High School Is the intersection of two planes a line?

[member 587159]

This is not a homework question. School year has ended for me and I'm doing some revision on my own.

I want to proof the following because in an exercise I had to find the equation of the line that passed through a given point and 2 given lines.

If a line r intersects with 2 given crossing lines a and b and passes through a given point P, then r is the intersection of the planes α(a,P) and β(b,P).

I started like this:

Let the intersections r∩a and r∩b be equal to S1 and S2. It's obvious that r = S1S2 = S1P = S2P. We need to show that r = S1S2 = α∩β. Since a ∈ α, S1 ∈ α too. But P ∈ α∩β, because it is both in α and β. P ∈ α. Thus, S1P ∈ α. Since S1P = S1S2 = r, r ∈ α. However, P ∈ β and S2 ∈ β. Therefore, S2P ∈ β. But S2P = S1S2 = r. So r ∈ β. Therefore, r ∈ α∩β and since r and α∩β are both lines, r = α∩β and this is what we wanted to show.

Is this a correct proof? Am I missing something?

 

[fresh_42]

Isn't it that this only shows $r \subseteq \alpha(a,P) \cap \beta(b,P)$ ? I think without additional conditions as $a \neq b \; , \; a \nparallel b \; , \; a$ and $b$ are skew lines or $P \notin a \cup b$ this cannot be done since the assumed situation can take place, e.g. on a single plane making it the entire intersection.

 

[member 587159]

Isn't it that this only shows $r \subseteq \alpha(a,P) \cap \beta(b,P)$ ? I think without additional conditions as $a \neq b \; , \; a \nparallel b \; , \; a$ and $b$ are skew lines or $P \notin a \cup b$ this cannot be done since the assumed situation can take place, e.g. on a single plane making it the entire intersection.

a and b are crossing lines. So, $a \neq b \;$ and $a \nparallel b \;$.
This is mentioned in what I wanted to proof. Maybe crossing is not the right translation for what I mean. English is not my native language. Thanks for all your help though :)

 

[fresh_42]

Ok, but what if all three lines are planar? Then this whole plane will be the intersection.

 

[micromass]

Ok, but what if all three lines are planar? Then this whole plane will be the intersection.

I think the OP means "crossing lines" to mean "skew lines".

 

[member 587159]

I think the OP means "crossing lines" to mean "skew lines".

I looked up the translation for skew lined in mathematical context and this is exactly what I meant. Sorry for that on my part. Is the proof correct then?

 

[fresh_42]

In such cases I use the following trick: I look up the Wiki page in my language and then switch to "english". It doesn't always work, but often. And I had to look up "windschief" here. So don't mind.

 

[member 587159]

In such cases I use the following trick: I look up the Wiki page in my language and then switch to "english". It doesn't always work, but often. And I had to look up "windschief" here. So don't mind.

Useful trick, thanks. Could you verify whether the proof is correct now?

 

[fresh_42]

The only weakness I saw, was the knowledge that $α∩β$ is a line which is true but not obvious. With it it looks ok to me.

 

[member 587159]

The only weakness I saw, was the knowledge that $α∩β$ is a line which is true but not obvious. With it it looks ok to me.

To show that α∩β is a line, we can proof this in the following way.

There are 3 possibilities.

1) α∩β = ∅

But P ∈ α∩β, contradiction

2) α∩β is a plane

If α∩β is a plane, then α = β. But this is impossible because a ∈ α and b ∈ β and a and b are skew lines and there is no plane that goes through two skew lines. Contradiction.

3) α∩β is a line

This is the only option left.

Thanks a lot for your help :)