I previously showed the derivation of how to determine the intersection of a plane and a cone. At the time I had to solve that equation, so after doing so I decided to publish it for anyone to use. Given the positive feedback, it seems this was useful, so I might as well continue with a few more.

Here is probably the most basic intersection: a ray and a plane. Solving it is straightforward, which I hope can be seen below. Like last time, I am using vector notation.

1. We define a ray with its origin $O$ and its direction as a unit vector $\hat{D}$.
   Any point $X$ on the ray at a signed distance $t$ from the origin of the ray verifies: $\overrightarrow{X}=\overrightarrow{O}+t\overrightarrow{D}$.
   When $t$ is positive $X$ is in the direction of the ray, and when $t$ is negative $X$ is in the opposite direction.
2. We define a plane with a point $S$ on that plane and the normal unit vector $\hat{N}$, perpendicular to the plane.
   The distance between any point $X$ and the plane is $d=|(\overrightarrow{X}–\overrightarrow{S})\cdot \overrightarrow{N}|$. If this equality is not obvious for you, you can think of it as the distance between $X$ and $S$ along the $\overrightarrow{N}$ direction. When $d=0$, it means $X$ is on the plane itself.
3. We define $P$ the intersection or the ray and the plane, and which we are interested in finding.

Since $P$ is both on the ray and on the plane, we can write: $$\{\begin{array}{l}\overrightarrow{P}=\overrightarrow{O}+t\overrightarrow{D}\\ |(\overrightarrow{P}–\overrightarrow{S})\cdot \overrightarrow{N}|=0\end{array}$$ Because the distance $d$ from the plane is $0$, the absolute value is irrelevant here. We can just write: $$\{\begin{array}{l}\overrightarrow{P}=\overrightarrow{O}+t\overrightarrow{D}\\ (\overrightarrow{P}–\overrightarrow{S})\cdot \overrightarrow{N}=0\end{array}$$ All we have to do is replace $P$ with $\overrightarrow{O}+t\overrightarrow{D}$ in the second equation, and reorder the terms to get $t$ on one side.
$$(\overrightarrow{O}+t\overrightarrow{D}–\overrightarrow{S})\cdot \overrightarrow{N}=0$$ $$\overrightarrow{O}\cdot \overrightarrow{N}+t\overrightarrow{D}\cdot \overrightarrow{N}–\overrightarrow{S}\cdot \overrightarrow{N}=0$$ $$t\overrightarrow{D}\cdot \overrightarrow{N}=\overrightarrow{S}\cdot \overrightarrow{N}–\overrightarrow{O}\cdot \overrightarrow{N}$$ $$t=\frac{(\overrightarrow{S}–\overrightarrow{O})\cdot \overrightarrow{N}}{\overrightarrow{D}\cdot \overrightarrow{N}}$$

A question to ask ourselves is: what about the division by $0$? Looking at the diagram, we can see that $\overrightarrow{D}\cdot \overrightarrow{N}=0$ means the ray is parallel to the plane, and there is no solution unless $O$ is already on the plane. Otherwise, the ray intersects the plane for the value of $t$ written above. That’s it, we’re done.

Note: There are several, equivalent, ways of representing a plane. If your plane is not defined by a point $S$ and a normal vector $\hat{N}$, but rather with a distance to the origin $s$ and a normal vector $\hat{N}$, you can notice that $s=\overrightarrow{S}\cdot \overrightarrow{N}$ and simplify the result above, which becomes: $$t=\frac{s–\overrightarrow{O}\cdot \overrightarrow{N}}{\overrightarrow{D}\cdot \overrightarrow{N}}$$

Signed distance to a plane

For the sake of simplicity, in the above we defined the distance to the plane as an absolute value. It is possible however to define it as a signed value: $d=(\overrightarrow{X}–\overrightarrow{S})\cdot \overrightarrow{N}$. In this case $d>0$ means $X$ is somewhere on the side of the plane pointed by $\overrightarrow{N}$, while $d<0$ means $X$ is on the opposite side of the plane.

Distances that can be negative are called signed distances, and they are a foundation of Signed Distance Fields (SDF).