Determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = kt is parallel to the plane with equation 4x + 3y – 3z -12 = 0.
r = <x, y, z> = <2 + 3t, -2 + 5t, kt>
Two points on the line r are
P = <2, -2, 0> (t = 0)
Q = <5, 3, k> (t = 1)
The vector PQ is parallel to the line r, and is given by
v = PQ = OQ – OP = < 3, 5, k>
The normal to the plane 4x + 3y – 3z = 12 is given by n = <4, 3, -3>
If the line r is parallel with the plane, then v is orthogonal with n, which means that their dot product is zero. i.e. n•v = 0.
Therefore, <4, 3, -3>•<3, 5, k> = 0
12 + 15 – 3k = 0
27 = 3k
k = 9