Shadow of a 3D object on a plane I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin. 
Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:
$$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$
where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.
A point on a ray-line has the equation: 
\begin{equation}
\mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}} 
\tag{1}
\end{equation}
being $\mathbf{r}_p$ a point on that line.
A point on a general plane is given by the equation:
\begin{equation}
\left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0
\tag{2}
\end{equation}
being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane. 
Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)
\begin{gather}
\lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}}
{\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\
\mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}}
{\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}}
\tag{3}
\end{gather}
Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ gives the shadow. 
A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane). 
Are there some conceptual mistakes in my reasoning?
Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors? 
$$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$
 A: I address here your second question, but I think that it also answers a little the first one.
Consider oblique projection where unit vertical vector $\vec{k}$ has for its shadow $\left(\begin{array}{c}a\\b\end{array}\right)$ on the $xOy$ plane.
Its matrix is:
$$\left(\begin{array}{ccc}1 & 0 & a\\0 & 1 & b\\0 & 0 & 0\end{array}\right)$$
This is how I have realized the following figure representing the projection (in black) of an helix (in red):

The explanation relies plainly on the way a transformation matrix is built: its successive columns are the images of vectors $\vec{i},\vec{j},\vec{k}$ (directing $x$,$y$,$z$ axis resp.)

*

*$\vec{i},\vec{j}$, being already on $x0y$ plane are projected on themselves.


*$\vec{k}$ is projected on its shadow $\left(\begin{array}{c}a\\b\\0 \end{array}\right)=\left(\begin{array}{c}\cos(\theta)\sin(\lambda)\\ \sin(\theta)\sin(\lambda)\\ 0 \end{array}\right)$
($\theta$ being the azimuth and $\lambda$ the latitude of the sun's direction).
Edit: Another way to define the third vector using the sun's direction $\vec{s}$ is  $\left(\begin{array}{c}a\\b\\0 \end{array}\right)=\left(\begin{array}{c}\left(\dfrac{\vec{s}.\vec{i}}{\vec{s}.\vec{k}}\right)\\\left(\dfrac{\vec{s}.\vec{j}}{\vec{s}.\vec{k}}\right)\\ 0 \end{array}\right).$
