Why do Clifford bivectors represent the orthogonal Lie algebra? It takes a long, painful, but straightforward calculation to see that the commutators of grade one elements $[e_i, e_j]$ of a Clifford algebra $\mathrm{Cl}(p,q)$ have exactly the same commutation relations as the rotations in the $(ij)$ plane that form the basis of $\mathfrak{so}(p,q)$. There is a different path to this result that I have yet to understand thoroughly, but on the surface it seems to involve a similarly unilluminating calculation to prove that $\mathrm{ad}_{[e_i, e_j]}$ maps vectors into vectors and (linear combinations of) bivectors into bivectors.
Is there a way to understand why it couldn’t be otherwise? (Not necessarily a complete proof, but at least some motivation to carry out the calculation knowing what to expect.)
 A: Reflection with respect to the hyperplane $u^\perp$ orthogonal to $u$ is $$v\mapsto v - 2(v,u)u/|u|^2,$$ or $$v\mapsto v + \{v,u\}u/|u|^2 = uvu/|u|^2 =-uvu^{-1}$$ in terms of the Clifford algebra product.
A rotation by $\alpha$ in the plane spanned by orthogonormal vectors $e$ and $f$, going from $e$ to $f$, can be expressed as two consecutive reflections, first with respect to $e^\perp$ and then to $[e\cos(\alpha/2) + f\sin(\alpha/2)]^\perp$:
$$v\mapsto rvr^{-1},\qquad r = -[e\cos(\alpha/2) + f\sin(\alpha/2)]e.$$
The minus sign does not change the result, but allows us to obtain for small $\alpha$
$$r = 1 + \alpha(ef/2) + o(\alpha),$$
i. e., the infinitesimal rotation in the $(ef)$ plane is represented by $ef/2 = [e,f]/4$.
This works for a euclidean metric; for a pseudo-euclidean one might need to write $\cosh$ and $\sinh$ sometimes, but that should not change anything. Not sure about degenerate bilinear forms or fields other than $\mathbf R$ or $\mathbf C$.
(This approach is tangentially mentioned on page 6 of Peter Woit’s notesPDF, with reference to Élie Cartan’s La théorie des spineurs. The last formula on the page seems to have a sign error.)
