Suppose we have a regular tetrahedron delimited by four points $A_{1}, A_{2}, A_{3}, A_{4}$. There are 24 permutations of vertices, but there are only two distinct terahedra that cannot be superimposed (chirality). I want to find a direct and computationally efficient method to check if two random permutations of vertices (also randomly oriented in a 3D space) are superimposable, given coordinates of their vertices.
One method I came up with is to rotate the tetrahedra so the points $A_{1}, A_{2}, A_{3}$ are located on x-y plane and $A_{4}$ is located above that plane. Then form vectors $\vec{A_{1}A_{2}}$ and $\vec{A_{2}A_{3}}$ and compute their cross product. If signs of the products are the same in both tetrahedra, then they are superimposable. The problem is that finding appropriate rotation angles and then transformation of coordinates is not very computationally efficient.
Is there a more direct method to check the chirality of a tetrahedron? Any ideas?
Examining the sign of $\vec{A_{1}A_{2}} \times \vec{A_{2}A_{3}} \cdot \vec{A_{1}A_{4}}$ (the scalar triple product) should give you chirality.