# Projective geometry: Intersection between two parallel planes

I have to show by example that two parallel planes intersect at the line on infinity.

To do that, I have chosen parallel planes $P$ and $Q$:

$P = (1, 1, 1, 1)^T$

$Q = (1, 1, 1, 2)^T$

My line $L$ can be calculated as $P Q^T - Q P^T$ which results in

$$L = \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & 0 \\ \end{matrix}$$

Sure, this is a Plücker matrix, but I would like to get a notation like $L = (0,0,0,1)$.

I couldn't find a way to do this. Am I doing something wrong here?

• May be by showing a division by 0? – Moti Apr 28 '18 at 16:03
• @Moti Not sure what you mean by that. Where to show division by zero? How? – Leta Apr 28 '18 at 16:24
• Attempt to calculate an intersection point of two parallel lines in the plans. – Moti Apr 28 '18 at 16:27
• @Moti I really have no idea how to do this with two Plücker matrices. Calculating a product between $L_1$ and $L_2^{-1}$ doesn't seem like the right way to me. Can you help me out? – Leta Apr 28 '18 at 17:25

Notice that $(x, y, z, w)$ is unique up to a non-zero common factor, leaving three degrees of freedom. That is enough for points or planes, but not for lines, which require four degrees of freedom. In a Plücker matrix there are six parameters, one relation between them, and a non-zero common factor, leaving the required number of free parameters.
For points on $L$, $$L \cdot (x, y, z, w) = (w, w, w, -x - y - z) = 0.$$ This shows that $L$ is a line at infinity, all points have $w = 0$.
You’re actually most of the way there for your proof. To get a bit picky about it, you have a dual Plücker matrix $\mathcal L^*$ for the line. A point $\mathbf x$ lies on the line iff $\mathcal L^*\mathbf x = 0$. The nullspace of $\mathcal L^*$ is spanned by $(1,-1,0,0)^T$ and $(1,0,-1,0)^T$, so it consists entirely of points at infinity.
You needn’t have constructed $\mathcal L^*$ to reach this conclusion, though. You can switch from the null space representation of the line to a span representation by computing the null space of $\small{\begin{bmatrix}P&Q\end{bmatrix}^T}$. This will give you the same two points as above, and the line is their join $\lambda(1,-1,0,0)^T+\mu(1,0,-1,0)^T$, which again consists entirely of points at infinity.