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Cross Product problem

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data
    i) Find all vectors v such that <1,2,1> X v = <3,1,-5>

    ii) Explain why there is no vector v such that <1,2,1> X v = <3,1,5>

    2. Relevant equations

    a X b = <a[itex]_{2}[/itex]b[itex]_{3}[/itex] - a[itex]_{3}[/itex]b[itex]_{2}[/itex], a[itex]_{3}[/itex]b[itex]_{1}[/itex] - a[itex]_{1}[/itex]b[itex]_{3}[/itex], a[itex]_{1}[/itex]b[itex]_{2}[/itex] - a[itex]_{2}[/itex]b[itex]_{1}[/itex])

    3. The attempt at a solution

    i)
    <1,2,1> X v
    = <2v[itex]_{3}[/itex] -v[itex]_{2}[/itex], v[itex]_{1}[/itex] - v[itex]_{3}[/itex], v[itex]_{2}[/itex] - 2v[itex]_{1}[/itex]>

    Which leaves me with three equations:
    2v[itex]_{3}[/itex] -v[itex]_{2}[/itex] = 3
    v[itex]_{1}[/itex] - v[itex]_{3}[/itex] = 1
    v[itex]_{2}[/itex] - 2v[itex]_{1}[/itex] = -5

    Now, every time I try to fiddle with them, I end up with 0 = 0.

    Any help?
     
  2. jcsd
  3. Sep 22, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Express v2 from the first equation and v1 from the second one. v3 cancels when you substitute for v1 and v2 in the third equation, and you get -5=-5. This means that the three equations are not independent, and you have two equations for three variables: one variable is arbitrary.

    ehild
     
  4. Sep 22, 2011 #3
    Ah, yes, I think I get it now. For some reason I was thinking that I needed to get actual numbers.

    Thanks.
     
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