Question

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?

Answer 1

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,
<b>u</b> = 4<b>j</b> + 4<b>k</b>
and
<b>v</b> = 3<b>i</b> + 4<b>k</b>
Now the normal is <b>w</b> = <b>u</b>×<b>v</b> = 16<b>i</b> + 12<b>j</b> - 12<b>k</b>

Now, find the vector equation of your line. You want k such that
<b>r</b>•<b>w</b> = 0

Answer 2

I see that I made it much too complicated.
You used the cross product, when you should have set the dot product = 0