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Solving a vector equation which seems to be indeterminate.

  1. Jun 19, 2013 #1
    I have a vector equation:

    [itex]\vec{A} \times \vec{B} = \vec{C} [/itex]. [itex]\vec{A}[/itex] and [itex]\vec{C}[/itex] are known, and [itex]\vec{B}[/itex] must be determined. However, upon trying to use Cramer's rule to solve the system of three equations, I find that the determinant we need is zero. I know now that I need to choose a "gauge" to proceed, but can someone outline what comes next? Thanks.
  2. jcsd
  3. Jun 19, 2013 #2


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    You know that AxB is perpendicular to both A and B, which means that B is a vector which is perpendicular to C. You have lots of choices for B because AxA = 0, so in particular
    [tex] A\times B = A\times(B+\kappa A) [/tex]
    for any choice of kappa. In particular you can pick kappa to make B perpendicular to A. Once you have B is perpendicular to A and C, all that's left is to find its magnitude (which you can do using the fact that |AxB| = |A||B| if they are perpendicular)
  4. Jun 19, 2013 #3
    Makes sense. Now I have |B|. To get the components of [itex]\vec{B}[/itex], I can use

    [itex]0 = A_{x}[/itex] [itex]B_{x}[/itex] + [itex]A_{y}[/itex][itex]B_{y}[/itex]+ [itex]A_{z}[/itex][itex]B_{z}[/itex]


    [itex] 0 = C_{x}[/itex] [itex]B_{x}[/itex] + [itex]C_{y}[/itex][itex]B_{y}[/itex]+ [itex]C_{z}[/itex][itex]B_{z}[/itex],


    [itex]B^{2} = B_{x}^{2} + B_{y}^{2} + B_{z}^{2}[/itex] .

    Does that sound right?
    Last edited: Jun 19, 2013
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