# Show that a vector is equal to the reciprocal of its reciprocal

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.

• What is the $(\vec u,\vec v,\vec w)$ notation in the denominators? – Henning Makholm Apr 15 at 19:58
• 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. – Henning Makholm Apr 15 at 20:07
• @Henning Makholm Yes, my bad, that's it. – Tyrone_87 Apr 15 at 20:09
• 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. – Henning Makholm Apr 15 at 20:17
• Your formulas for $v'$ and $w'$ are the same! – Andrei Apr 15 at 20:28

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 $$\vec u'' = \vec u$$.
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 $$\frac{u'\times v'}{(u',v',w')}=\frac{\frac{w}{(u,v,w)}}{\frac{1}{(u,v,w)}}=1$$
• 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. – Misha Lavrov Apr 15 at 20:53
• I prefer the presentation without the \vecs, since they interfere with the 's. – Blue Apr 15 at 21:06
• In the final equation, I think you want $u'\times v'$ in the numerator on the left and $w$ after the last $=$. – David K Apr 15 at 21:13