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## Homework Statement

Find the point(s) of intersection (if any) of the plane and the line. Also determine whether the line lies in the plane.

[tex]2x-2y+z=12, x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}[/tex]

## Homework Equations

[tex]2x-2y+z=12[/tex]

[tex]x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}[/tex]

## 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:

[tex]x=t+\frac{1}{2}[/tex]

[tex]y=-t-\frac{3}{2}[/tex]

[tex]z=2t-1[/tex]

I put the x, y, and z into the equation for the plane and solved for t:

[tex]2(t+\frac{1}{2})-2(-t-\frac{3}{2})+2t-1=12[/tex]

[tex]t=\frac{3}{2}[/tex]

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

[tex]\vec{l}\bullet\vec{n}= <2,-2,1>\bullet<1,-1,2> = 2(1) + (-2)(-1) + 1(2) = 6[/tex]

So the line does not lie in the plane.

Is this assumption and my math correct? If not, where did I go wrong?

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