2d projection of a 3d image I am having a problem where I have a $2$D object which can move in $3$-dimensional space about a fixed point (the origin). I want to rotate this object using Euler angles and axes of rotation. 
If you think of the object as a square with co-ordinates:
$(0, 1, 0)$, $(1, 0, 0)$, $(0, -1, 0)$ and $(-1, 0, 0)$. 
What I am trying to do is to rotate this square first by rotating around the $x$ axis ($y$ axis and $z$ axis will move by the given angle $\theta$ giving $y'$ and $z'$) then around the new axis $y'$ with an angle $\phi$, giving $x'$ and $z''$ and finally around $z''$ with an angle $\psi$ giving at last $x''$, $y''$ and $z''$.
What I need after rotating the axes is the angle between the $z''$ axis and the $x''$ axis to be a given angle $\alpha$ when projected onto a $2$D plane. 
The angle will always be $90$ degrees but when projected as a $2$D image it will appear to have a different angle. I need it to be around $80$ degrees. 
 A: $\newcommand{\NEG}{\phantom{-}}$Not positive I understand in detail what you want, but here are a couple of factlets that should suffice to piece together a solution. First, rotations by $\theta$ about the three Cartesian coordinate axes are accomplished by
$$
R_{x}(\theta) = \left[\begin{array}{@{}ccc@{}}
    1 & 0 & 0 \\
    0 & \cos\theta & -\sin\theta \\
    0 & \sin\theta & \NEG\cos\theta \\
  \end{array}\right],\
R_{y}(\theta) = \left[\begin{array}{@{}ccc@{}}
    \NEG\cos\theta & 0 & \sin\theta \\
    0 & 1 & 0 \\
   -\sin\theta & 0 & \cos\theta \\
  \end{array}\right],\
R_{z}(\theta) = \left[\begin{array}{@{}ccc@{}}
    \cos\theta & -\sin\theta & 0 \\
    \sin\theta & \NEG\cos\theta & 0 \\
    0 & 0 & 1 \\
  \end{array}\right].
$$
Second, if $P$ is a rotation matrix carrying the axes $(x, y, z)$ to $(x', y', z')$, then the transpose $P^{t}$ carries $(x', y', z')$ to $(x, y, z)$, and (for example)
$$
R_{x'}(\theta) = P R_{x}(\theta) P^{t}
$$
is rotation by $\theta$ about the $x'$ axis.
Thus, for example, to rotate by $\theta$ about $x$ and then to rotate by $\phi$ about $y'$, the image of the $y$ axis under the first rotation, you'd form the product of $P = R_{x}(\theta)$ and $R_{y'}(\phi) = P R_{y}(\phi) P^{t}$, i.e.,
$$
R_{y'}(\phi)R_{x}(\theta) = R_{x}(\theta) R_{y}(\phi).
$$
Once you've computed the composition of all your rotations, the first column is a set of direction cosines for $x''$ (with respect to the original $(x, y, z)$ coordinate system) and the third column is a set of direction cosines for $z''$. To project to the $(x, y)$ plane, discard the third component of each vector; then take the dot product to get the cosine of the angle between the "shadows".
