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How to find the intersection of two vectors

  1. Feb 3, 2010 #1
    L1: r(t) = (-5 + 2t)i + (5 + t)j
    L2: r(t) = (3 + 4t)i + (4 - 8t)j

    I know that they are perfendicular but how do I go about finding the point of intersection?
     
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

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    At any point of intersection, the x-coordinates have to be equal and the y-coordinates have to be equal.
     
  4. Feb 3, 2010 #3
    I'm sorry but could you walk me through this step?
     
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    It would be a good idea to have a different parameter for the second vector function, say s.
    L2: r2(s) = (3 + 4s)i + (4 - 8s)j

    While we're at it let's give different names to the two functions so we can tell them apart.

    For each value of s, r2(s) gives you a different vector. This vector extends from the origin to a point in the plane. What are the coordinates of that point?

    Similarly, for each value of t, r1(t) likewise gives you a different vector. This vector extends from the origin to a point in the plane. What are the coordinates of that point?

    At any point of intersection the coordinates of the point on L1 have to be equal to the coordinates of the point on L2.
     
  6. Feb 3, 2010 #5
    So I've been trying to set the i value of L1 to theat of L2 and likewise for the j value. I can't seem to get the answers though. I solve for the variable (in L1's case "T", correct?).
     
  7. Feb 3, 2010 #6

    Mark44

    Staff: Mentor

    What did you get?
     
  8. Feb 3, 2010 #7
    t=-4 for i and t=-1/9 for j
     
  9. Feb 3, 2010 #8
    Anyone else have any help???
     
  10. Feb 3, 2010 #9

    LCKurtz

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    Show your work for solving for t and s and we can see where you went astray.
     
  11. Feb 3, 2010 #10
    -5+2t=3+4t

    solved for t to equal -4

    5+t=4-8t

    solved for t to equal -1/9

    I'm not really sure if I went about that right but I'm stuck.
     
  12. Feb 3, 2010 #11

    LCKurtz

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    You need to heed Mark44's advice about using different parameters for different curves in the same problem. Use s for one and t for the other and try again.
     
  13. Feb 3, 2010 #12
    LCKurtz, is there a way we could IM?

    I set the second set with an "s"

    -5+2t = 3+4s

    solved for s = -2+(1/2)t

    Do I then plug that in for the s in the original to solve?
     
  14. Feb 3, 2010 #13

    LCKurtz

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    Now you are making progress. Remember the j components have to be equal too. So you should get two equations in two unknowns s and t. See why you need different letters?

    Once you get s and t figured out you can plug them in their equations to check they are at the same point. I assume you can take it from here.
     
  15. Feb 3, 2010 #14
    Okay, I pluged them back in and solved t to equal (7/2). I'm at a brain block and now do not know what to do with this. Please help. It's due in less than an hour. I feel like I'm right there but just cant get it to click.
     
  16. Feb 3, 2010 #15

    LCKurtz

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    Two equations, two unknowns. Solve them carefully -- I'm not going to do that for you because I know you can do it. I promise you it will work. I've gotta hit the sack now.
     
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