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

[Fifthman]

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?

 

[tiny-tim]

Welcome to PF!

Hi Fifthman! Welcome to PF!

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

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

(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)

 

[Architeuthis Dux]

It seems like you are on the right track with your approach. Solving the system of equations by subtracting the second equation from the first is a good method. This will eliminate the A term and leave you with a simpler equation to solve for Y. From there, you can use the fact that A and B are perpendicular to find the magnitude of Y and then solve for the components of Y using the given equation. It might also be helpful to remember that the cross product of two vectors is perpendicular to both of them, so you can use that fact to simplify your equations. Keep working with the equations and see if you can come up with a complete solution. Good luck!