Jacobi's Identity and Lagrange's Identity

Theorem 6.9 (Jacobi's Identity). For any three vectors $\vec a,\vec b,\vec c$:

Proof. Expand each of the three terms with Theorem 6.8: $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$, $\vec b\times(\vec c\times\vec a)=(\vec b\cdot\vec a)\vec c-(\vec b\cdot\vec c)\vec a$, $\vec c\times(\vec a\times\vec b)=(\vec c\cdot\vec b)\vec a-(\vec c\cdot\vec a)\vec b$. Adding all three and using that the dot product is commutative ($\vec a\cdot\vec c=\vec c\cdot\vec a$, etc.), every term cancels against an equal-and-opposite partner, leaving $\vec 0$. Jacobi's identity is the deepest structural fact about the vector triple product: it says the three "cyclic" vector triple products of any three vectors always balance to zero — this is exactly the property that makes the cross product, together with Jacobi's identity, the defining bracket of a Lie algebra in more advanced mathematics.

Theorem 6.10 (Lagrange's Identity). For any FOUR vectors $\vec a,\vec b,\vec c,\vec d$:

$(\vec a\times\vec b)\cdot(\vec c\times\vec d)=\begin{vmatrix}\vec a\cdot\vec c&\vec a\cdot\vec d\\ \vec b\cdot\vec c&\vec b\cdot\vec d\end{vmatrix}=(\vec a\cdot\vec c)(\vec b\cdot\vec d)-(\vec a\cdot\vec d)(\vec b\cdot\vec c).$ …