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I'm trying to get the third vertex of triangle in the $3D$ space $xyz$ and I have:

  • $2$ points

  • Normal of the plane where the triangle is

  • I know the triangle has an angle of $90^{\circ}$ that is on the third unknown point

Does anyone have any idea on how to solve this? I've been searching for a solution to this, but I didn't have any success...

Any help would be greatly appreciated!

Nathaniel Bubis
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Wach
  • 21
  • You could've provided more details, otherwise it's too vague. What are coordinates of given two vertices? Angle at which vertex is $90^\circ$? For example, if the right angle is at third unknown vertex, then the problem has infinite number solutions – all points on the circle, with given two points as diameter. – Kaster Oct 22 '13 at 23:06
  • Sorry I just added more info. Regarding the 2 points, I'm programming a system that solves this problem, so it can be any 2 points in the xyz space. As for the infinite number of solutions I think that there should only be 2 possible solutions as I know the plane of the triangle. But I don't know the math to do this :S – Wach Oct 22 '13 at 23:56

1 Answers1

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Use the fact that for a plane passing through a point $\mathbf{x_1}$ with normal $\mathbf{n}$, all points $x$ on it satisfy the equation: $$\mathbf{n}\cdot(\mathbf{x} - \mathbf{x_0}) =0 $$ You have $\mathbf{n},\mathbf{x_1},\mathbf{x_2}$ and want to find $\mathbf{x_3}$.

Since the angle at $\mathbf{x_3}$ is $90^\circ$, we have two equations: $$(\mathbf{x_3} - \mathbf{x_1})\cdot(\mathbf{x_3}-\mathbf{x_2})=0$$ $$ \mathbf{n}\cdot(\mathbf{x} - \mathbf{x_3}) =0 $$ But we have three unknowns! this means the system is underdetermined, and that there are an infinite number of solutions, which indeed there are: $$ $$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

Nathaniel Bubis
  • 33,018