I have two triangles in $R^3$:

  1. $p_1$, $p_2$, $p_3$
  2. $p_1$, $p_2$, $p_4$

The triangles share points $p_1$ and $p_2$ and thus edge $p_2 - p_1$.

I would like to rotate $p_4$ such that it will be diametrically with respect to $p_3$, i.e. the angle between $p_3$ and $p_4$ should be $180$ degrees or $\pi$.

I can derive the current angle between $p_3$ and $p_4$:

$d_1 = (p_3 - p_1) \times (p_2 - p_1)$

$d_2 = (p_4 - p_1) \times (p_2 - p_1)$

$rad = \arccos(d_1 / |d_1| \cdot d_2 / |d_2|)$

The next step would be to rotate $p_4$ around edge ($p_2 - p_1$) by $\pi - rad$. However I do not know how to rotate a corner of a triangle around the opposite edge. Therefore I was wondering if anybody would know how to accomplish this.


You have the angle $\phi$ and the axis of rotation $\hat{n} = \frac{p_2-p_1}{|p_2-p_1|}$. You only need Rodrigues' rotation formula: $$p_{\text{rot}} = p_4 \cos\phi + (\hat{n}\times p_4) \sin\phi + \hat{n}(\hat{n}\cdot p_4)(1-\cos\theta).$$


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