T/F: An equation of the plane containing the points (-8,3,3), (-18,5,1), (-8,-1,-3) is 2x + 3y + 2z - 2 =0 ?

Answer 1

False.

Define two vectors, u and v, that lie in the plane.

u = PQ = Q - P = <-18+8, 5-3, 1-3> = <-10, 2, -2>
v = PR = R - P = <-8+8, -1-3, -3-3> = <0, -4, -6>

The normal vector of the plane n, is the cross product of the two vectors u and v.

n = u X v = <-10, 2, -2> X <0, -4, -6> = <-20, -60, 40>

Any non-zero multiple of n is also a normal vector to the plane. Divide by -20.

n = <1, 3, -2>

With the normal vector and one point we can write the equation of the plane. Let's choose P(-8, 3, 3).

1(x + 8) + 3(y - 3) - 2(z - 3) = 0
x + 8 + 3y - 9 - 2z + 6 = 0
x + 3y - 2z + 5 = 0

This is not the given plane so the answer is false.

Answer 2

Plug in the points and see if they satisfy the equation. It is not that hard.