# Does line belongs to plane?

• Sep 15th 2006, 02:03 AM
OReilly
Does line belongs to plane?
I have to prove this:

If p and q are two parallel lines, line p doesn't intersect plane alpha, and line q and plane alpha have at least one common point, prove that line q belongs to plane alpha.

My proof:

Lines p and q determine one plane beta.
Then planes alpha and beta have one common point A (A belongs to line q).
Then there is common line q' for alpha and beta.
Line p is parallel to line q' (otherwise it would intersect plane alpha) but then is q'=q beacuse only one line is parallel to line p which goes throught point A, so q belongs to plane alpha.

Is my proof correct?
• Sep 15th 2006, 11:03 AM
Plato
“Lines p and q determine one plane beta.
Then planes alpha and beta have one common point A (A belongs to line q).
Then there is common line q' for alpha and beta.
Line p is parallel to line q' (otherwise it would intersect plane alpha) but then is q'=q because only one line is parallel to line p which goes through point A, so q belongs to plane alpha.”

It seems to me as if there is problem with “Line p is parallel to line q' (otherwise it would intersect plane alpha) but then is q'=q”. Just because two lines do not intersect does not make them parallel.
Skew lines have that property.

You can say that both p & q’ are in plane beta. Because q’ is a subset of alpha it cannot intersect p. Therefore in plane beta we have two non-intersecting lines so they must be parallel.

Now the rest of your proof works.
• Sep 15th 2006, 01:46 PM
OReilly
Quote:

Originally Posted by Plato
“Lines p and q determine one plane beta.
Then planes alpha and beta have one common point A (A belongs to line q).
Then there is common line q' for alpha and beta.
Line p is parallel to line q' (otherwise it would intersect plane alpha) but then is q'=q because only one line is parallel to line p which goes through point A, so q belongs to plane alpha.”

It seems to me as if there is problem with “Line p is parallel to line q' (otherwise it would intersect plane alpha) but then is q'=q”. Just because two lines do not intersect does not make them parallel.
Skew lines have that property.

You can say that both p & q’ are in plane beta. Because q’ is a subset of alpha it cannot intersect p. Therefore in plane beta we have two non-intersecting lines so they must be parallel.

Now the rest of your proof works.

Of course that if two lines do not intersect does not make them parallel automaticaly.

From "Lines p and q determine one plane beta" and "Then there is common line q' for alpha and beta" follows that p and q' belongs to beta.
So if p wouldn't be parallel to q' it would intersect q' and then p would have common point with alpha which is not given assumption so they must be parallel.
• Sep 15th 2006, 01:51 PM
Plato
Quote:

Originally Posted by OReilly
From "Lines p and q determine one plane beta" and "Then there is common line q' for alpha and beta" follows that p and q' belongs to beta. So if p wouldn't be parallel to q' it would intersect q' and then p would have common point with alpha which is not given assumption so they must be parallel.

Yes now it is very clear. The whole proof works.