Find the Cartesian and vector equations of a plane passing through the point (1, 2, 3) and perpendicular to a line with direction ratios 2, 3, -4.
Given :
A = (1, 2, 3)
Direction ratios of perpendicular vector = (2, 3, -4)
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 = (1, 2, 3), position vector is
Vector perpendicular to the plane with direction ratios (2, 3, -4) is
Now, 
= 2 + 6 – 12
= - 4
Equation of the plane passing through point A and perpendicular to vector 
is
As 
= 2x + 3y – 4z
Therefore, equation of the plane is
2x + 3y – 4z = -4
Or
2x + 3y – 4z + 4 = 0