find the vector equation of a plane which is defined by A102

find the vector equation of a plane which is defined by A(-1,0,2) B(2,3,5) C(2,4,6)

We can get two vectors in the plane by subtracting pairs of points in the plane:

A - B = ( -3 , -3, -3) ; C - B = ( 0 , 1 , 1)

cross product of these two vectors will be in the unique direction orthogonal to both, and hence in the direction of the normal vector to the plane.

( -3 , -3, -3) x ( 0 , 1 , 1) = (0, 3, -3)

The equation of a plane is ax+by+cz = d, where

plug a=0 ; b = 3; c = -3 ; 3y -3z =d

Plug (2, 3, 5) to find d:

3*3 -5*3 = d ; d = -6

So, 3y -3z = -6 (equation of Plane)