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

1. Oct 9, 2009

### Fifthman

1. The problem statement, all variables and given/known data

Consider the equation $\mathbf{A}\mathbf{\times Y}=\mathbf{B}$ 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.)

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

3. The attempt at a solution

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

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

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

$\mathbf{A}\mathbf{\times Y}=\mathbf{B}$
$\mathbf{A}\mathbf{\times Y_{o}}=\mathbf{B}$

which gave me

$\mathbf{A}\mathbf{\times(Y-Y_{o})}=\mathbf{0}$

What am I missing that could help me tie this together?

Last edited: Oct 9, 2009
2. Oct 9, 2009

### 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)