What I have:

  • a plane given by its normal$\ n $ and a point on the plane$\ p $.
  • a 3D-point$\ v $.
  • a direction$\ d$.

What I need:

  • the projection of$\ v$ along$\ d$ onto the plane.

How can I calculate this? Thank you!

P.S.: I need to implement that in GLSL, in case this is of any information. I would be fine with the formula so that I can work out the implementation myself, anyways. :-)


1 Answer 1


To answer your question we just need to write it as linear algebra equations and solve them. Although your question doesn't state it, I assume that $v$ and $d$ are unit vectors. Let's call the projected point $x$.

First, because the projected point is in the direction $d$, we can write: $$\vec{vx} = \lVert\vec{vx}\rVert d $$

Second, because $p$ and $x$ are on the plane and $n$ is the plane normal, we can write: $$\vec{vx}\cdot n = \vec{vp}\cdot n$$

We can now work on the equation until we get a definition for $\vec{vx}$: $$\begin{align} \vec{vx}\cdot n & = \vec{vp}\cdot n \\ \lVert\vec{vx}\rVert d\cdot n & = \vec{vp}\cdot n \\ \lVert\vec{vx}\rVert & = \frac{\vec{vp}\cdot n}{d\cdot n} \\ \vec{vx} & = \frac{\vec{vp}\cdot n}{d\cdot n}d \\ \end{align}$$

Or written differently: $$x = v + \frac{(p-v)\cdot n}{d\cdot n}d$$

The result is undefined when the scalar product $d \cdot v$ is $0$, which happens when $d$ and $n$ are orthogonal, when $d=0$, or when $n=0$. The first case means $d$ is parallel to the plane and the projection doesn't have a point solution, and the two other cases mean that either the plane or the projection are not defined.


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