Let a be a vector depending on time t, and Ω is a constant vector. The vector a obeys the equation of motion $\frac{da}{dt} = \Omega \times a $
Show that $a . \frac{da}{dt} = 0 $ and what this tells us about the motion of A.
Show $\frac{da^2}{dt} = 0$ and that $d(a.\Omega)/dt=0$ and what this tells us about the motion of A.
From this, sketch the motion of a.
How many initial conditions are required to get a unique solution for a?
By definition of the cross product, $\Omega\times a$ is a vector perpendicular to $a$ and $\Omega$, this implies that $a.(\Omega\times a)=a.{{da}\over{dt}}=0$.
We have ${d\over{dt}}\|a\|^2=2a.{{da}\over{dt}}$, we deduce that $\|a\|$ is constant. The motion is in a sphere.
${{d^2a}\over{dt}}={d\over{dt}}(\Omega\times a)={{d\Omega}\over{dt}}\times a+\Omega\times {{da}\over{dt}}$
$=\Omega\times(\Omega\times a)=0$ anticommutativity
${{d(a.\Omega)}\over{dt}}={{da}\over{dt}}.\Omega+a.{{d\Omega}\over{dt}}$
$=(\Omega\times a).\Omega=0$ by definition, $\Omega\times a$ is orthogonal to $\Omega$. The motion is in the circle orthogonal to $\Omega$.