I am trying to solve the following equation for x, in plain algebra this was easy
$ y = x - \frac{1}{ x} $
$ x^{2} - yx - 1 = 0 $
$ x = \frac{-y \pm \sqrt (y^{2} + 4)}{2} $
However, throwing arrows on top of this solution leads to a nonsense result. attempting to do this with vectors has me confused
$ \vec y = \vec x - \frac{\vec x}{||\vec x ||^{2}}$
$ \vec y ||\vec x ||^{2} = \vec x ||\vec x ||^{2} - \vec x$
what do I do with the magnitude of the unknown vector?
On
It might help if you introduce a new unit vector $$ \vec{u} = \frac{\vec{x}}{|\vec{x}|} $$
Then we make real coefficients, $$ \vec{x} = x \; \vec{u}, $$ $$ \vec{y} = y \; \vec{u}. $$ This means we have chosen $x = |\vec x|$
Next, you wanted $$ \vec y = \vec x - \frac{\vec x}{||\vec x ||^{2}},$$ $$ y \vec u = x \; \vec u - \frac{ x \;\vec u}{x^{2}},$$ $$ y \; \vec u = \left(x - \frac{1}{x} \right) \; \vec u .$$ $$ y = \left(x - \frac{1}{x} \right) .$$
$\vec x$ is a scalar multiple of $\vec y$, so you really just need to find their relative magnitudes, i.e., solve for $\|x\|$ in terms of $\|y\|$. That leads right back to your original equation and solution.