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How to determine whether two lines are parallel or perpendicular to eachother

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Let l1:x=3+t, y=1-t, z=2t and l2:x=-1+s, y=2s, z=1+kt (not sure if this one is just a typo, in which t should actually be s, or whether this is fundamental to the problem) be two lines in R3.
    a) Find all value(s) of k, (if any) for which l1 and l2 are parallel. If not possible then explain why.
    a) Find all value(s) of k, (if any) for which l1 and l2 are perpendicular. If not possible then explain why.


    2. Relevant equations
    These two lines are given in parametric forms, from which I should be able to get the directional vector, and the point associated with each one.

    3. The attempt at a solution
    After noticing that I could grab a little information about each line, I decided to take a look at their directional vectors.
    [tex]l_1: <1,-1,2>[/tex]
    [tex]l_2: <1,2,k>[/tex]
    I'm assuming I'm supposed to be solving for k in this problem?

    after looking at these, I can't see them being proportional to each other, so I don't think they are parallel, granted, I'm not sure how to necessarily show this mathematically.

    My biggest problem is finding all k in which these may be perpendicular, the only idea I've been playing around with would be that if I took the Dot Product between the directional vectors of line 1 and line 2, it would need to equal 0 in order for these two lines to be perpendicular, but I'm not sure if this is the right track of thinking, since I'm not sure what to do about k.

    Is my thinking somewhere along the right lines, or have I made a horrible mistake?

    Any help would be great.
     
  2. jcsd
  3. Sep 28, 2011 #2
    Yes, they are orthogonal if the dot product is 0 and parallel if the dot product is 1
     
  4. Sep 28, 2011 #3

    micromass

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    Not true at all. They are parallel if the directional vectors are a nonzero multiple of eachother.

    To the OP: your track of thinking is great, continue like that.
     
  5. Sep 28, 2011 #4
    Oh wait I meant the cosine between them is 1.
     
  6. Sep 28, 2011 #5

    micromass

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    Not true either. It could be -1 as well.
     
  7. Sep 28, 2011 #6
    But that's anti-parallel
     
  8. Sep 28, 2011 #7

    micromass

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    There is no such thing.
     
  9. Sep 28, 2011 #8
    Yeah when you are 180 degrees, i mean the vectors, not the lines...
     
  10. Sep 28, 2011 #9
    You can take the absolute value of its dot product and the do what is stated by the replies above.Otherwise, you can easily find a vector that is perpendicular to the first vector and "dot" with the second.
     
  11. Sep 28, 2011 #10

    micromass

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    If vectors make an angle of 180 degrees, then they are parallel.
     
  12. Sep 28, 2011 #11
    BUt they point in opposite directions, so they are ANTI- parallel!
     
  13. Sep 28, 2011 #12
    Alright, good to know my approach isn't a bust. So I'll try taking the dot product and solve for k.

    [tex]<1,-1,2> . <1,2,k> = 0[/tex]
    [tex]1(1)+2(-1)+2(k) = 0[/tex]
    [tex]1-2+2k = 0[/tex]
    [tex]-1+2k = 0[/tex]
    [tex]2k = 1[/tex]
    [tex]k = \frac{1}{2}[/tex]

    how does this look?
     
  14. Sep 28, 2011 #13

    micromass

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    That's good!!

    No, they're not.
     
  15. Sep 28, 2011 #14
    Alright, thanks for the tips. Just one last thing, I wanted to know if my statement about these lines not being parallel is valid.

    Since I can't see the directional vector of l1 and l2 being proportional for any value of k, they can't be parallel. This is because if their proportional, that would mean that they both point in the same direction, but have different magnitudes correct?

    And also, what if these lines WERE parallel for some value of k, is there a way to mathematically solve for these values of k?

    Thanks for all the help so far everyone.
     
  16. Sep 28, 2011 #15

    micromass

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    Yes, your solution for checking parallelism is ok.

    But, let's say that our exercise was actually the following:
    (1,-1,2)
    (2,-2,k)

    Then the lines are parallel for a certain k. Indeed, they are a multiple of eachother if there is a t such that

    1=2t
    -1=-2t
    2=kt

    The first and second equation give us that t=1/2. Plugging in in the second gives us k=4. So this is when the lines are parallel!!!!
     
  17. Sep 28, 2011 #16
    Beautiful:cry:

    Thanks, this helped a lot.
     
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