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Help With Vector Addition

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\vec{A}[/tex]=(6.0[tex]\hat{i}[/tex] - 8.0[tex]\hat{j}[/tex])
    [tex]\vec{B}[/tex]=(-8.0[tex]\hat{i}[/tex] + 3.0[tex]\hat{j}[/tex])
    [tex]\vec{C}[/tex]=(26.0[tex]\hat{i}[/tex] + 19.0[tex]\hat{j}[/tex])

    If aA+bB+C=0, what are the magnitudes of a and b?

    2. Relevant equations

    3. The attempt at a solution

    a(6.0[tex]\hat{i}[/tex] - 8.0[tex]\hat{j}[/tex]) + b(-8.0[tex]\hat{i}[/tex] + 3.0[tex]\hat{j}[/tex]) + (26.0[tex]\hat{i}[/tex] + 19.0[tex]\hat{j}[/tex]) = 0

    I know thats the set up of the equation, but I have no idea how to solve for a and b. Shouldn't there be a second equation to give the relationship between a and b or else there's an infnite number of solutions? Just by looking at it for a few periods, I saw that a = 5 and b = 7 works, but I can't figure out how to reach that answer using algebra!
  2. jcsd
  3. Oct 2, 2007 #2


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    Well you will get two equations. To get the zero vector both i and j components have to be zero.
  4. Oct 2, 2007 #3
    OK, but the thing is what method would be used here to find those two variables? What strategy would be used?
  5. Oct 2, 2007 #4
    Nevermind, I got it
  6. Oct 2, 2007 #5


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    You have:

    [tex] a(6\mathbf{\hat{i}}-8\mathbf{\hat{j}})+b(-8\mathbf{\hat{i}}+3\mathbf{\hat{j}})+(26\mathbf{\hat{i}}+19\mathbf{\hat{j}}) = (6a-8b+26)\mathbf{\hat{i}} + (-8a+3b+19)\mathbf{\hat{j}}=0[/tex]

    For that to equal zero both i and j components must equal 0, so you have 2 simultaneous equations.
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