I have an arbitrary triangle in $\Bbb R^3$, i.e. it's scalene. I know the lengths of all sides ($l_0$, $l_1$ and $l_2$) and the coordinates of the barycentre ($O$ in the diagram) and one vertex ($v_1$). I know that this does not uniquely define a single triangle but I also know the normal vector for the triangle, ie for the plane in which it lies which I believe should define a unique triangle.
I've spent some time trying to derive a closed form solution for the coordinates of vertices $v_0$ and $v_2$ without success.
For example, it seems to me that the intersection of two spheres, one centred at $v_1$ with radius $l_0$ the other at $X$ with radius $\frac{l_1}{2}$, and the plane in which the triangle lies ought to give me $v_2$ but I have been unable to express this in a way which lets me plug in the unknowns to get the result.
I'm aware that I could try rotating the whole frame into the XY plane and trying to solve there but it feels like there ought to be a simpler solution.
Am I correct in thinking the information I have defines a unique triangle?
Is there an efficient solution for locating the two unknown vertices?
Image showing triangle with vertices and barycentre labelled
The problem is essentially two-dimensional, as you know on which plane it lies. Assume w.l.o.g. that this is a horizontal plane, and we will use just two coordinates. Also assume w.l.o.g. that the barycenter has coordinates $(0,0)$ and $v_i=(x_i,y_i)$, $i=1,2,3$, and you are given $(x_1,y_1)$. Then you immediately get $x_2+x_3=-x_1$; $y_2+y_3=-y_1$; you also are given $l_1=\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$, $l_2=\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}$, $l_3=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Thus you have 5 equations and only 4 unknowns, so unless the data are inconsistent you can solve these equations and get a finite number of possible solutions (possibly, more than one, see the diagram for two possibilities with the same side lengths).