Question

Find an equation of the plane through the point (6,3,2) and perpendicular to the vector <2, 1, 5>

Answer 1

Since the vector n = <3, -3, 1> is perpendicular to the plane it is the normal vector to the plane. The point P(-2, 2, 3) is also in the plane.

With the normal vector of the plane and a point in the plane we can write the equation of the plane. Remember, the normal vector of the plane is orthogonal to any vector that lies in the plane. And the dot product of orthogonal vectors is zero. Define R(x,y,z) to be an arbitrary point in the plane. Then vector PR lies in the plane.

n • PR = 0
n • <R - P> = 0
<3, -3, 1> • <x + 2, y - 2, z - 3> = 0
3(x + 2) - 3(y - 2) + 1(z - 3) = 0
3x + 6 - 3y + 6 + z - 3 = 0
3x - 3y + z + 9 = 0

3x - 3y + z = -9