Why is it that we can represent vectors using the even part of the Clifford algebra? You can represent a vector by a quaternion with no scalar part, and you can also represent the rotation itself as a quaternion.  Then the rotation is applied to the vector by conjugation.  The situation is similar in 2D, where you can represent both the vector and the rotation by complex numbers -- but in this case, the vector includes the "scalar" (real) part of the complex number.  Furthermore, there is no need for conjugation in 2D, and in fact if you tried to conjugate you'd just get the identity rotation due to the commutativity I think.
Now, I know that these cases are both really the even components of their respective Clifford algebras, in which vectors have their own proper representation, and rotations are always applied by conjugation.  But it's interesting that representations of vectors can be finagled from within the even subalgebras.  For the quaternions, I think it might just be because the vectors and bivectors in 3D can effectively be identified with each other (which is why we can call the cross product a vector).  But I'm not sure exactly how to explain why it works, and it's even less clear why it should work for the complex numbers.  And I'm also not sure whether you could pull off a vector representation in higher dimensions; I'm guessing not.
Can anyone shed some light on this topic?
 A: Usually, in this context "vectors" refer to elements of the base inner product space $V$. This canonically embeds into $\text{Cl}(V)$ as $\text{Cl}^1(V)\cong V$. This explains what you're seeing  in dimension $d=3$, since there is an isomorphism (canonical up to sign) between $\text{Cl}^1(V)$ and $\text{Cl}^{d-1}(V)$.
The $d=2$ case, on the other hand, is more exceptional, and the two notions of rotations don't really coincide. While it is true that $\text{Cl}^{\text{even}}(\mathbb{R}^2)\cong\mathbb{C}$ as algebras, this is not an isomorphism of representations.
In the context of Clifford algebras, "rotations" usually refers to the conjugation actions by the pin group $\text{Pin}(V)\subset\text{Cl}(V)$. On $\text{Cl}^1(V)\cong V$ this action factors through $O(V)$ and gives the standard double covering (likewise for the subgroups $\text{Spin}$ and $\text{SO}$). The isomorphisms $\text{Cl}^1(V)\cong\text{Cl}^{d-1}(V)$ is equivariant w.r.t. this action, but the action of $\text{Pin}(2)$ on $\text{Cl}^{\text{even}}(\mathbb{R}^2)$ is trivial, and so not isomorphic to the standard action of $\text{Pin}(2)$ on $\mathbb{C}$.
