I need to find the missing point of a tetrahedron side length $x$, with three points being $$v_1=(0,0,0) \quad v_2=\left(\frac{1}{2}x,0,\frac{\sqrt3}{2}x\right) \quad v_3=(x,0,0)$$ I can't seem to find the 4th point.
Can anyone help? I just need the equation.
On
In a plane you have these points forming an equilateral triangle.
If you know that medians trisect one another, you know the centroid of this triangle is $\frac{1}{3}$ the way along the indicated line. If you didn't know that, you could calculate the midpoint of either side, derive the line equation, and calculate the intersection, to arrive at the same centroid of $(\frac{x}{2},0,\frac{\sqrt{3}}{6})$.
The fourth vertex lies above or below this at whatever distance forms a side of length $x$:
$x^2(\frac{1}{4}+z^2+\frac{3}{36})=x^2 \Rightarrow z=\pm \frac{2}{\sqrt{6}}$
On
We have
$$ \cases{ v_1=(0,0,0) \\ v_2=\left(\frac{1}{2}x,0,\frac{\sqrt3}{2}x\right)\\ v_3=(x,0,0) } $$
the triangle with vertices on $v_1,v_2,v_3$ has the baricentre given by
$$ g=\frac 13(v_1+v_2+v_3) = \left(\frac{x}{2},0,\frac{x}{2 \sqrt{3}}\right) $$
and the normal to the plane containing $v_1,v_2,v_3$ is $\vec n=(0,1,0)$ then we can propose as fourth vertex $v_4 = g + \gamma\vec n$. Now the tetrahedron baricentre is given by
$$ G=\frac 14(v_1+v_2+v_3+v_4) = \left(\frac{x}{2},\frac{\gamma}{4},\frac{1}{4} \left(\frac{\sqrt{3} x}{2}+\frac{x}{2 \sqrt{3}}\right)\right) $$
now as $\|v_1-G\|^2 = \|v_4-G\|^2$ we have
$$ \frac{\gamma^2}{2}-\frac{x^2}{4}-\frac{1}{16} \left(\frac{\sqrt{3} x}{2}+\frac{x}{2 \sqrt{3}}\right)^2+\left(\frac{x}{2 \sqrt{3}}-\frac{1}{4} \left(\frac{\sqrt{3} x}{2}+\frac{x}{2 \sqrt{3}}\right)\right)^2=0 $$
and solving for $\gamma$ we got
$$ \gamma = \pm\sqrt{\frac{2}{3}} x $$
and thus the generic regular tetrahedron is defined by the four vertices
$$ \cases{ v_1=(0,0,0) \\ v_2=\left(\frac{1}{2}x,0,\frac{\sqrt3}{2}x\right)\\ v_3=(x,0,0)\\ v_4 = \left(\frac{x}{2},\pm\sqrt{\frac{2}{3}} x,\frac{x}{2 \sqrt{3}}\right) } $$
That triangle lives in the $xz$ plane. Find its center $P=(a,0,b)$ (which is $1/3$ the sum of the vertices). There are two possible fourth vertices, each of the form $(a,t,b)$ with $t$ chosen so that the point is $x$ units from the origin, since that is the side length of your regular tetrahedron.