# Find the point of intersection of the plane and line. Determine if line lies in plane

by raytrace
Tags: determine, intersection, lies, line, plane, point
 P: 9 1. The problem statement, all variables and given/known data Find the point(s) of intersection (if any) of the plane and the line. Also determine whether the line lies in the plane. $$2x-2y+z=12, x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$ 2. Relevant equations $$2x-2y+z=12$$ $$x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$ 3. The attempt at a solution I can seem to find the point of intersection just fine. What I'm having difficulty figuring out is how to determine whether the line lies in the plane or not. I solved all the x, y, and z equations to t: $$x=t+\frac{1}{2}$$ $$y=-t-\frac{3}{2}$$ $$z=2t-1$$ I put the x, y, and z into the equation for the plane and solved for t: $$2(t+\frac{1}{2})-2(-t-\frac{3}{2})+2t-1=12$$ $$t=\frac{3}{2}$$ Plug t=3/2 back into the equations for x, y, and z and I end up with the point of intersection at (2, -3, 2) Now here is where I get stuck. Shoot, while typing this I may have just came up with the solution but I'll ask and see if anyone can confirm. Do I calculate the vector of the line and do a dot product to the normal of the plane. If the result is 0 then that means that the line lies on the plane? Therefore $$\vec{l}\bullet\vec{n}= <2,-2,1>\bullet<1,-1,2> = 2(1) + (-2)(-1) + 1(2) = 6$$ So the line does not lie in the plane. Is this assumption and my math correct? If not, where did I go wrong?