# Determine if projection of 3D point onto plane is within a triangle

In 3D, given three points $P_1$, $P_2$, and $P_3$ spanning a non-degenerate triangle $T$. How to determine if the projection of a point $P$ onto the plane of $T$ lies within $T$?

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $$\vec{u}=P_2-P_1$$, $$\vec{v}=P_3-P_1$$, $$\vec{n}=\vec{u}\times\vec{v}$$, $$\vec{w}=P-P_1$$. We then have directly the barycentric coordinates of the projection $$P'$$ of $$P$$ onto $$T$$ as

• $$\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$$
• $$\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$$
• $$\alpha=1-\gamma-\beta$$

The coordinates of the projected point is

• $$P'=\alpha P_1+\beta P_2 +\gamma P_3$$

The point $$P'$$ lies inside $$T$$ if

• $$0\leq\alpha\leq 1$$,
• $$0\leq\beta\leq 1$$, and
• $$0\leq\gamma\leq 1$$.
• Are there any corner cases where this should fail? The paper doesn't say anything to it Nov 29, 2017 at 11:41
• Doesn't this only work if point P already lies on the plane created by the triangle [P1,P2,P3]? Oct 25, 2018 at 8:59
• @MasterHD The point is projected onto the plane, so it does not have to lie on the plane. Oct 25, 2018 at 9:11
• Quick question... in the equations for gamma and beta, what on earth is this "[" and this "]". I imagine it must mean "signed magnitude" and explicitly not just magnitude (for the author would have written | or || , plus Craig's answer below codes it as such), but I can't find something online to confirm what [ and ] are meaning here. Dec 8, 2019 at 5:07
• @Xenial They are just the ordinary grouping brackets. Note that they are not necessary in this case. Dec 8, 2019 at 8:10

Thanks to Håkon Hægland for asking and answering this question, and especially for providing the key information from the hard-to-obtain paper: Wolfgang Heidrich, 2005, Computing the Barycentric Coordinates of a Projected Point, Journal of Graphics Tools, pp 9-12, 10(3).

I was web searching for exactly this information. Among the top hits were a question on the gamedev.stackexchange.com site, and this one. After implementing the technique presented here by Håkon, I posted that code and a link to this question at the game dev site: Easy way to project point onto triangle (or plane). Later it occurred that the C++/Eigen implementation might be useful to readers of this question:

bool pointInTriangle(const Eigen::Vector3f& query_point,
const Eigen::Vector3f& triangle_vertex_0,
const Eigen::Vector3f& triangle_vertex_1,
const Eigen::Vector3f& triangle_vertex_2)
{
// u=P2−P1
Eigen::Vector3f u = triangle_vertex_1 - triangle_vertex_0;
// v=P3−P1
Eigen::Vector3f v = triangle_vertex_2 - triangle_vertex_0;
// n=u×v
Eigen::Vector3f n = u.cross(v);
// w=P−P1
Eigen::Vector3f w = query_point - triangle_vertex_0;
// Barycentric coordinates of the projection P′of P onto T:
// γ=[(u×w)⋅n]/n²
float gamma = u.cross(w).dot(n) / n.dot(n);
// β=[(w×v)⋅n]/n²
float beta = w.cross(v).dot(n) / n.dot(n);
float alpha = 1 - gamma - beta;
// The point P′ lies inside T if:
return ((0 <= alpha) && (alpha <= 1) &&
(0 <= beta)  && (beta  <= 1) &&
(0 <= gamma) && (gamma <= 1));
}