In geometric algebra, for deriving the rotations formula in $ \mathbb R^3 $ I see the following steps ($ \mathbf i $ is the plane of rotation, $ \mathbf u $ is the vector to be rotated, with components $ \mathbf u_{\parallel} $ in the plane of rotation and $ \mathbf u_{\bot} $ perpendicular to it, $ \mathbf v $ is the resultant vector):
$$ \mathbf v = \mathbf u_{\parallel} e^{\mathbf i\theta} + \mathbf u_{\bot} $$ $$ = \mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}} + \mathbf u_{\bot}e^{{-\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}} $$ $$ = e^{{-\mathbf i\theta}/{2}}\mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}} + e^{{-\mathbf i\theta}/{2}}\mathbf u_{\bot}e^{{\mathbf i\theta}/{2}} $$ $$ = e^{{-\mathbf i\theta}/{2}}\mathbf u e^{{\mathbf i\theta}/{2}} $$
I am unable to understand step (3).The hint given says that the geometric product of two orthogonal vectors is anti-commutative, which is clear enough, but how does that extend to this case where we have a product of a vector and a multivector ($e^{{\mathbf i\theta}/{2}}$)?
edit: I was able to verify the equality by taking $ \mathbf i = e_1e_2 $ and $ \mathbf u_{\parallel} = ae_1 + be_2 $, but stll not able to understand it geometrically.
This can be seen by breaking the exponential down into trig functions.
$$u_\parallel e^{i\theta/2} = u_\parallel (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$$
Since $u_\parallel$ lies entirely in the plane $i$, it should be clear that $i u_\parallel = -u_\parallel i$. Either way, this is a vector orthogonal to $u_\parallel$ lying in the plane $i$, but the order of multiplication determines whether you get a particular vector or its negative (based on what wedge product of $u_\parallel$ and $i u_\parallel$ yields a bivector with the same orientation as $i$).
When you do that, you get the following:
$$ (\cos \frac{\theta}{2} - i \sin \frac{\theta}{2})u_\parallel = e^{-i \theta/2}u_\parallel$$
Welcome to the wide world of geometric algebra; I hope you continue to find answers to your questions here.