Projection of Rectangle from $\mathbb{R}^3$ into $\mathbb{R}^2$ Lets have two known vectors $x_1'$ and $x_2'$ in XY ($\mathbb{R}^2$) representing the orthogonal projection of two unknown vectors $x_1$ and $x_2$ in $\mathbb{R}^3$, with known lengths $|x_1|=l_1$, $|x_1|=l_2$, and $x_1\cdot x_2=0$. 
Let be the orthogonal projection made through $\hat z$, thus projecting orthogonally onto XY.
Consider the simple case of $x_1$ lying on X, i.e. $x_1'=l_1 \hat x$. 
Trivially in this case:
$$x_1=(l_1,0,0)\\
x_2=(0,|x_2'|,\sin \arccos \frac{|x_2'|}{l_2})
$$
How should i calculate the $x_1$ and $x_2$ vectors in the general case, for any given $x_1'$ and $x_2'$ image vectors, any in the sense they were indeed projected from a rectangle with sides $l_1$ and $l_2$?.
 A: (I’m going to use slightly different notation from yours so as not to confuse coordinates with vectors.)  
For each of the vectors $\mathbf v_1'=(x_1,y_1,0)$ and $\mathbf v_2'=(x_2,y_2,0)$ there are two possible pre-images, even in your simple example: $\mathbf v_1=\left(x_1,y_1,\pm\sqrt{l_1^2-x_1^2-y_1^2}\right)$ and $\mathbf v_2=\left(x_2,y_2,\pm\sqrt{l_2^2-x_2^2-y_2^2}\right)$, respectively. This is a simple application of the Pythagorean theorem. That gives you four possibilities for the original rectangle. The additional constraint $$\begin{align} \mathbf v_1\cdot\mathbf v_2 &= x_1x_2+y_1y_2+z_1z_2 \\ &= x_1x_2+y_1y_2\pm\sqrt{l_1^2-x_1^2-y_1^2}\sqrt{l_2^2-x_2^2-y_2^2} \\ &= \mathbf v_1'\cdot\mathbf v_2'\pm\sqrt{l_1^2-\|\mathbf v_1'\|^2}\sqrt{l_2^2-\|\mathbf v_2'\|^2} \\ &=0 \end{align}$$ will allow you to eliminate some of these combinations, but in general you’ll end up with two possibilities for the original rectangle. If the angle between $\mathbf v_1'$ and $\mathbf v_2'$ is acute, then the signs of $z_1$ and $z_2$ must be different, but they can be either $+/-$ or $-/+$. Similarly, if the angle is obtuse, then the signs of the two $z$-coordinates are the same, but there’s no way to decide with the information at hand whether they should be positive or negative. (If $\mathbf v_1'\cdot\mathbf v_2'=0$, then the rectangle is parallel to the $x$-$y$ plane.)
A: This is not an answer to your question, but rather just pointing
out (tangentially), as reflected in 
this post,
that projecting a $90^\circ$ rectangle angle
can result in a much smaller planar angle:

          


          

An angle is $68^\circ$ incident to $v$ in 3D,
but it projects to $45^\circ$ in 2D.


