Find the scalar equation for the plane that contains point (3,-3,0) and line x = 2, y=3+t, z=-4-2t?

Answer 1

The plane contains

Point P(3,-3,0) and
Line L defined by the parametric equations:
x = 2 + 0t
y = 3 + t
z = -4 - 2t

The line contains a directional vector.
v = <0,1,-2>

If we set t = 0 we can get a second point.
Q(2,3,-4)

Now we can get a second directional vector QP.
u = QP = <3-2,-3-3,0-(-4)> = <1,-6,4>

To get the normal vector n to the plane take the cross product of u and v.

n = u X v = <1,-6,4> X <0,1,-2> = 8i + 2j + k

Taking point P and n we can write the equation of the plane.

8(x - 3) + 2(y - (-3)) + 1(z - 0) = 0
8x - 24 + 2y + 6 + z + 0 = 0
8x + 2y + z - 18 = 0