determine cube orientation given one side in a perspective projection Suppose that we are given an arbitrary quadrilateral T that does not have any parallel edges. I want to draw a cube in a three-point perspective projection such that T is one of its sides. The projection parameters are not given. Is the solution always unique? How to find it? 
 A: Yes, it is always possible, and no, the solution is not unique. There is a simple construction that will give the locus of solutions.
The quadrilateral $T$ will determine two vanishing points, $L$ and $R$. The key is then to look for the vertical axis point, $\textit{VAP}$, from which all angles to the line through $LR$ are true angles. (This is analogous to the station point in one-point perspective.) Observe that $\textit{VAP}$ will lie on the circle with diameter $LR$, for the angle subtended by $LR$ from $\textit{VAP}$ must be a right angle.
Next observe that since $T$ is square, the same applies to the diagonals. In other words, extending the diagonals to the horizon through $LR$ gives two additional vanishing points, and $\textit{VAP}$ must be on the circle having those points as diameter.
It follows that $\textit{VAP}$ is on the intersection of these two circles. Now drop a perpendicular from $LR$ through $\textit{VAP}$ and choose a third vanishing point $V$ on that line; this is the locus of solutions. ($V$ should of course be far enough away such that $LVR$ is an acute triangle, which is to say that it should be on the far side of $\textit{VAP}$.) You may then proceed to construct the remainder of the cube.

A: Two opposite sides of the quadrilateral, if prolonged, meet at a vanishing point. So two of your three perspective points are easy to find and the third one should be arbitrary. But it is not true that every quadrilateral can be the face of a cube, see for instance the image below. If $ABCD$ is your quadrlateral, $E$ and $F$ are two vanishing points, while $H$ and $I$ are the corresponding measuring points ($FH=FG$ and $EI=EG$). But for the edges $BC$ and $CD$ to be equal, you should have $KC=JC$, which is not true.

