Find the vector and Cartesian equations of a plane which is at a distance of from the origin and whose normal vector from the origin is
Given :
To Find : Equation of a plane
Formulae :
1) Unit Vector :
Let be any vector
Then the unit vector of is
Where,
2) Dot Product :
If are two vectors
then,
3) Equation of plane :
Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is
Where,
For given normal vector
Unit vector normal to the plane is
Equation of the plane is
This is a vector equation of the plane.
Now,
= (x × 2) + (y × (-3)) + (z × 4)
= 2x - 3y + 4z
Therefore equation of the plane is
2x - 3y + 4z = 6
This is the Cartesian equation of the plane.