# Finding the Intersection of a Line and Plane

1. Jan 11, 2014

### pandalolwut

Say you have a point on one surface. You know the normal vector of the surface at this point. You have a triangle somewhere else in space defined by it's three vertices. How do you find the intersection - if any - between the normal vector at the point on the surface with the triangle?

I found this article on Wikipedia, and I've been using the formula at the very bottom. Assuming it is correct - and please tell me if it is - I want to understand the process more. Could someone please explain to me what is going on here? What do each of the dot products yield and if you had to abstract this process into words what would you say?

More practically, here is the exact formula that I'm using right now; could someone check if it is correct?
I have the three vertices of the triangle defined as v1,v2,v3.
The normal vector of this triangle is norm.
The point on the surface is p1, and p2 is the point gained by adding the normal vector of the surface onto p1.
intPt is the intersection point.

Then am I right in saying that:
intPt = p1 + $\frac{(v1-p1) \bullet norm}{(p2-p1) \bullet norm}$ $\times$ (p2-p1)

Please tell me if anything needs to be clarified and thank you for any help at all in advance!

Last edited: Jan 11, 2014
2. Jan 16, 2014

### Office_Shredder

Staff Emeritus
If you have a plane with normal vector norm, then the plane is defined as the set of all vectors x such that
$$x\cdot norm = \alpha$$
where alpha is some number. If I start at a point p1 and I start moving in direction p2-p1, then my position looks like
$$p_1 + t(p_2-p_1)$$
as t varies, and what I am interested in what is
$$(p_1 + t (p_2-p_1))\cdot norm$$
and for what value of t is this equal to $\alpha$.

In this case we haven't been told what $\alpha$ is explicitly. But we know that $v_1$ is on the plane, so
$v_1 \cdot norm = \alpha$

From here you should be able to solve for t explicitly and get your final answer.