How can one obtain (5.12) from (5.11)?
The relation
$$\tag{5.11}
\left( \mathbf{a} \times \mathbf{b}\right) \cdot \left( \mathbf{c} \times \mathbf{d}\right)
= \left( \mathbf{a} \cdot \mathbf{c}\right) \left( \mathbf{b} \cdot \mathbf{d}\right) –
\left( \mathbf{a} \cdot \mathbf{d}\right) \left( \mathbf{b} \cdot \mathbf{c}\right)
$$
is called the Identity of Langrange. From ($5.11$) follows
$$\tag{5.12}
\left(\mathbf{a} \times \mathbf{b} \right) \times \mathbf{c} = \left(\mathbf{a}\cdot \mathbf{c} \right)\mathbf{b} – \left( \mathbf{b}\cdot \mathbf{c} \right)\mathbf{a}
$$
Can someone give a hint?
Best Answer
A key observation is that if $\mathbf{A}$ and $\mathbf{B}$ are two vectors such that $\mathbf{A}\cdot \mathbf{C} = \mathbf{B}\cdot \mathbf{C}$ for all vectors $\mathbf{C}$, then $\mathbf{A} = \mathbf{B}$. This fact will be used here.
If $\mathbf{d}$ is an arbitrary vector, then $$[(\mathbf{a}\times \mathbf{b})\times \mathbf{c}]\cdot \mathbf{d} = (\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{d})$$ Use (5.11) to obtain the equivalent expression $$[(\mathbf{a}\cdot \mathbf{c})\,\mathbf{b} - (\mathbf{b}\cdot \mathbf{c})\,\mathbf{a}]\cdot \mathbf{d}$$ Since $\mathbf{d}$ was arbitrary, we deduce (5.12).