Find the vector and Cartesian equation of the plane which passes through the point (5, 2, -4) and perpendicular to the line with direction ratios 2, 3, -1.

Given: The plane is passing through P(5, 2, -4) and perpendicular to the line having 2, 3, -1 as the direction ratios.

To find: the equation of the plane

Let the position vector of this point P be

And it is also given the plane is normal having 2, 3, -1 as the direction ratios.

Then

We know that the vector equation of a plane passing through the point and perpendicular/normal to the vector is given by

Substituting the values from eqn(i) and eqn(ii) in the above equation, we get

(by multiplying the two vectors using the formula )

is the vector equation of a required plane.

Let

Then, the above vector equation of the plane becomes,

Now multiplying the two vectors using the formula, we get

2x + 3y - z = 20

This is the Cartesian form of the equation of the required plane.