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Hestenes's "New Foundations for Classical Mechanics" book (page 47, 1.1c) sets a problem to show:

$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot B=\mathbf{a} \left( \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B \right)+\mathbf{b} \left( \left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B \right)+\mathbf{c} \left( \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot B \right).\end{aligned}$

I'm having trouble proving this. Using an antisymmetric expansion of the wedge within a grade one selection, we have

$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot B&=\frac{1}{{6}}{\left\langle \left( \mathbf{a} \mathbf{b} \mathbf{c}+\mathbf{b} \mathbf{c} \mathbf{a}+\mathbf{c} \mathbf{a} \mathbf{b}-\mathbf{a} \mathbf{c} \mathbf{a}-\mathbf{b} \mathbf{a} \mathbf{c}-\mathbf{c} \mathbf{b} \mathbf{a}\right)B \right\rangle}_1 \\ &=\frac{1}{{3}}{\left\langle \mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right) B+\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a}\right) B+\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b}\right) B\right\rangle}_1 \\ &=\frac{1}{{3}}\left(\mathbf{a} \left( \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B \right)+\mathbf{b} \left( \left( \mathbf{c} \wedge \mathbf{a}\right) \cdot B \right)+\mathbf{c} \left( \left( \mathbf{a} \wedge \mathbf{b}\right) \cdot B \right)\right) \\ &\quad +\frac{1}{{3}}\left(\mathbf{a} \cdot {\left\langle \left( \mathbf{b} \wedge \mathbf{c} \right) B \right\rangle}_2+\mathbf{b} \cdot {\left\langle \left( \mathbf{c} \wedge \mathbf{a}\right) B \right\rangle}_2+\mathbf{c} \cdot {\left\langle \left( \mathbf{a} \wedge \mathbf{b}\right) B \right\rangle}_2\right).\end{aligned}$

The first three terms have the desired form, but are off by a factor of 3. My attempts to eliminate the latter three terms have all gone in circles. Any tips, or suggestions for a different approach?

Peeter Joot
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1 Answers1

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I figured it out. The key is trying not to be so clever, instead just expanding in successive dot products and then regrouping. With $B = \mathbf{u} \wedge \mathbf{v}$

$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\cdot B&={\left\langle{{ \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \wedge \mathbf{v} \right) }}\right\rangle}_{1} \\ &={\left\langle{{ \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \mathbf{v} - \mathbf{u} \cdot \mathbf{v} \right) }}\right\rangle}_{1} \\ &=\left( \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{u} \right) \cdot \mathbf{v} \\ &= \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \mathbf{v} \left( \mathbf{c} \cdot \mathbf{u} \right)+\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot \mathbf{v} \left( \mathbf{b} \cdot \mathbf{u} \right)+\left( \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{v} \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a} \left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right)-\mathbf{b} \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{c} \left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right)-\mathbf{a} \left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{b} \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right)-\mathbf{c} \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) - \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{b}\left( \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) - \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \right) \\ &\quad +\mathbf{c}\left( \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) - \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \right) \\ &=\mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right)\cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &\quad +\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a} \right)\cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &\quad +\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &=\mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B+\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B+\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot B.\end{aligned}$

Peeter Joot
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