj^+Ck^

 

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

(r→−a→).N→=0

 

((xi^+yj^+zk^)−(x1i^+y1j^+z1k^)).Ai^+Bj^+Ck^=0

 

[(x−x1)i^+(y−y1)j^+(z−z1)k^](Ai^+Bj^+Ck^)=0

 

A(x−x1)+B(y−y1)+C(z−z1)=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.

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