If the line drawn from (4, -1, 2) meets a plane at right angles at the point (-10,5,4), find the equation of the plane.

It means the plane passes through the point B (-10, 5, 4). Therefore the position vector of this point is,

And also given the line segment joining points A(4, -1, 2) and B (-10, 5, 4) and is at right angle to it.

Then

Position vector of - position vector of

We know that vector equation of a plane passing through 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

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