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Suppose I have an isometry $f:\Bbb R^3 \rightarrow \Bbb R^3$. In an exercise given in my geometry notes it asks to determine whether isometries preserve the cross product, so that If I have two vectors

$a = (a_1,a_2,a_3)$ and $b=(b_1,b_2,b_3)$ that it is true that $u \times v = f(u) \times f(v)$.

Then I know that $a \times b = [ (a_2b_3-a_3b_2), - (a_1b_3-a_3b_1), a_1b_2-a_2b_1)$ but I am having difficulty preceeding.

Edit: It is also pointed out in the comments that $a \times b = ||a||b|| sin \theta n$ where $\theta$ is the angle between $a,b$ and $n$ is a unit vector perpendicular to the plane containing $a$ and $b$.

So If I have points in $P,Q,R, \in \Bbb R^3$ and let $u= Q-P, v= R-P, u'=f(Q)-f(P), v' = f(R)-f(P)$ then $|u|=|P-Q|=|f(P)-f(Q)|=|u'|$ and similarly $|v| = |v'|$, and since isometries preserve angles then

$|u||v|sin \theta = |u'||v'|sin \theta $ but what about the unit vector perpendicular to the plane spanned by $u,v$ and $u',v'$?

Any hints or insights much appreciated.

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    Hint: do you know the _geometric_ characterization of the cross product? That will be more useful than the algebra. – 2017-01-28
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    Hint 2: Consider the isometry $(x,y,z)\mapsto(x,y,-z)$ and try your goal equation for a trial pairs of $u$, $v$ ... – 2017-01-28
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    @EthanBolker Check out the edit if you have time. – 2017-01-28
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    The edit is good and gets you most of the way to the answer. There's another comment that finishes the job. – 2017-01-28
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    @DietrichBurde this is for general isometries not just rotations though? – 2017-01-28
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    @JamesDickens Yes, but if it is not true for reflections, you are done. And this is explained in the duplicate (see comment). – 2017-01-28

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