Find the vector and Cartesian equations of a plane which is at a distance of 6 units from the origin and which has a normal with direction ratios 2, -1, -2.
Given :
d = 6
direction ratios of are (2, -1, -2)
To Find : Equation of 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 vector equation of the plane.
Now,
= (x × 2) + (y × (-1)) + (z × (-2))
= 2x - y – 2z
Therefore equation of the plane is
This is Cartesian equation of the plane.