Vector cross product identity for $(a\times b)\cdot(c \times d)$

Question

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.

Answer 1

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:

$$(a\times b)\cdot(c\times d)=(c\cdot a)(d\cdot b)-(c\cdot b)(d\cdot a).~~\blacksquare$$

Answer 2

$$(a\times b)\cdot c=[(c\cdot b)a-(c\cdot a)b$$ $$(a\times b)\cdot (c\times d)$$ Taken $c\times d=x$ $$=[(a\times b)\cdot x]_i$$ $$=x_i\epsilon_{ijk}a_jb_k$$ $$=a_j\epsilon_{jik}x_ib_k$$ $$=[a\cdot (x\times b)]_j$$ $$=[a\cdot ((c\times d)\times b]_j$$ $$=[a\cdot[(b\cdot d)c-(b\cdot c)d]]_j$$ $$=[(a\cdot c)(b\cdot d)-(b\cdot c)(a\cdot d)]_j$$