Find the vector and Cartesian equations of the plane passing through the point (2, -1, 1) and perpendicular to the line having direction ratios 4, 2, -3.
Given :
A = (2, -1, 1)
Direction ratios of perpendicular vector = (4, 2, -3)
To Find : Equation of a plane
Formulae :
1) Position vectors :
If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,
2) Dot Product :
If 
are two vectors
then,
3) Equation of plane :
If a plane is passing through point A, then the equation of a plane is
Where, 
For point A = (2, -1, 1), position vector is
Vector perpendicular to the plane with direction ratios (4, 2, -3) is
Now, 
= 8 – 2 – 3
= 3
Equation of the plane passing through point A and perpendicular to vector 
is
As 
= 4x + 2y – 3z
Therefore, the equation of the plane is
4x + 2y – 3z = 3
Or
4x + 2y – 3z - 3 = 0