common point at infinity of two parallel planes

Question

How do I find the common points at infinity of two parallel planes:

• $\alpha: x-y+2z-t=0$
• $\beta: 3x-3y+6z-7t=0$

My understanding is that the planes are not only going to have common points, but they are going to have infinitely many common points at infinity since they have common line at infinity.

Is it alright if I just find two infinity points satisfying the condition that they just intersect through the first plane (since both planes have the same line at infinity) and construct a line parametric equation of a line going through those two points? In this case that would be:

$M(1,-1,-1,0) \\N(3,1,-1,0) \\ u_g: \begin{cases} x=\lambda+3\mu & \text{} \\ y=-\lambda+\mu & \text{} \\ z=-\lambda-\mu & \text{} \\ t=0 \end{cases}$

So, the common points at infinity of the two planes are all the points on the line $u_g.$

Is this correct? If no, how do I find such points?

Answer 1

Your solution method is fine. In fact, representing lines in $\mathbb{RP}^3$ is a bit problematic, and one of the ways to do so is as the intersection of a pair of planes $\mathbf p_1$ and $\mathbf p_2$. This can be packaged up into matrix form as $$L=\begin{bmatrix}\mathbf p_1^T\\\mathbf p_2^T\end{bmatrix}.$$ The line is the join of the null vectors of this matrix, which is exactly what you’ve produced in $u_g$ by solving the homogeneous system of linear equations represented by $L$. This works for any pair of planes, by the way, not just for parallel ones.

Answer 2

We want to find the intersection of these two hyperplanes of $\mathbb{R^4}$. So we have to solve the following system

$$\begin{cases} x-y+2z-t=0 & \\ 3x-3y+6z-7t=0 & \end{cases}$$

the equation of the solution space is

$$\begin{cases} x-y+2z=0 & \\ t=0& \end{cases}$$

Solution space is a 2 dimensional vector subspace, projectivetily speaking is the line at infinity that you are looking for; the points at infinity are the directions of the lines on this 2 dimensional space.