Find the equation of a plane which is at a distance of 3√3 units from the origin and the normal to which is equally inclined to the coordinate axes.
Given :
To Find : Equation of plane
Formulae :
1) Distance of plane from the origin :
If is the vector normal to the plane, then distance of the plane from the origin is
Where,
2)
Where
3) Equation of plane :
If is the vector normal to the plane, then equation of the plane is
As
⇒ l = m = n
Therefore equation of normal vector of the plane having direction cosines l, m, n is
= √1
= 1
Now,
distance of the plane from the origin is
Therefore equation of required plane is
This is the required equation of the plane.