How to calculate this scalar triple product? I have been trying for over an hour to calculate this scalar triple product but I just can't succeed, I always get some crazy long expression that I can't do anything with.
We have $$\overrightarrow{b_1}=\frac{\overrightarrow{a_2}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_2}=-\frac{\overrightarrow{a_1}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_3}=\frac{\overrightarrow{a_1}\times \overrightarrow{a_2}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$
I have to calculate $\overrightarrow{b_1}\cdot(\overrightarrow{b_2}\times\overrightarrow{b_3})$.
The solution given in the book is $$\frac{1}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$
 A: Notation, $\vec a \cdot \vec b \times \vec c = [\vec a \, \vec b \, \vec c]$
We use vector quadruple product formula
$$(\vec a \times \vec b)\times(\vec c \times \vec d)=[\vec a \, \vec b \, \vec d]\vec c - [\vec a \, \vec b \, \vec c]\vec d$$
to obtain $$\vec b_2 \times \vec b_3 = \dfrac{-(\vec a_1 \times \vec a_3)\times(\vec a_1 \times \vec a_2)}{[\vec a_1\, \vec a_2 \, \vec a_3 ]^2}$$
$$ = \dfrac{-[\vec a_1\, \vec a_3 \, \vec a_2 ] \, \vec a_1}{[\vec a_1\, \vec a_2 \, \vec a_3 ]^2}$$
$$ = \dfrac{\vec a_1}{[\vec a_1\, \vec a_2 \, \vec a_3 ]}$$
Hence $$[\vec b_1 \, \vec b_2 \, \vec b_3] = \dfrac{(\vec a_2 \times \vec a_3 \cdot \vec a_1)}{[\vec a_1 \, \vec a_2 \, \vec a_3]^2}$$
$$ = \dfrac{1}{[\vec a_1 \, \vec a_2 \, \vec a_3]}$$
A: $$\vec u=\frac{\vec a \times \vec b}{[\vec a, \vec b, \vec c]}, ~\vec v=\frac{\vec b \times \vec c}{[\vec a, \vec b, \vec c]},~\vec w=\frac{\vec c \times \vec a}{[\vec a, \vec b, \vec c]}~~~~(1)$$
$$\vec u. (\vec v \times \vec w)=\frac{1}{[\vec a, \vec b, \vec c]^3}[(\vec a \times \vec b).(\vec b \times \vec c)\times (\vec c \times \vec a)]~~~~(2)$$
Let $(\vec b \times \vec c)=\vec p $,
Then$$(\vec b \times \vec c)\times (\vec c \times \vec a)=\vec p \times (\vec c \times \vec a)=(\vec p .\vec a)\vec c-(\vec p. \vec c)\vec a=[\vec b, \vec c, \vec a]\vec c-[\vec b, \vec c, \vec c]\vec a=[\vec a, \vec b, \vec c]\vec c~~~(3)$$
Using this (2) becomes:
$$\vec u. (\vec v \times \vec w)=\frac{[\vec a, \vec b, \vec c]^2}{[\vec a, \vec b, \vec c]^3}=\frac{1}{[\vec a, \vec b, \vec c]}.$$
A: First, note that $$b_2\times b_3=\frac 1{[a_1,a_2,a_3]^2}((a_1\times a_2)\times (a_1\times a_3))$$
since $(-y)\times x=x\times y$ (cross product is anticommutative) and $(\alpha u\times\beta v)=\alpha\beta(u\times v)$ where $\alpha,\beta$ are scalars and $u,v$ are vectors.
We have, by a property of cross-product, $$(a_1\times a_2)\times (a_1\times a_3)=(a_1\cdot(a_2\times a_3))a_1=[a_1,a_2,a_3]a_1$$
So, we have $b_2\times b_3=\frac 1{[a_1,a_2,a_3]}a_1$ and finally,
$$[b_1,b_2,b_3]=b_1\cdot \frac 1{[a_1,a_2,a_3]}a_1=\frac 1{[a_1,a_2,a_3]}(a_1\cdot b_1)=\frac 1{[a_1,a_2,a_3]}$$
since $a_1\cdot b_1=\dfrac {[a_1,a_2,a_3]}{[a_1,a_2,a_3]}=1$
