- #1
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?
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?
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