Proving a couple of lines do not intersect a plane in ℝ4

[JulienB]

Homework Statement

Hi! I'm not sure the title I gave properly reflects my problem, but I did not know how to describe it in another way:

(roughly translated from German)
"In three dimensions, we can divide the space in two parts through a plane. In four dimensions, that does not work anymore. Given the plane in ℝ⁴:

E = { (x₁, x₂, 0, 0) | x₁, x₂ ∈ ℝ}

Show that two arbitrary points in ℝ⁴ not belonging to E can be joined by a maximum of two lines without intersecting with E."

Homework Equations

What gives me a headache is that we have not even started to talk about higher dimensions than 3 yet. Our teacher encourages us through the homework to make some research and learn on our own, but I could not find any relevant information about how to deal with such a problem.

However, we have mentioned scalar multiplication in ℝⁿ and the fact that two coordinates of the plane are 0 makes me think that this problem might be easier than it seems.

The Attempt at a Solution

I have no idea where to start :( So far I only tried to formulate all I know in an algebraic way to avoid trying to "imagine" what four spatial dimensions might imply.

So I first defined two points:
P₁, P₂ ∈ ℝ⁴ ∧ P₁, P₂ ∉ E
For two lines to exist (and forget about the trivial case where one line would be enough), there must be a third point:
P₃ ∈ ℝ⁴, ∉ E
And for ∀ P_i ∈ ℝ⁴, the lines A (P₁, P₂) and B (P₂, P₃) ∉ E

Unfortunately that didn't lead me anywhere... Any advice or suggestion of what to search for would be very much appreciated.Thank you in advance for your answers.J.

 

[haruspex]

Since the given points are not in E, you know that each is nonzero in one or other of the last two ordinates.
You can get from a point P to a point Q on a straight line by writing $\lambda P+(1-\lambda)Q$ and varying lambda from 1 to 0.
Your problem is that if P1 and P2 have opposite signs in their third ordinates, doing this will pass through a point which is zero in the third ordinate, so risks being in E.
So think in terms of making two transitions, one to get the third ordinate matching and one to get the fourth ordinate matching, that doesn't risk both those being zero at some intermediate point.

 

[JulienB]

To begin, thank you very much for your answer. I already get a better idea of what's the direction to take. I give it a go:

We define three points P, Q, S ∈ ℝ⁴ ∧ P, Q, S ∉ E, which means that their coordinates cannot be equal to the corresponding coordinates of the plane:
P (p₁, p₂, p₃, p₄) with p₁ ≠ x₁, p₂ ≠ x₂, p₃ ≠ 0, p₄ ≠ 0
Q (q₁, q₂, q₃, q₄) with q₁ ≠ x₁, q₂ ≠ x₂, q₃ ≠ 0, q₄ ≠ 0
S (s₁, s₂, s₃, s₄) with s₁ ≠ x₁, s₂ ≠ x₂, s₃ ≠ 0, s₄ ≠ 0

You mentioned that we can define a straight line from P to Q by: λP+(1−λ)Q with 0 ≤ λ ≤ 1. Since I have never seen and even less used this formula before, I'm unsure as to what it really means and how to use it. In order for that line to never intersect E, I would assume that the coordinates of the line only need to never be equal to {x₁, x₂, 0, 0}, right?

If so, then the line equation would give me a linear system of equations with four unknowns:
λp₁ + (1 - λ)q₁ ≠ x₁
λp₂ + (1 - λ)q₂ ≠ x₂
λp₃ + (1 - λ)q₃ ≠ 0
λp₄ + (1 - λ)q₄ ≠ 0

Then of course the same for the second line from Q to S, but...That's too beautiful to be true, I'm sure I messed up somewhere!

 

[JulienB]

After reading the last line of your post again, I'm pretty sure that what I just wrote is completely wrong. I have to focus on the 3rd and 4th coordinates, so that both of them are never equal to 0 at the same time. Let me work on it for a while, and I'll come back again :)

 

[JulienB]

I just have an extra question to feel more sure: if I have let's say a point P defined by (x₁, x₂, 7, 0). Does my point belong to E when three coordinates are equal to the ones of the plane?

 

[HallsofIvy]

A point belongs to a plane if and only if its coordinates satisfy the equations defining the plane. Here those equations are x_3= 0 and x_4= 0. In the case of (x_1, x_2, 7, 0), x_1= 7\ne 0 so is NOT in the plane.