Linear Algebra Problem: How do I find the plane which...?

Question

The Plane Ax+By+Cz=36 contains the vector the planes of
3x+2y-3z=-3
2x+1y+3z=-11
and the point
(0, -6, -1)
How would I go about finding A, B, and C?

Answer 1

I assume you mean the plane contains the * normal * vectors of the planes. It is the only unique vector to a plane.

The normal vectors of the two given planes are:

u = <3, 2, -3>
v = <2, 1, 3>

Since these vectors lie in the desired plane, the normal vector of the desired plane is perpendicular to both of them. Take the cross product.

n = u X v = <3, 2, -3> X <2, 1, 3> = <9, -15, -1>

With the normal vector n, of the desired plane and a point on the plane we can write the equation of the desired plane.

9(x - 0) - 15(y + 6) - 1(z + 1) = 0
9x - 15y - 90 - z - 1 = 0
9x - 15y - z - 91 = 0

Answer 2

I'm not this far along in math... but it might have something to do with dot products or matrices?