How can one prove that when two lines are parallel in space ($\mathbb R^3$) and they are projected from a point onto a plane (not parallel to the lines), they are no longer parallel?
It seems something easy to visualize but I have not been able to find an easy way to prove it.
To be precise, I'm considering the projection from a point ($\overrightarrow{P_0}$) onto a plane ($\alpha$) of a point
($\overrightarrow{P}$) in space as the point obtained when intersecting the line $\overrightarrow{P_0} + t(\overrightarrow{P}-\overrightarrow{P_0})$ and the plane $\alpha$.
So far I understand that if we establish the coordinate system $(\overrightarrow{O},\overrightarrow{\xi_1},\overrightarrow{\xi_2})$ where $\overrightarrow{O} = \overrightarrow{P_0} + d\overrightarrow{n}$, with $d$ the distance between the plane and $\overrightarrow{P_0}$, $\overrightarrow{n}$ the unit vector perpendicular to the plane, and $(\overrightarrow{\xi_1},\overrightarrow{\xi_2})$ is an orthonormal basis in $\alpha$, the point $\overrightarrow{P}$, when projected, is going to end up in the point $(u,v)$ where:
$$
\left\{
\begin{array}{}
u = \dfrac{\bigg\langle\overrightarrow{P}-\overrightarrow{P_0},\overrightarrow{\xi_1}\bigg\rangle}{\bigg\langle\overrightarrow{P}-\overrightarrow{P_0},\overrightarrow{n}\bigg\rangle} \\
v = \dfrac{\bigg\langle\overrightarrow{P}-\overrightarrow{P_0},\overrightarrow{\xi_2}\bigg\rangle}{\bigg\langle\overrightarrow{P}-\overrightarrow{P_0},\overrightarrow{n}\bigg\rangle}
\end{array}
\right.
$$
Any ideas on how to go about this? My only idea so far has been to try to compute the vector product of the projection of two segments of lines from each line in space and somehow verify that it is somehow different than zero, but it sounds like an exhausting computation.
I appreciate the help, thanks in advance!
Best Answer
Here's a fleshed-out geometric argument as given in the comments. It's not an algebraic argument along the lines of the question, but may help construct such an argument.
Let's call our lines $\ell_{1}$ and $\ell_{2}$. Denote by $\alpha_{1}$ the plane containing $P_{0}$ and $\ell_{1}$, and similarly for $\alpha_{2}$.
If $\alpha$ is a plane, the image of $\ell_{i}$ under projection from $P_{0}$ to $\alpha$ is the intersection of planes $\alpha_{i} \cap \alpha$: Map each point $P$ on $\ell_{i}$ to $\overline{P_{0}P} \cap \alpha$. The set of lines $\overline{P_{0}P}$ as $P$ runs over $\ell_{i}$ sweeps out $\alpha_{i}$ (except for the line through $P_{0}$ and parallel to $\ell_{i}$).
Now, the planes $\alpha_{1}$ and $\alpha_{2}$ intersect along a line $\ell$ because $P_{0}$ lies on both. If $\alpha$ intersects $\ell$ in a point (i.e., $\alpha$ is not parallel to the plane containing $\ell_{1}$ and $\ell_{2}$), then $\ell_{1}'$ and $\ell_{2}'$ intersect at $\ell \cap \alpha$. That is, the images of parallel lines under projection from a point need not be parallel.