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Given perpendicular vectors A and B, solve A x Y = B for Y

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the equation [itex]$\mathbf{A}\mathbf{\times Y}=$\mathbf{B}$ [/itex] for perpendicular vectors A and B.

    Derive a general solution for Y.

    2. Relevant equations

    The solution was actually given to us, and I plugged it in to make sure it works. (It does.)

    [itex]
    \textbf{$\mathbf{Y=\frac{1}{\left|A\right|^{2}}}(c\mathbf{A}-\mathbf{A\times}\mathbf{B})$}
    [/itex]

    3. The attempt at a solution

    The solution, conceptually, is the set of all vectors Y perpendicular to B such that
    [itex]
    $\left|\mathbf{Y}\right|sin\theta=\mathbf{\frac{|B|}{|A|}}$
    [/itex]

    As an aside, I tried taking
    [itex]
    \mathbf{A}\mathbf{\times(A\times B})=\mathbf{A(A}\cdot\mathbf{B)}-\mathbf{B|A|^{2}}
    [/itex]
    noting that A and B are perpendicular.

    The instructor, as a hint, suggested solving the system:

    [itex]
    $\mathbf{A}\mathbf{\times Y}=$\mathbf{B}$
    [/itex]
    [itex]
    $\mathbf{A}\mathbf{\times Y_{o}}=$\mathbf{B}$
    [/itex]

    which gave me

    [itex]
    $\mathbf{A}\mathbf{\times(Y-Y_{o})}=$\mathbf{0}$
    [/itex]

    What am I missing that could help me tie this together?
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 9, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Fifthman! Welcome to PF! :smile:

    (try using the B and X2 tags just above the Reply box :wink:)

    If A x (Y - Y0) = 0, then (Y - Y0) must be a multiple of A. :wink:

    (Alternatively, you could have said that Y must be a linear combination of A B and A x B, and just plugged that into the original equation to find the coefficients)
     
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