curve with perpendicular derivative We are given a $3 \times 3$ anti-symmetric matrix $A$ and a curve $\gamma :\mathbb{R} \to \mathbb{R}^3$ such that $\gamma' (t)=A\gamma(t)$ Prove that $\gamma$ is a circle.
My attempt:It is a well known fact that $3 \times 3$ transformation matrix can be repaced by a cross product of some vector $a$.So we can write $\gamma'(t)=a\times \gamma(t)$.Now it is clear that $\gamma$ and $\gamma '$ are perpendicular in this case so their inner product is $0$.
$\gamma(t) . \gamma '(t) =0\Rightarrow \gamma(t) .\gamma(t)=const$
It is clear from here that the curver lies on an sphare.I also attempted to show that $\gamma(t) \times \gamma '(t)$ is always in the same direction by multiplying sided of equation by $\gamma(t)$ and using bac-cab identity but I didn't achieve anythin.Any hints to continue?
 A: The solution of the differential equation can actually explicitly given as
$$
\gamma(t) = aa^T\gamma_0
+(I-aa^T)\gamma_0 \cos(\omega t)
+(a\times \gamma_0) \sin(\omega t)
$$
with $\gamma'(t) = \omega\,a \times \gamma(t)$ , $\|a\|=1$ and
$\omega\in\mathbb{R}^{+}_0$
Defining $\gamma$ as shown above, we get
\begin{eqnarray}
\gamma'(t) & = & - (I-aa^T) \,\gamma_0 \, \omega  \,\sin(\omega t)
+(a\times \gamma_0)  \,\omega \,\cos(\omega t) \\
& = & \phantom{+} \omega \, (a\times \gamma_0) \,\cos(\omega t)
- \omega \, (I-aa^T)\, \gamma_0 \,\sin(\omega t)
\end{eqnarray}
and
\begin{eqnarray}
\omega\,a \times \gamma(t) & = &
\phantom{+} \; \omega\, (a \times a) (a^T \gamma_0) \\
& & + \;\omega\, (a \times \gamma_0) \,\cos(\omega t)-\omega\,
(a \times a) (a^T \gamma_0) \cos(\omega t) \\
& & + \;\omega\, a\times (a\times \gamma_0)\,\sin(\omega t) \\
& = & \phantom{+} \; \omega \, (a\times \gamma_0) \,\cos(\omega t)
- \omega \, (I-aa^T)\, \gamma_0 \,\sin(\omega t)
\end{eqnarray}
because
$$
a\times (a\times \gamma_0) = a(a^T \gamma_0)-\gamma_0(a^T a) = 
aa^T\gamma_0 -\gamma_0 = - (I-aa^T)\gamma_0
$$
Now it can be easily shown that $(I-aa^T)\gamma_0$ and $(a\times \gamma_0)$ are two vectors of the same length, which are perpendicular to each other. This means that $\gamma$ is a circle, if $a\times \gamma_0 \neq 0$. Otherwise, it is a single point.
