Consider the following picture:
There are two planes $\pi, \pi'$, and two associated (violet) euclidean coordinate frames, with $z-$axis orthogonal to the respective planes. We are given a point $x' \in \pi'$, in coordinates with respect to the right frame of reference. Thus, $x'=(x'_1,x'_2,1)$.
Task: show that the line $l: d \mapsto d \cdot x'$ (in coordinates of the right frame), when projected into $\pi$, is a 2D line
By projection of the point $(x,y,z)$ (in left coordinates) onto $\pi$ I mean $(x/z, y/z)$.
Here is my idea: find a change of coordinates $(R,T)$ from right to left frame, and write $l$ in left coordinates, i.e. $l: d\mapsto Rx'+T$. This is obviously still a line, but when projecting onto $\pi$ I get a non linearity. What am I doing wrong?
Let's examine the 3D line specified by the vector-valued function $$\mathbf{p}(\lambda) = (1 - \lambda) \mathbf{p}_0 + \lambda \mathbf{p}_1 = \mathbf{p}_0 + \lambda (\mathbf{p}_1 - \mathbf{p}_0)$$ Without loss of generality, we use a coordinate system where the projection plane is perpendicular to the $z$ axis at $z = 1$. Then, the projection simplifies to $$\left\lbrace \begin{aligned} x^\prime &= \frac{x}{z} \\ y^\prime &= \frac{y}{z} \\ z^\prime &= \frac{z}{z} = 1 \\ \end{aligned} \right.$$ The projection of points $$\mathbf{p}_0 = \left[ \begin{matrix} x_0 \\ y_0 \\ z_0 \end{matrix} \right], \quad \mathbf{p}_1 = \left[ \begin{matrix} x_1 \\ y_1 \\ z_1 \end{matrix} \right]$$ to the projection plane are $$\mathbf{q}_0 = \left[ \begin{matrix} \frac{x_0}{z_0} \\ \frac{y_0}{z_0} \\ 1 \end{matrix} \right], \quad \mathbf{q}_1 = \left[ \begin{matrix} \frac{x_1}{z_1} \\ \frac{y_1}{z_1} \\ 1 \end{matrix} \right]$$ and the projection of $\mathbf{p}(\lambda_q)$, the vector-valued function specifying the line, is $$\mathbf{q}(\lambda_q) = \left[\begin{matrix} \displaystyle\frac{x_0 + \lambda_q (x_1 - x_0)}{z_0 + \lambda_q (z_1 - z_0)} \\ \displaystyle\frac{y_0 + \lambda_q (y_1 - y_0)}{z_0 + \lambda_q (z_1 - z_0)} \\ 1 \end{matrix}\right]$$ whereas the vector-valued function specifying the line between the two projected points is $$\mathbf{d}(\lambda_d) = \left[\begin{matrix} \displaystyle\frac{x_0}{z_0} \\ \displaystyle\frac{y_0}{z_0} \\ 1 \end{matrix}\right] + \lambda_d \left[\begin{matrix}\displaystyle\frac{x_1}{z_1} - \frac{x_0}{z_0} \\ \displaystyle\frac{y_1}{z_1} - \frac{y_0}{z_0} \\ 0 \end{matrix}\right] = \left[\begin{matrix} \displaystyle\frac{x_0 z_1 + \lambda_d (x_1 z_0 - x_0 z_1)}{z_0 z_1} \\ \displaystyle\frac{y_0 z_1 + \lambda_d (y_1 z_0 - y_0 z_1)}{z_0 z_1} \\ 1 \end{matrix}\right]$$ The two are collinear if their vector cross product is the zero vector: $$\begin{aligned} \left( \mathbf{q}(\lambda_q) - \mathbf{q}(0) \right) \times \left( \mathbf{d}(\lambda_d) - \mathbf{d}(0) \right) &= \left[ \begin{matrix} \displaystyle \frac{x_0 + \lambda_q (x_1 - x_0)}{z_0 + \lambda_q (z_1 - z_0)} - \frac{x_0}{z_0} \\ \displaystyle \frac{y_0 + \lambda_q (y_1 - y_0)}{z_0 + \lambda_q (z_1 - z_0)} - \frac{y_0}{z_0} \\ 0 \end{matrix} \right] \times \left[\begin{matrix} \displaystyle \lambda_d \left( \frac{x_1}{z_1} - \frac{x_0}{z_0} \right) \\ \displaystyle \lambda_d \left( \frac{y_1}{z_1} - \frac{y_0}{z_0} \right) \\ 0 \end{matrix}\right] \\ ~ &= \left[ \begin{matrix} 0 \\ 0 \\ \displaystyle\frac{\lambda_q ( x_1 y_0 - x_0 y_1 )}{ \lambda_q z_0 z_1 + (1 - \lambda_q) z_0^2 } \end{matrix} \right] \\ ~ &= \left[ \begin{matrix} 0 \\ 0 \\ \displaystyle\frac{\lambda_q ( x_1 y_0 - x_0 y_1 )}{ z_0^2 + \lambda_q z_0 ( z_1 - z_0 )} \\ \end{matrix} \right] \\ \end{aligned}$$ i.e., if and only if $x_1 y_0 = x_0 y_1$.
This means that if in the coordinate system where the projection plane is perpendicular to the $z$ axis,
the line intersects with the $z$ axis (in which case there is a parametrisation for the line with $x_0 = y_0 = 0$ or $x_1 = y_1 = 0$),
or
the line is in the $x z$ plane (in which case $y_0 = y_1 = 0$),
or
the line is in the $y z$ plane (in which case $x_0 = x_1 = 0$),
the projected line is also a straight line. (Technically, the first case also includes the other two cases: the projected line is a straight line if and only if the 3D line intersects the $z$ axis, when the projection plane is perpendicular to the $z$ axis.)