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Given three vectors $\vec u$, $\vec v$, $\vec w$, and their reciprocals $$ \vec u^{\,\prime} := \frac{\vec v \times \vec w}{(\vec u,\vec v, \vec w)} \qquad \vec v^{\,\prime} := \frac{\vec w \times \vec u}{(\vec u,\vec v, \vec w)} \qquad \vec w^{\,\prime} := \frac{\vec u \times \vec v}{(\vec u,\vec v, \vec w)}$$ where ($(\vec u,\vec v, \vec w) := \vec u \cdot (\vec v \times \vec w)$ is the scalar triple product), show that the reciprocals of $\vec u^{\,\prime}$, $\vec v^{\,\prime}$, and $\vec w^{\,\prime}$ are $\vec u$, $\vec v$, and $\vec w$.

Taking $\vec u$ for instance, which is equal to $$\frac{\vec v^{\,\prime} \times \vec w^{\,\prime}}{(\vec u^{\,\prime},\vec v^{\,\prime}, \vec w^{\,\prime})}$$ what identities can I use to simplify the fraction?

This feels like it'd be somewhat of a trivial proof, but I still couldn't find anything to get closer to a solution.

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    $\begingroup$ What is the $(\vec u,\vec v,\vec w)$ notation in the denominators? $\endgroup$ – Henning Makholm Apr 15 at 19:58
  • $\begingroup$ Hmm, things appear to make some sense if it is the scalar triple product. In particular, then $$ \frac{\vec v\times \vec w}{(\vec u,\vec v,\vec w)} = \frac{\vec v \times \vec w}{\vec u\cdot(\vec v\times \vec w)}$$ which all plays out in the plane spanned by $\vec u$ and $\vec v\times \vec w$ -- and similarly for the other coordinates. $\endgroup$ – Henning Makholm Apr 15 at 20:07
  • $\begingroup$ @Henning Makholm Yes, my bad, that's it. $\endgroup$ – Tyrone_87 Apr 15 at 20:09
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    $\begingroup$ One idea (but I haven't checked if it works) might be to note that it is at least true when $(\vec u,\vec v,\vec w)=(\mathbf e_1,\mathbf e_2,\mathbf e_3)$, and then verify that the identity is preserved if you add a multiple of one of the vectors to one of the others, or multiply one of the vectors by a nonzero constant. $\endgroup$ – Henning Makholm Apr 15 at 20:17
  • $\begingroup$ Your formulas for $v'$ and $w'$ are the same! $\endgroup$ – Andrei Apr 15 at 20:28
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First of all, $\vec u''$ is a multiple of $\vec v'\times \vec w'$, so it's a multiple of $(\vec u \times \vec w) \times (\vec u \times \vec v)$. We know that the $1$-dimensional subspace of $\mathbb R^3$ orthogonal to $\vec x$ and $\vec y$ is spanned by $\vec x \times \vec y$. Well, $\vec u \times \vec w$ and $\vec u \times \vec v$ are both orthogonal to $\vec u$, so $(\vec u\times \vec w) \times (\vec u \times \vec v)$ is a multiple of $\vec u$, and this means that $\vec u''$ is a multiple of $\vec u$.

To see that it's actually equal to $\vec u$, notice that $$\vec u \cdot \vec u'= \vec u \cdot \frac{\vec v\times \vec w}{\vec u \cdot (\vec v \times\vec w)} =\frac{\vec u \cdot (\vec v\times \vec w)}{\vec u \cdot (\vec v \times\vec w)} = 1$$ and also $$\vec u'' \cdot \vec u' = \vec u' \cdot \frac{\vec v' \times \vec w'}{\vec u' \cdot (\vec v' \times \vec w')} = \frac{\vec u' \cdot (\vec v' \times \vec w')}{\vec u' \cdot (\vec v' \times \vec w')} = 1.$$ So $\vec u \cdot \vec u' = \vec u'' \cdot \vec u'$. In general, if we had $\vec u'' = c \vec u$, then $\vec u'' \cdot \vec u'$ would be $c (\vec u \cdot \vec u')$, so in this case $c=1$ and therefore $\vec u'' = \vec u$.

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I'll skip the vector sign, so I can type faster. Lets's calculate $u'\times v'$: $$u'\times v'=\frac{v\times w}{(u,v,w)}\times\frac{w\times u}{(u,v,w)}$$ We can use the formula for triple product $$a\times(b\times c)=(ac)b-(ab)c$$ Let's have $a=v\times w$, $b=w$ and $c=u$. Then $$u'\times v'=\frac{[(v\times w)u]w-[(v\times w)w]u}{(u,v,w)^2}$$ The second term in the numerator is zero, since $v\times w$ is perpendicular to $w$. So $$u'\times v'=\frac{[(v\times w)u]w}{(u,v,w)^2}=\frac{(u,v,w)w}{(u,v,w)^2}=\frac{w}{(u,v,w)}$$ Now let's calculate $(u',v',w')$ $$(u',v',w')=(u'\times v')w'=\frac{w}{(u,v,w)}w'=\frac{w(u\times v)}{(u,v,w)^2}=\frac{1}{(u,v,w)}$$

From these, you get $$\frac{u'\times v'}{(u',v',w')}=\frac{\frac{w}{(u,v,w)}}{\frac{1}{(u,v,w)}}=1$$

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  • $\begingroup$ You can go back and put in some \vec's everywhere. You don't need to be fast, it's not a race. Also, if it were a race, I'd have beaten you by four minutes. $\endgroup$ – Misha Lavrov Apr 15 at 20:53
  • $\begingroup$ I prefer the presentation without the \vecs, since they interfere with the 's. $\endgroup$ – Blue Apr 15 at 21:06
  • $\begingroup$ In the final equation, I think you want $u'\times v'$ in the numerator on the left and $w$ after the last $=$. $\endgroup$ – David K Apr 15 at 21:13
  • $\begingroup$ @MishaLavrov True, your reply was faster. But putting vector signs everywhere will not add anything more to this answer, except that I would need more time to type it. $\endgroup$ – Andrei Apr 15 at 21:19

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