Physlib.Mathematics.CrossProduct

For any two vectors $v, w \in \mathbb{R}^3$ and any index $i \in \{0, 1, 2\}$, the $i$-th component of the triple cross product $v \times (w \times v)$ is given by the identity: $$(v \times (w \times v))_i = \left(\sum_k v_k^2\right) w_i - \left(\sum_j v_j w_j\right) v_i$$ where $\sum_k v_k^2$ corresponds to the squared magnitude $\|v\|^2$ and $\sum_j v_j w_j$ corresponds to the dot product $v \cdot w$.

Let $R$ be a commutative ring. For any two three-dimensional vectors $v, w \in R^3$, the dot product of $w$ with the vector triple product $v \times (w \times v)$ satisfies the identity: $$w \cdot (v \times (w \times v)) = (w \times v) \cdot (w \times v)$$ In the specific case where $R = \mathbb{R}$, this identity shows that contracting $w$ with $v \times (w \times v)$ yields the squared length of the cross product, $|w \times v|^2$.

For a fixed vector $\omega \in \mathbb{R}^3$, the function $f: \mathbb{R}^3 \to \mathbb{R}$ defined by $v \mapsto (\omega \times v) \cdot (\omega \times v)$ is smooth (of class $C^\infty$). The expression $(\omega \times v) \cdot (\omega \times v)$ represents the squared magnitude $\|\omega \times v\|^2$ of the cross product of $\omega$ and $v$.