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Linear: Finding all scalars for given vector equation

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Vectors
    u=(-2, 9, 6)
    v=(-3, 2, 1)
    w=(1, 7, 5)


    2. Relevant equations
    Show that there is no scalar (Cn) such that:
    C1(u) + C2(v) + C3(w) = (0, 5, 4)

    I'd also like to know where would I start to find all the scalars if there were any, since I'm pretty sure this problem will come eventually, and there's nothing in the textbook about it!

    Thanks
     
  2. jcsd
  3. Oct 8, 2008 #2

    Dick

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    Equate the two sides component by component. That's three equations in three unknowns. The first one is C1*(-2)+C2*(-3)+C3*1=0.
     
  4. Oct 8, 2008 #3
    Thanks! Here's what I tried:

    1: C1(-2) + C2(-3) + C3(1) = 0
    2: C1(9) + C2(2) + C3(7) = 5
    3: C1(6) + C2(1) + C3(5) = 4

    I've eliminated C2 and ended up with 2 new equations:

    C1(-7) + C3(-3) = -3
    C1(16) + C3(16) = 12

    I guess from here I could solve for either C1 or C3, then go back to the original equations, plugging it in, and solving again (3 equations, with 2 unknown this time), is that right? Also, at what point will I reach the end (ie. when will it be clear that there's no solution)

    Another thing, seeing how this is for a linear class, I was tempted to use matrices and start with the following:

    [-3 4 6 | 2]
    [ 1 0 -1 | 0]
    [ 2 -8 -4 | 4]

    Would that work? If so, should I try to solve for [1 0 0], [0 1 0], [0 0 1] system, and whenever I get to the point where I can do anything, I "proved" that this problem has no solution?

    Thanks!
     
  5. Oct 8, 2008 #4

    Dick

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    You are on the right track. Except I don't think the equations in C1 and C3 are right. Check that. Once you think you've got it right, try and eliminate, say C1. You should wind up with an inconsistent equation. Like 0=1. That would show there are no solutions. You can do the same thing with matrices as well. Though I don't see where you got that matrix from.
     
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