Find an equation of a plane containing the three points 3 1

Solution

Direction vector from point (-1, 8, 0) to (2, 4, -3) = <3, -4, -3> Direction vector from point (-1, 8, 0) to (-1, 9, 2) = <0, 1, 2> Vector normal to plane is <3, -4, -3> x <0, 1, 2> = <-5, -6, 3> Vector -1 * <-5, -6, 3> = <5, 6, -3> is also normal to plane Using normal vector <5, 6, -3> and point (2, 4, -3), we get equation of plane: 5 (x-2) + 6 (y-4) - 3 (z+3) = 0 5x - 10 + 6y - 24 - 3z - 9 = 0 5x + 6y - 3z = 43