equation of the plane?

Question

Let P_1 be the plane defined by the points:

P(-2,-3,-2), Q(-4,-1,-4) R(-1,-6,-5).

Find the equation of the plane perpendicular to P_1 and containing the line:

[x,y,z]^T = [ 3,0,0 ]^T + t [ -3,3,3 ]^T.

Your answer should be in the form Ax + By + Cz - D = 0.
The plane is: ? = 0

Answer 1

Plane P_1 is defined by the three points

P(-2,-3,-2), Q(-4,-1,-4) R(-1,-6,-5)

Create two vectors from the points.

PQ = = <-4+2, -1+3, -4+2> = <-2, 2, -2>
PR = = <-1+2, -6+3, -5+2> = <1, -3, -3>

The normal vector n1, to plane P_1 is the cross product of the vectors.

n1 = PQ X PR = <-2, 2, -2> X <1, -3, -3> = <-12, -8, 4>

Any non-zero multiple will also be a normal vector to the plane. Divide by -4.

n1 = <3, 2, -1>

The desired plane contains the normal vector n1, to plane P_1 and the directional vector v, of the line. Its normal vector n2, will be normal to both of them. Take the cross product.

n2 = n1 X v = <3, 2, -1> X <-3, 3, 3> = <9, -6, 15>

Any non-zero multiple will also be a normal vector to the plane. Divide by 3.

n2 = <3, -2, 5>

The normal vector n2 to the plane and a point on the plane
(3, 0, 0) is enough to write the equation of the plane.

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

Answer 2

It's hard, I'm only in Grade 9. Good luck though! =)