Prove that
$(a\times b)\cdot(c \times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)$
I would appreciate some hints on how to solve this. I assume there is a method which does involve equating LHS and RHS by break them into components which would be extremely tedious to say the least.
$\endgroup$2 Answers
$\begingroup$It might be helpful if you first introduce a new symbol to refer to one of the vector cross-products as a whole. E.g., let's define $(a\times b)=:x$. Using the cyclic property of the scalar triple product, we equate the scalar quadruple product to the dot-product of one of the vectors with the vector triple product of the other three:
$$\begin{align} (a\times b)\cdot(c\times d) &=x\cdot(c\times d)\\ &=d\cdot(x\times c)\\ &=d\cdot\left[(a\times b)\times c\right]. \end{align}$$
Expanding the triple product via Lagrange's formula,
$$(a\times b)\times c=(c\cdot a)b-(c\cdot b)a,$$
we can substitute this back into the formula above. Distributing the dot product with $d$, we find:
$$(a\times b)\cdot(c\times d)=(c\cdot a)(d\cdot b)-(c\cdot b)(d\cdot a).~~\blacksquare$$
$\endgroup$ $\begingroup$$$(a\times b)\cdot c=[(c\cdot b)a-(c\cdot a)b$$$$(a\times b)\cdot (c\times d)$$Taken $c\times d=x$$$=[(a\times b)\cdot x]_i$$$$=x_i\epsilon_{ijk}a_jb_k$$$$=a_j\epsilon_{jik}x_ib_k$$$$=[a\cdot (x\times b)]_j$$$$=[a\cdot ((c\times d)\times b]_j$$$$=[a\cdot[(b\cdot d)c-(b\cdot c)d]]_j$$$$=[(a\cdot c)(b\cdot d)-(b\cdot c)(a\cdot d)]_j$$
$\endgroup$
