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

[Fifthman]

Homework Statement

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

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

The Attempt at a Solution

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

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

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

<br /> $\mathbf{A}\mathbf{\times Y}=$\mathbf{B}$<br />
<br /> $\mathbf{A}\mathbf{\times Y_{o}}=$\mathbf{B}$<br />

which gave me

<br /> $\mathbf{A}\mathbf{\times(Y-Y_{o})}=$\mathbf{0}$<br />

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)