Lagrange's Identity

The vector identity

$$({\bf A}\times {\bf B})\cdot ({\bf C}\times {\bf D}) = ({\b... ...B}\cdot {\bf D})-({\bf A}\cdot {\bf D})({\bf B}\cdot {\bf C}).$$ (1)

This identity can be generalized to $n$-D,

$$({\bf a}_1\times\cdots\times {\bf a}_{n-1})\cdot({\bf b}_1\t... ..._1 & \cdots & {\bf a}_{n-1}\cdot{\bf b}_{n-1}\cr}}\right\vert,$$ (2)

where $\vert{\hbox{\sf A}}\vert$ is the Determinant of A, or

$$\left({\,\sum_{k=1}^n a_kb_k}\right)^2=\left({\,\sum_{k=1}^n... ...{b_k}^2}\right)- \sum_{1\leq k\leq j\leq n} (a_kb_j-a_jb_k)^2.$$ (3)

See also Vector Triple Product, Vector Quadruple Product

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, p. 1093, 1979.