How on earth....Vectors are hard?

Question

Please show that the equation of the plane that contains three given points, whose position vectors are A,B,and C, can be written in symetrical form

(3r-A-B-C).(AxB+BxC+CxA)=0

Answer 1

First note that (A-B)x(A-C) is a vector that is orthogonal to the plane.

(A-B)x(A-C) = Ax(A-C) - Bx(A-C)

= AxA - AxC - BxA + BxC

= 0 + CxA + AxB + BxC

= AxB + BxC + CxA

Now, (r-A), (r-B), and (r-C) are all vectors parallel to the plane, so if we sum them, we'll get a vector that is parallel to the plane.  That sum is: (3r-A-B-C).

Since (3r-A-B-C) is parallel to the plane, and AxB+BxC+CxA is orthogonal to the plane, they're orthogonal to one another and therefore their dot product is zero.