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Finding the point of intersection between two curves. (Vectors)

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    At what point do the curves r1(t) = (t, 4-t, 63+t^2) and r2(s)= (9-s, s-5, s^2) intersect?
    Answer in the form: (x,y,z) = ____

    Find the angle of intersection theta to the nearest degree.

    2. Relevant equations



    3. The attempt at a solution

    i: t=9-s
    j: 4-t=s-5
    k: 63+t^2=s^2

    i/j: t-9s
    k: 63+(9-s)^2=s^2
    "Solving for "s""
    s=8
    t=1
    ...
    I know not what to do from here. :-(
     
  2. jcsd
  3. Feb 10, 2013 #2

    cepheid

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    Welcome to PF Jaqsan,

    It sounds like you want to find the "direction" of each curve at the point of intersection. Can you think of a vector that describes this, and how to compute that vector given r(t)?
     
  4. Feb 10, 2013 #3
    I don't understand. The question asks for the point of intersection of the two curves. Do I find the derivative, dot product? Just point me in the right direction please.
     
  5. Feb 10, 2013 #4

    I don't understand. The question asks for the point of intersection of the two curves. Do I find the derivative, dot product? Just point me in the right direction please.
     
  6. Feb 10, 2013 #5

    haruspex

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    What is the vector r1 when t=1?
     
  7. Feb 10, 2013 #6

    Dick

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    You already found the intersection point correctly. So the two tangent vectors are r1'(1) and r2'(8). What are they? Then the angle between two vectors a and b is a.b/(|a||b|). Use that.
     
  8. Feb 10, 2013 #7
    Okay, so I got
    r1'(1) = <1,-1,2>
    r2'(8)= <-1,1,16>
    So do I dot them to get them in one (x,y,z) form?
     
  9. Feb 10, 2013 #8

    Dick

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    Ummm, to answer the (x,y,z)=___ you just substitute. To answer the angle question you want to form a dot product.
     
  10. Feb 10, 2013 #9
    I thought I already did the substitution. r1' = <1,-1,2> r2' = <-1,1,16> What I'm trying is it looks like there are two numbers for each value <x,y,z>
     
  11. Feb 10, 2013 #10

    Dick

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    You substituted correctly into the derivatives. That's fine. When they are asking for the intersection point (x,y,z) you should substitute t=1 into r1(t) or s=8 into r2(s). That's what you calculated for the intersection point, yes? They had better both be the same.
     
    Last edited: Feb 10, 2013
  12. Feb 10, 2013 #11
    Thanks. I figured it out. I was just being retarded. My answers are (1,3,64) and 40degrees
     
    Last edited: Feb 10, 2013
  13. Feb 10, 2013 #12

    cepheid

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    I think you mean (1,3,64), right?
     
  14. Feb 10, 2013 #13
    My bad. Exhibiting my retardness once again. (1,3,64) and 40 degrees.
     
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