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Showing skew lines lie in parallel planes

  1. Jul 16, 2009 #1
    Im givin these two lines..
    L1= x=4+5t y=5+5t z=1-4t
    L2= x=4+s y=-6+8s z=7-3s

    What i tried doing was taking the directional vector of both lines <5 5 -4> <1 8 -3>, and crossing them to find the normal vector. I hav enough information to find 1 equation of a plane, but how can I find the other. Can someone please point me in the right direction. Thanks in advance!
     
  2. jcsd
  3. Jul 16, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi mikemichiel! Welcome to PF! :wink:
    You have the normal …

    can't you find the plane from that?

    And why do you need more than 1 equation for a plane? :smile:
     
  4. Jul 16, 2009 #3

    HallsofIvy

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    To find the equation of a plane, you need a normal vector and a point in the plane. You have two lines that presumably lie in each line in the planes you want. Any point on the line will do.
     
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