{"id":1516,"date":"2020-07-03T02:11:45","date_gmt":"2020-07-03T02:11:45","guid":{"rendered":"http:\/\/lousodrome.net\/blog\/light\/?p=1516"},"modified":"2020-09-20T18:28:38","modified_gmt":"2020-09-20T18:28:38","slug":"intersection-of-a-ray-and-a-plane","status":"publish","type":"post","link":"http:\/\/lousodrome.net\/blog\/light\/2020\/07\/03\/intersection-of-a-ray-and-a-plane\/","title":{"rendered":"Intersection of a ray and a plane"},"content":{"rendered":"\n

I previously showed the derivation of how to determine the intersection of a plane and a cone<\/a>. 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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

  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: $\\vec{X} = \\vec{O} + t\\vec{D}$.
    When $t$ is positive $X$ is in the direction of the ray, and when $t$ is negative $X$ is in the opposite direction.<\/li>
  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 $d = \\lvert (\\vec{X} – \\vec{S}) \\cdot \\vec{N} \\rvert$. If this equality is not obvious for you, you can think of it as the distance between $X$ and $S$ along the $\\vec{N}$ direction. When $d=0$, it means $X$ is on the plane itself. <\/li>
  3. We define $P$ the intersection or the ray and the plane, and which we are interested in finding.<\/li><\/ol>\n\n\n\n
    \"\"<\/figure><\/div>\n\n\n\n

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

    A question to ask ourselves is: what about the division by $0$? Looking at the diagram, we can see that $\\vec{D} \\cdot \\vec{N} = 0$ 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.<\/p>\n\n\n\n

    Note:<\/strong> 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 $s = \\vec{S} \\cdot \\vec{N}$ and simplify the result above, which becomes: $$ t = \\frac{s – \\vec{O} \\cdot \\vec{N}}{ \\vec{D} \\cdot \\vec{N} } $$ <\/p>\n\n\n\n

    \n\n\n\n

    Signed distance to a plane<\/h2>\n\n\n\n

    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: $d = (\\vec{X} – \\vec{S}) \\cdot \\vec{N}$. In this case $d>0$ means $X$ is somewhere on the side of the plane pointed by $\\vec{N}$, 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).
    <\/p>\n","protected":false},"excerpt":{"rendered":"

    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. 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