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.