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Math Help - I have a question

  1. #1
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    Exclamation I have a question

    How do I do this:
    Determine whether the lines L1 and L2 given by the vector equations are parallel, perpendicular, or neither.
    L1: r(t) = (-3 + 3t)i + (3 + t)j
    L2: r(t) = (4 + 4t)i + (2 - 12t)j

    If the lines are not parallel, find their point of intersection.
    (__,__)

    This is a practice problem, that we can try to solve before getting to the actual online hw. But due to the fact that I have not had a professor since starting school, I've turned to asking this forums for help, and so far have been the most helpful. Thanks once again
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Consider the direction vectors


    L1: r(t) = (-3 + 3t)i + (3 + t)j v1 = 3 i + j


    L2: r(t) = (4 + 4t)i + (2 - 12t)j v2 = 4 i - 12 j

    If they are parallel then they are scalar multiples i.e. v1 = k v2
    for some k --they are not since to match the j components you would have to mutiply v1 by -12 but then the i components wouldn't match

    If they are perpindicular the dot dt product would be 0

    v1*v2 = 12 - 12 = 0 so they are perpindicular

    to find the point of intersection then it occurs at time t1 and t2

    such that the x and y components match:

    1) -3 +3t1 = 4 + 4t2

    2) 3 +t1 = 2- 12t2

    Mutiply 1) by3 and add to 2) -6 + 10t1 = 14 t1 = 2 use this in L1

    the pt is (3,5)

    To check note -1 = 4t2 t2 = -1/4 use this in L2

    the point is (3,5)
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  3. #3
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    I dont think I quiet understand, can you show me with this other practice problem.

    Determine whether the lines L1 and L2 given by the vector equations are parallel, perpendicular, or neither.
    L1: r(t) = (-3 + 2t)i + (4 + t)j
    L2: r(t) = (3 + 3t)i + (4 - 6t)j

    If the lines are not parallel, find their point of intersection.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    I'm going to let you work on this one yourself.

    However I will elaborate a little.

    you can think of the lines as representing the trajectories of 2 particles.

    They could collide meaning they reach the point of intersection at the same time. Or more generally they reach the point of intersection at different times as in the previous example which is why I used 2 different times t1 and t2.

    A particle traveling on L1 reached the pt of int at t = 2 while the other hit the same pt at t =-1/4
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  5. #5
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    This is where I got lost though
    "Mutiply 1) by3 and add to 2) -6 + 10t1 = 14 t1 = 2 use this in L1"
    Like could you show the work for this because I was trying it and didn't see how you got this
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  6. #6
    MHF Contributor Calculus26's Avatar
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    1) -3 +3t1 = 4 + 4t2

    2) 3 +t1 = 2- 12t2

    After multiplying 1) by 3 we have the system


    1) - 9 + 9t1 = 12 + 12t2
    2) 3 +t1 = 2- 12t2

    Add them

    -6 +10t1 = 14

    10t1 = 20

    t1 = 2

    the particle on L1 reaches the point of intersection at t= 2

    plug this into r(t) = (-3 + 3t)i + (3 + t)j

    what do you get for the x and y components?
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  7. #7
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    What I dont unerstand is why you selected to mutiply by 3 is that a rule or something. Thats what I needed explained
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  8. #8
    MHF Contributor Calculus26's Avatar
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    Have you solved 2 equations in 2 unknowns before ?

    We had to eliminate one variable. We could have multiplied 2) by -3 and added to 1) to eliminate t1 instead.
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  9. #9
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    Okay I understand now, and to answer yor question no, my professor is at a conference and so has not shown up to 1 class, just assign us questions online.
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  10. #10
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    Well I tried and major fail
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