I just read some proof, where the identity
\begin{equation} a^T (|b|^2 \text{id}) a - a^T (b \otimes b)a = |b \times a|^2\ \ \ \forall a,b \in \mathbb{R}^3\setminus\{0\}, \end{equation}
is used, but I do not see why this is true. When I try to compute it I get
\begin{align} a^T (|b|^2 \text{id}) a - a^T (b \otimes b)a &= \left( \begin{array}{c} a_1 |b|^2\\ a_2 |b|^2\\ a_3 |b|^2\\ \end{array} \right) \cdot a - \left( \begin{array}{c} \sum _{i=1}^3 a_i |b|^2\\ \sum _{i=1}^3 a_i|b|^2\\ \sum _{i=1}^3 a_i |b|^2\\ \end{array} \right) \cdot a \nonumber \\ &= \sum _{i=1}^3 a_i^2 |b|^2 - \sum _{j=1}^3\sum _{i=1}^3 a_ia_j |b|^2. \end{align}
In particular, this becomes negative if all $a_i$ are positive. So, is the above identity wrong or am I missing something here?
We can see this without computing components, though since you don't give which definition of the cross product you're using, it's unclear from context whether you have available the needed facts:
Hint The first term is just $|a|^2 |b|^2$, the second is $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 \theta$, where $\theta \in [0, \pi]$ is the angle between $a, b$, and $|a \times b|^2 = |a|^2 |b|^2 \sin^2 \theta$.