Find the distance the point P(-2, 2, 3), is to the plane through the three points Q(-1, 0, 2), R(1, -2, 5), and S(-3, -2, 4).

Question

Find the distance the point P(-2, 2, 3), is to the plane through the three points 

Q(-1, 0, 2), R(1, -2, 5), and S(-3, -2, 4). 

Answer 1

Let get vectors: QR = <2, - 2, 3>; QS = <- 2, - 2, 2>

Normaql vector to plane N = <a, b, c>. N·QR = 2a - 2b + 3c = 0; N·QS = - 2a - 2b + 2c = 0

c = a + b; 2a - 2b + 3a + 3b = 0; b = - 5a; c = a - 5a = - 4a.

So, normal vector N = <1, - 5, - 4> (we set a = 1>.

Equation of plane: 1(x + 1) - 5(y - 0) - 4(z - 2) = 0;

x - 5y - 4z + 9 = 0

Q(-1, 0, 2), R(1, -2, 5), and S(-3, -2, 4). 

Now, normal equation of plane: √1 + 25 + 16 = √42;

(x - 5y - 4z + 9)/(- √42). So, distance from P(- 2, 2, 3) to this plane d = I(- 2 - 10 - 12 + 9)/(- √42)I = 15/√42