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

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Fifthman
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Homework Statement



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

Derive a general solution for Y.

Homework Equations



The solution was actually given to us, and I plugged it into 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]

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?
 
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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)