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Math Help - Skew Lines

  1. #1
    Member billa's Avatar
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    Skew Lines

    Could I get a hint on how to go about this problem

    Let L1 and L2 be skew lines; let P1, and P2 be any two distinct points on L1 and let Q1 and Q2 be any two distinct points on L2; let M be a plane parallel to both L1 and L2. Show that the intersections of M with the four lines determined by the line-segments P1Q1, P1Q2,P2Q1, and P2Q2 form the vertices of a parallelogram.

    I tried to solve this for awhile. I started writing things that I knew about the situation down, and I found that the (shortest) distance between the plane and either line was constant (though a different constant for both lines). I figured out a few more things but I am not sure if there are necessary for the argument.


    Thanks, even the smallest piece of advice would be helpful :-)
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  2. #2
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    Very hard to see, but let me give it a try. But beware: how you draw the figure can make it difficult to see.

    I'll make more definitions:
    Let distance of L1 from M be p.
    Let distance of L1 from M be q.
    Let normal from P1 intersects M at A1.
    Let normal from P2 intersects M at A2.
    Let normal from Q1 intersects M at B1.
    Let normal from Q2 intersects M at B2.

    Let P1Q1 intersects M at X.
    Let P1Q2 intersects M at Y.
    Let P2Q1 intersects M at Z.
    Let P2Q2 intersects M at T.
    Clear? Are you still here?

    Now (for point P2)
    |ZB1|/|ZA2| = q/p (similar triangles)
    |TB2|/|TA2| = q/p (similar triangles)
    => triangles: A2B1B2 & A2ZT are similar.
    => ZT is parallel to B1B2. (1)

    Do the same thing for point A1:
    |XB1|/|XA1| = q/p (similar triangles)
    |YB2|/|YA1| = q/p (similar triangles)
    => triangles: A1B1B2 & A1XY are similar.
    => XY is parallel to B1B2. (2)

    (1) and (2) => ZT is parallel to XY.

    Now I will stop here. You try to see:
    ZX || A1A2 and TY || A1A2

    Beware: how you draw the figure can make it difficult to see.

    Do you need a figure? I won't promise if I can do it. But I may try.

    -O believes - deeply.
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  3. #3
    Member billa's Avatar
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    I get the definitions you made(even tho I think you switched forgot to use L2 for q), and I understand what you are trying to show, but I don't get why q/p and |ZB1|/|ZA2| are parts of similar triangles. I am still working on it, but if you happen to read this before I figure it out then maybe you could help. Thanks for the post

    EDIT:

    Alright I think I see why they are similar triangles - you were right about drawing the figure correctly
    When you draw the in a segment from B1 to L1 normal to M and a segment from A2 to L2 that is also normal to M, and you draw in ZA1 and ZA2, there are two right triangles with a vertical angle. I think...

    Still working lol...
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  4. #4
    Member billa's Avatar
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    Ok I think I understand what you wrote enough to do it myself now - but I have one last question

    isn't it
    |ZB1|/|ZA2| = p/q (similar triangles)
    |TB2|/|TA2| = p/q (similar triangles)
    instead of = q/p

    because p is part of the triangle that contains ZB1 and q is part of the triangle that contains ZA2?

    And also, when I finish showing that all the sides are parallel, I think I still need to show that the distances of at least 1 pair of opposite sides are equal - I have to stop working on it tonight tho so I will try again tomorrow.

    Thanks very much again
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  5. #5
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    Depends on how you draw the figure. Please see that attached picture. One line is thicker to give you the feeling of perspective.

    |ZB1|/|ZA2| = q/p (similar triangles)
    |TB2|/|TA2| = q/p (similar triangles)

    Can you see: TZ || B1B2 ?

    Please note the figure show only 25% of the solution. Then you have to look at point P1.
    Then Do the above 2 steps for line A1A2.

    Good luck.

    -O sleeps if need be.
    Attached Thumbnails Attached Thumbnails Skew Lines-8foxes.com_88_skew-lines.png  
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