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This problem requires some explanation, maybe because it's very special or I just don't know the right terminology. In any way, please bear with me as I try to explain this as best as I can.

Given are the five vectors $a$, $b$, $c$, $n$ and $r$. The vectors $a$, $b$ and $c$ each describe points on the same plane.

I want to define a local coordinate system by using the plane defined by the three vectors $a$, $b$ and $c$ for the $X/Y$-plane of the coordinate system and the normal of that plane as the $Z$-axis. The origin of that coordinate system should lie in the point described by $n$.

I construct the vectors $x$, $y$ and $z$, that define that coordinate system, as follows (note that they'll each have to be converted to unit-vectors as well to keep the scale of the global coordinate system, I've omitted that step here for simplicities sake):

  • $z = (a - b) \times (c - b)$
  • Let $u$ be the vector $(0,1,0)$ of the global coordinate system (the global "up")
  • $x = z \times u$
  • $y = x \times z$

Now this is where I run into a problem. This coordinate system works as is, but each axis-defining-vector could also be its inverse. To prevent ambiguity the local coordinate system should also fulfill the following criteria (where $r$ defines a point that isn't on the plane defined by $a$, $b$ and $c$):

  1. Given an observer in $r$, the $z$ of the local coordinate system should point out of the plane towards the observer
  2. The $y$ of the local coordinate system should point in the direction, where it is "closest" to the global "up". That is, given both alternatives for $y$ you should choose the one pointing "upwards".
  3. Given that same observer in $r$ as in criteria 1, the $x$ should point to the right if the observer were to look at the plane.

I have no idea how I can check if my calculated $x$, $y$ and $z$ fulfill these criteria and therefore I also don't know if I need to invert them. I'd be thankful for any idea in this regard as I've really struggled to find a solution for this.

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1 Answer 1

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All of this is easily accomplished via dot and cross products. Assuming that $a$, $b$, $c$ and $n$ are coplanar and that $r$ does not lie on this plane,

  1. Examine the sign of $z\cdot(r-n)$. If it is positive, then $r$ is in the half-space into which $z$ points. If negative, it’s on the opposite side of the plane, so flip $z$. In other words, to point $z$ in the desired direction, multiply $z$ by the sign of $z\cdot(r-n)$ regardless.

  2. Similarly, multiply $y$ by the sign of $y\cdot u$, i.e., by the sign of its own second coordinate. If that’s zero, $y\perp u$ and neither $y$ nor $-y$ is “closer” to pointing upwards.

  3. Here, I’ll assume that this observer is oriented with her “up” aligned with global “up.” Once you’ve aligned $y$ and $z$ correctly, this criterion is satisfied by having $x$ be clockwise of $y$ when looking toward the plane from $r$, so set $x=y\times z$, normalized, of course. Again, if $y\perp u$, $x$ will end up pointing “upward” or “downward,” but this will at least ensure that the $x$-direction is in some sense to the right of the $y$-direction in that case.

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