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Perpendicular Lines in Three Space

  1. Mar 31, 2005 #1

    Please bear with me...my brain is in vapor lock.

    I have a line L1 given by the following parametric equations:

    x = 2+3t, y = -1+5t, z = 8+2t

    I need to find the equation of a line L2 passing through point B = (1,2,5) and perpendicular to L1.

    For the life of my tired, worn out brain...I cannot figure out how to determine the direction numbers that would make L2 perpendicular to L1 (with direction numbers (3,5,2)).

    Can someone please give my brain a kick start.

    Thanks a bunch.

  2. jcsd
  3. Mar 31, 2005 #2
    Any vector [itex](x, y, z)[/itex] such that [itex](3, 5, 2) \circ (x, y, z) = 0[/itex] has direction perpendicular to that of your line.
    Last edited: Mar 31, 2005
  4. Mar 31, 2005 #3
    That makes sense, but I can see a couple of soulutions to that:

    (a,b,c) = (1,-1,1), (-1,1,-1),(1,-3,6),...

    If I'm on the right track, how do I determine which is correct?

    Thanks again.
  5. Mar 31, 2005 #4
    Any of those would work. Pick one and figure out the intercepts.
  6. Mar 31, 2005 #5
    okay, my brain is starting to show some activity...thanks all

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