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$l_{1}, l_{2}, l_{3} $ are 3 pairwise orthogonal lines in $\mathbb{E_3}$

Prove that there are exactly 8 isometries $F$ with $F(l_{1})=l_{2}$, $F(l_{2})=l_{3}$, $F(l_{3})=l_{1}$ and $l_{1} \cap l_{2} \cap l_{3} $ is a fixed point.

I have no idea how to prove it. Any hints?

bob
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  • Consider first the case where $l_1$ is the $x$-axis, $l_2$ is the $y$-axis and $l_3$ is the $z$-axis. – levap Jan 20 '16 at 15:21
  • I still don't know it.I have to do a rotation or something, but I don't know how exactly. – bob Jan 20 '16 at 15:35
  • The vector $F(e_1)$ should lie in $l_2$ and, since $F$ is isometry, the only options are $F(e_1) = \pm e_2$. Similarly, $F(e_2) = \pm e_3$ and $F(e_3) = \pm e_1$. This gives you eight isometries. – levap Jan 20 '16 at 16:03

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