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 us find equation of the plane the three points , Q , R, and S define.

Vector RQ = < -2, 2, -3 > and RS = < -4, 0, -1> and their cross product

RQ x RS = < -2, 2, -3 > x < -4, 0, -1> = < -1 , 5, 4> = n is a vector normal to the plane.

Then the equation of the plane is ( -1 )( x +1 ) + 5y +4 ( x-2) = 0

or equivalently - x +5y +4z = 9.

Now let us take a point on the plane say Ω ( -9, 0, 0)

Then vector ΩP = <-7, 3, 3 >.

Hence the distance of point P from the plane is the length of the projection of vector ΩP on the normal vector of the plane n.

d = || proj _n ΩP|| = |n• ΩP| / || n || = 34 / √42