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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?

danny
  • 946

1 Answers1

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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).

kobe
  • 43,217