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For what values of a are teh vectors parallel

  1. Sep 14, 2013 #1
    For what values of "a" are teh vectors parallel

    1. The problem statement, all variables and given/known data
    v = 5a i -3 j
    w = a^2 i -6 j

    2. Relevant equations

    3. The attempt at a solution

    My first thought was too get the unit vectors of both and set them equal to each other and solve for a, but that got messy.

    (5a i- 3j)/sqrt(25a^2 +9) = (a^2i -6j)/sqrt(a^4 +36)

    I tried solving for a separately and got a=5, but its obvious that is wrong because i end up with
    25 i -3 j =25 i -6 j

    I'm not supposed to use the dot product for this problem.
     
  2. jcsd
  3. Sep 14, 2013 #2

    ehild

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    For two vectors parallel, one is a scalar multiple of the other. wu.

    Write it out in components.

    ehild
     
  4. Sep 14, 2013 #3
    If you can't use the dot product, how about the cross product instead?
     
  5. Sep 14, 2013 #4
    Oh right. That seems to work out. So a=-10. The two also seem to be parallel if a=0 since you would end up with
    -3j=6j
    but that isn't something that results from doing what you told me. Is that a trivial case i should just notice or is there an algebraic way of solving that? Thanks for your help thus far.

    I can't use the cross product either.
     
  6. Sep 14, 2013 #5
    Using the proportional method suggested by ehild, you do get both answers analytically:

    If you do the math right, after solving for λ and substituting, you should end up with an equation that looks like $$ a^2 = 10 a$$

    This has solutions ## a={10, 0}##. Perhaps you divided out the ##a## earlier on before you noticed the zero solution, which prevents that division?
     
  7. Sep 14, 2013 #6
    Oh I see. Yeah I did do that division a bit earlier. That makes much more sense. Thanks very much for your help guys.
     
  8. Sep 14, 2013 #7
    Sure thing!
     
  9. Sep 14, 2013 #8

    ehild

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    No, that is wrong. Check.


    Both solution result from the method I suggested.

    ehild
     
  10. Sep 14, 2013 #9
    The vectors are conviniently in a 2D world so you can work with slopes - assume any of the i or j unit vectors as x or y or what notation you might prefer. Tan(slope angle) is the vector's ordinate/abscissa so you get 5a/-3 = a^2/-6 or reverse the denoms and numerators since it's called the main property of an equality or something where a/b = c/d is the same as b/a = d/c. The slopes are equal because the vectors are parallel and you arrive at the same solution a = {0, 10}
     
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