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?
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).