# Vertices of a tetrahedron with origin as vertex, given two other known vertices in the tetrahedron

I am tying to find the the last vertex in a tetrahedron, where the origin is one of the vertices and I already know two of the other vertices (which form an equilateral triangle). I understand I could just try to find the centre point of this triangle, but how do i find its 3-Dimensional coordinates?

Let the known vertices be $\vec{v}_1 = (x_1, y_1, z_1)$ and $\vec{v}_2 = (x_2, y_2, z_2)$. The centroid of the known face will be at $\vec{c} = \frac{1}{3} (\vec{v}_1 + \vec{v}_2)$. The height of a tetrahedron is $h = \frac{\sqrt{6}}{3} a$, where $a = |\vec{v}_1| = |\vec{v}_2|$ is the length of one of the sides. The fourth vertex will therefore be displaced from $\vec{c}$ by a distance of $h$, in a direction orthogonal to both $\vec{v}_1$ and $\vec{v}_2$. We can define a unit vector that is orthogonal to both of these by taking the cross product and normalizing it appropriately: $$\hat{e} = \frac{\vec{v}_1 \times \vec{v}_2}{|\vec{v}_1 \times \vec{v}_2|} = \frac{\vec{v}_1 \times \vec{v}_2}{|\vec{v}_1| |\vec{v}_2| \sin \frac{\pi}{3}} = \frac{2}{\sqrt{3} |\vec{v}_1|^2} \vec{v}_1 \times \vec{v}_2.$$

Thus, the remaining vertex will be at $$\vec{v}_3 = \vec{c} \pm h \hat{e} = \frac{1}{3} (\vec{v}_1 + \vec{v}_2) \pm \left( \frac{\sqrt{6}}{3} |\vec{v}_1| \right) \left( \frac{2}{\sqrt{3} |\vec{v}_1|^2} \vec{v}_1 \times \vec{v}_2 \right) = \boxed{ \frac{1}{3} \left[ \vec{v}_1 + \vec{v}_2 \pm \frac{2 \sqrt{2}}{|\vec{v}_1|} \vec{v}_1 \times \vec{v}_2 \right].}$$

The plus-or-minus symbol is present because there are two possible points that can complete the tetrahedron, one on each "side" of the known face.

• Thank you this is very insightful and laid out great! – J. Doe Dec 16 '16 at 16:20

Given: Three vertices of a triangle. $\{O(0,0,0);p_0(x_0,y_0;z_0);p_1(x_1,y_1;z_1)\}$
Test to make sure that the three points you have actually constitute an equilateral triangle. Does $\overline{Op_0}=\overline{Op_1}=\overline{p_0p_1}?$

Set $d=\overline{Op_0}.$

Solve
$\left\{\begin{array}{l} x^2+y^2+z^2=d^2\\ (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=d^2\\ (x-x_1)^2+(y-y_1)^2+(z-z_1)^2=d^2 \end{array}\right.$

There are two xyz-triads as solutions.

• I was a little concerned that this method would require the solution of a complicated quartic equation. But after actually writing out the equations, I think it can be solved only using the quadratic formula. – Michael Seifert Dec 16 '16 at 17:42
• @Michael Seifert \ Since the equation of a sphere, which all three of these are, does not have an xy, yz, zx, or xyz term, the system never gets complicated. – Senex Ægypti Parvi Dec 16 '16 at 19:16