# the conditions vec(r)* vec(s) =0, and vec(r) * vec(x) =c, and vec(r) xx vec(x) = vec(s) . Find x in each of the three mutually orthogonal directions, vec(r), vec(s) , and vec(r) xx vec(s).

The conditions $\stackrel{\to }{r}\cdot \stackrel{\to }{s}=0$, and $\stackrel{\to }{r}\cdot \stackrel{\to }{x}=c$, and $\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}$ . Find x in each of the three mutually orthogonal directions, $\stackrel{\to }{r}$ , $\stackrel{\to }{s}$ , and $\stackrel{\to }{r}×\stackrel{\to }{s}$
So far $\stackrel{\to }{{x}_{r}}=\frac{\stackrel{\to }{x}\cdot \stackrel{\to }{r}}{|r|}\frac{\stackrel{\to }{r}}{|r|}=\frac{c}{|r{|}^{2}}\stackrel{\to }{r}$ , and $\stackrel{\to }{{x}_{s}}=\frac{\stackrel{\to }{x}\cdot \stackrel{\to }{s}}{|s|}\frac{\stackrel{\to }{s}}{|s|}=\frac{\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)}{|s|}\frac{\stackrel{\to }{s}}{|s|}=0$ Since $\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)$ is the volumn of a degenerate parallelepiped. Where I'm having most difficulty is in...
$\stackrel{\to }{{x}_{\stackrel{\to }{r}×\stackrel{\to }{s}}}=\frac{\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{s}\right)}{|\stackrel{\to }{r}×\stackrel{\to }{s}{|}^{2}}\stackrel{\to }{r}×\stackrel{\to }{s}$ . Since r and s are orthogonal does that mean $|\stackrel{\to }{r}×\stackrel{\to }{s}|=|r||s|$. And also can I calculate $\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{s}\right)$using the triple product to be $|r||s||\stackrel{\to }{{x}_{s\perp }}|=|r||s||\stackrel{\to }{x}-\stackrel{\to }{{x}_{s}}|=|r||s||\stackrel{\to }{x}|$. Is there a simpler simplification of this?
So does $\stackrel{\to }{{x}_{\stackrel{\to }{r}×\stackrel{\to }{s}}}=\frac{|\stackrel{\to }{x}|}{|\stackrel{\to }{r}||\stackrel{\to }{s}|}\stackrel{\to }{x}$?
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Lorenzo Acosta
Just to simplify the notation a bit, let $\stackrel{\to }{t}=\stackrel{\to }{r}×\stackrel{\to }{s}$
Because ${\stackrel{\to }{x}}_{\stackrel{\to }{s}}=0$, we know $\stackrel{\to }{x}={\stackrel{\to }{x}}_{\stackrel{\to }{r}}+{\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ . Therefore
$\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{r}×\left({\stackrel{\to }{x}}_{\stackrel{\to }{r}}+{\stackrel{\to }{x}}_{\stackrel{\to }{t}}\right)=\stackrel{\to }{s}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{r}}+\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}.$
But $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{r}}=0$, so we conclude that $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}$ . Moreover, $\stackrel{\to }{r}$ and ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ are orthogonal, so $|\stackrel{\to }{r}||{\stackrel{\to }{x}}_{\stackrel{\to }{t}}|=|\stackrel{\to }{s}|$
This tells us $|{\stackrel{\to }{x}}_{\stackrel{\to }{t}}|=\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}|}$, so ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}=±\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}|}\frac{\stackrel{\to }{t}}{|\stackrel{\to }{t}|}=±\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}||\stackrel{\to }{t}|}\stackrel{\to }{t}$. We also know $|\stackrel{\to }{t}|=|\stackrel{\to }{r}||\stackrel{\to }{s}|$, further simplifying ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ to $±\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{t}$
Which sign is it? Well, we want $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}$ , but $\stackrel{\to }{r}×\stackrel{\to }{t}$ is proportional to $-\stackrel{\to }{s}$. Therefore ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}=-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{t}$
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Marcus Bass
consider $\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}$
taking cross product from both sides
$\stackrel{\to }{r}×\left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)=\stackrel{\to }{r}×\stackrel{\to }{s}$
you can use vector triple cross product formula
then
$\left(\stackrel{\to }{r}.\stackrel{\to }{x}\right)\stackrel{\to }{r}-\left(\stackrel{\to }{r}.\stackrel{\to }{r}\right)\stackrel{\to }{x}=\stackrel{\to }{r}×\stackrel{\to }{s}$
$\left(\stackrel{\to }{r}.\stackrel{\to }{x}\right)\stackrel{\to }{r}-|\stackrel{\to }{r}{|}^{2}\stackrel{\to }{x}=\stackrel{\to }{r}×\stackrel{\to }{s}$
rearranging
$\stackrel{\to }{x}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}×\stackrel{\to }{s}$
since $\stackrel{\to }{r}$ ,$\stackrel{\to }{s}$ and $\stackrel{\to }{r}×\stackrel{\to }{s}$ are mutually orthogonal they are linearly independent therefore you can get $\stackrel{\to }{x}$ , as a linear combination of these $\stackrel{\to }{r}$ ,$\stackrel{\to }{s}$ and $\stackrel{\to }{r}×\stackrel{\to }{s}$ vectors
$\stackrel{\to }{x}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{\left(}r×\stackrel{\to }{s}\right)-0\stackrel{\to }{s}$
so you can take component as
${\stackrel{\to }{x}}_{\stackrel{\to }{r}}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}$
${\stackrel{\to }{x}}_{\stackrel{\to }{r}×\stackrel{\to }{s}}=\frac{-1}{|\stackrel{\to }{r}{|}^{2}}$
and
${\stackrel{\to }{x}}_{\stackrel{\to }{s}}=0$