a line parallel to a plane will be perpendicular to the plane's normal.
So, pick a point in the plane, say (0,0,-4). Two other points might be (0,4,0) and (3,0,0)
That will give you two vectors in the plane,
u = 4j + 4k
and
v = 3i + 4k
Now the normal is w = u×v = 16i + 12j - 12k
Now, find the vector equation of your line. You want k such that
r•w = 0
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.
I got the cross product of (2,-2,0) and (4,3,-3) which is (6,6,14), but that's not adding up with the given equation.Where did I mess up?
2 answers
I see that I made it much too complicated.
You used the cross product, when you should have set the dot product = 0
You used the cross product, when you should have set the dot product = 0