(Geometric algebra) Acceleration of a particle with constant speed as a bivector-vector inner product I've been working on (self-studying) Geometric Algebra for Physicists which, sadly, has no solutions manual. This is not a problem in general, but I feel like one of my solutions for a question asked in the textbook is incomplete.
The question is:

A particle in three dimensions moves along a curve $x(t)$ such that $|v|$ is constant. Show that there exists a bivector $Ω$ such that
  $$
\dot v = Ω\cdot v
$$
  and give an explicit formula for $Ω$. Is this bivector unique?

My solution:

Since we're in three dimensions, we can construct a vector $\dot v$ with the following property:
  $$
v^2=v_0^2 \implies \dot v v+v \dot v = 0 \implies v\cdot \dot v = 0
$$
  (This is obvious from elementary multivariable calculus)
As $\dot v$ must always be perpendicular to $v$, we can always come up with such a vector by forming a plane with it and some arbitrary vector and then take this resulting bivector's dual. That is,
  $$
\dot v = I(v\wedge b)
$$
  Where $I=e_1e_2e_3$ is the unit pseudoscalar. We can re-write this in the following form:
  $$
I(v\wedge b) = v\cdot (Ib)=-(Ib)\cdot v
$$If we allow $b$ to absorb the constant, then we can claim:
  $$
\Omega = Ib(t)
$$
  Where $b(t)$ is any vector-valued function of $t$. Clearly, then, the bivector is not unique.

Is this all? It seems like the question implies there should be a more restrictive condition on $\Omega$, but I haven't been able to find any (and, intuitively, it doesn't seem like it could be made more restrictive).
Thank you.
 A: For the sake of completeness, I'll just post up the full answer, which is relatively simple after ahmetselcuk's response.
Since $v^2 = v_0^2$, then we can describe $v$ purely by means of rotations. That is, for some unit vector $u$, we have:
$$
v = v_0RuR^\dagger
$$
where all of the information of motion is purely on the rotor $R$.
Then, we must have 
$$\dot v = v_0\frac{d}{dt}(RuR^\dagger)=v_0\dot RuR^\dagger+v_0Ru\dot R^\dagger.$$
Since $u = \frac{1}{v_0}R^\dagger v R$, it follows that
$$
\dot v = \dot RR^\dagger vRR^\dagger +RR^\dagger v R\dot R^\dagger =\dot RR^\dagger v+vR\dot R^\dagger. 
$$
We know that $\dot RR^\dagger = -R\dot R^\dagger$. Hence, $\dot RR^\dagger$ is a pure bivector. Therefore, the above reduces to:
$$
\dot v = (2\dot RR^\dagger)\cdot v = \Omega \cdot v.
$$
Clearly, $R$ depends on the choice of $u$, thus $\Omega$ is similarly dependent, yet $v$ is not. Hence $\Omega$ is not unique.
A: The restriction comes from the fact that $v$ is on a curve. If you look at the Rotating frames section in the same chapter you will find the answer.
