# Solving a vector equation which seems to be indeterminate.

1. Jun 19, 2013

### plasmoid

I have a vector equation:

$\vec{A} \times \vec{B} = \vec{C}$. $\vec{A}$ and $\vec{C}$ are known, and $\vec{B}$ must be determined. However, upon trying to use Cramer's rule to solve the system of three equations, I find that the determinant we need is zero. I know now that I need to choose a "gauge" to proceed, but can someone outline what comes next? Thanks.

2. Jun 19, 2013

### Office_Shredder

Staff Emeritus
You know that AxB is perpendicular to both A and B, which means that B is a vector which is perpendicular to C. You have lots of choices for B because AxA = 0, so in particular
$$A\times B = A\times(B+\kappa A)$$
for any choice of kappa. In particular you can pick kappa to make B perpendicular to A. Once you have B is perpendicular to A and C, all that's left is to find its magnitude (which you can do using the fact that |AxB| = |A||B| if they are perpendicular)

3. Jun 19, 2013

### plasmoid

Makes sense. Now I have |B|. To get the components of $\vec{B}$, I can use

$0 = A_{x}$ $B_{x}$ + $A_{y}$$B_{y}$+ $A_{z}$$B_{z}$

and

$0 = C_{x}$ $B_{x}$ + $C_{y}$$B_{y}$+ $C_{z}$$B_{z}$,

and

$B^{2} = B_{x}^{2} + B_{y}^{2} + B_{z}^{2}$ .

Does that sound right?

Last edited: Jun 19, 2013