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B Proof in RxRxR

  1. Jul 15, 2016 #1

    Math_QED

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    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?
     
    Last edited: Jul 15, 2016
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  3. Jul 15, 2016 #2

    fresh_42

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    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.
     
  4. Jul 15, 2016 #3

    Math_QED

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    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 :)
     
    Last edited: Jul 15, 2016
  5. Jul 15, 2016 #4

    fresh_42

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    Ok, but what if all three lines are planar? Then this whole plane will be the intersection.
     
  6. Jul 15, 2016 #5

    micromass

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    I think the OP means "crossing lines" to mean "skew lines".
     
  7. Jul 15, 2016 #6

    Math_QED

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    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?
     
  8. Jul 15, 2016 #7

    fresh_42

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    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.
     
  9. Jul 15, 2016 #8

    Math_QED

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    Useful trick, thanks. Could you verify whether the proof is correct now?
     
  10. Jul 15, 2016 #9

    fresh_42

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    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.
     
  11. Jul 15, 2016 #10

    Math_QED

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    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 :)
     
    Last edited: Jul 15, 2016
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