Find the equation of the plane?

Question

Find the equation of the plane which contains the point (2; 1;−1) and is perpendicular to the planes 3x + y − z = 6 and 2x+2y+z=4.

Answer 1

Find the equation of the plane which contains the point
(2; 1;−1) and is perpendicular to the planes 3x + y − z = 6 and 2x+2y+z=4.

Since the plane is perpendicular to both given planes, its normal vector n, is perpendicular to the normal vectors of the given planes. Take the cross product of the normal vectors of the given planes.

3x + y − z = 6
2x + 2y + z = 4

n = <3, 1, -1> X <2, 2, 1> = <3, -5, 4>

With the normal vector and the point in the plane (2; 1;−1), we can write the equation of the desired plane.

3(x - 2) - 5(y - 1) + 4(z + 1) = 0
3x - 6 - 5y + 5 + 4z + 4 = 0
3x - 5y + 4z + 3 = 0