I need help with this problem: Topic: Lines and Planes (Calculus 3)

Question

Find the distance the point P(4, 1, -6), is to the plane through the three points

Q(2, 2, -4), R(-1, 4, -3), and S(6, 4, -2).

Answer 1

Vectors: QR = <- 3, 2, 1>, QS = <4, 2, 2>:

Normal vector to plane QRS: N = QR x QS = [(i j k) (-3 2 1 ) (4 2 2)] = 2i + 10j - 14k = <2 10 - 14>

Equation of Plane: 2(x - 2) + 10(y - 2) - 14(z + 4); x + 5y - 7z - 40 = 0

Normal equation of plane (x + 5y - 7z - 40)/√(1 + 25 + 49) = 0; (x + 5y - 7z - 40)/(5√3)

Distance D = (4 + 5 + 42)/(3√5) = 17/√5