Show that the line is parallel to the plane
The direction of the line is: .
The normal of the plane is: .
If the line is parallel to the plane, then: .
Show that the line x = t(2 1 3) is parallel to the plane 3x - 3y - z = 2.
I'm not sure how to check if it's parallel. I converted the plane to parametric form, but the line doesn't seem to be a scalar multiple of either directional vectors of the plane. How do I solve this?
Here's another way to do it:
A line is parallel to a plane if and only if it does not intersect that plane. So what happens if you try to find where that line intersects the plane?
The line is x= 2t, y= t, z= 3t and the plane is 3x- 3y- z= 2. Putting those expressions for x, y, and z into the equation of the plane, 3(2t)- 3(t)- (3t)= 0t= 2. Obviously, that cannot be solved: the line and plane do not intersect. Of course the left hand side is essentially the dot product of the two vectors <2, 1, 3> and <3, 3, -1>, the one in the direction of the line, the other perpendicular to the plane. Since the two vectors have 0 dot product, they are perpendicular which means that the line is parallel to the plane.