= Solution
Combine the diagonal and cyclic products from <11/b> into $\sigma_i\sigma_j=\delta_{ij}I+i\sum_k\epsilon_{ijk}\sigma_k$. Distribute the coefficients in the two Pauli-vector factors:
$$
\begin{aligned}
\sigma(\mathbf a)\sigma(\mathbf b)
&=\sum_{i,j}a_ib_j\sigma_i\sigma_j\\
&=\left(\sum_i a_ib_i\right)I
+i\sum_k\left(\sum_{i,j}\epsilon_{ijk}a_ib_j\right)\sigma_k.
\end{aligned}
$$
The inner sums are precisely the components of $\mathbf a\times\mathbf b$, so
$$
\boxed{\sigma(\mathbf a)\sigma(\mathbf b)
=(\mathbf a\cdot\mathbf b)I+i\sigma(\mathbf a\times\mathbf b).}
$$
Interchanging the two factors shows that their <commutator> is $2i\sigma(\mathbf a\times\mathbf b)$. The three <Pauli matrices> are linearly independent: the entries of $\sigma(\mathbf c)$ are $c_z,c_x-ic_y,c_x+ic_y,-c_z$, which vanish together only when $\mathbf c=0$. Thus
$$
\boxed{[\sigma(\mathbf a),\sigma(\mathbf b)]=0
\iff\mathbf a\times\mathbf b=0
\iff\mathbf a,\mathbf b\text{ are linearly dependent}.}
$$
For nonzero real <vectors>, this means parallel or antiparallel directions. The dot and cross products here are bilinear; no complex conjugates are inserted if complex coefficients are considered.
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