enter image description hereThe vector algebra gets heavy after few steps. Can someone give me a hint as to how to start proving that because I dont think I am doing it right.
Prove that $$ (A\cdot B\times C)(a\cdot b \times c) = \left| \begin{matrix} A\cdot a & A\cdot b & A \cdot c \\ B\cdot a & B\cdot b & B\cdot c \\ C\cdot a & C\cdot b & C\cdot c \end{matrix}\right|. $$
That s my try.
The left-hand expression is linear and fully antisymmetric (that is, an alternating multilinear form) in $A$, $B$ and $C$. We know from the theory of determinants that this implies that it's a multiple of the determinant of the matrix formed by $A$, $B$ and $C$. The same is true for $a$, $b$ and $c$, and the same is true for the right-hand side. Thus it follows that the two sides are multiples of each other, and the factor is readily checked to be $1$ by substituting the canonical basis for both triples.