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Prove the following:

$a⋅(b\times c)=(a\times b)⋅c$

all letters are vectors, the x is cross product, the ⋅ is dot product

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    This looks like a homework question. Mind sharing what you have tried so far? – 2017-02-08
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    Hint $a \cdot (b \times c) = det(a,b,c)$ – 2017-02-08

2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \vec{a}\cdot\pars{\vec{b}\times\vec{c}} & = \sum_{i}a_{i}\pars{\vec{b}\times\vec{c}}_{i} = \sum_{i}a_{i}\pars{\sum_{jk}\color{#f00}{\epsilon_{jki}}b_{j}c_{k}} = \sum_{k}\pars{\sum_{ij}\color{#f00}{\epsilon_{ijk}}a_{i}b_{j}}c_{k} \\[5mm] & = \sum_{k}\pars{\vec{a}\times\vec{b}}_{k}c_{k} = \bbx{\ds{\vec{a}\times\vec{b}\cdot\vec{c}}} \end{align}

Note that $\ds{\epsilon_{ijk} = \epsilon_{jki}}$.

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My favorite formulation of this is that if $a,b,c$ are row vectors, then

$$a \cdot (b \times c) = \det \left( \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right) $$

This is particularly easy to see by comparing to the mnemonic for the cross product

$$ b \times c = \det \left( \begin{matrix} \mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right) $$

By permuting the rows around, we get the matrix computing $c \cdot (a \times b) = (a \times b) \cdot c$.

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    p.s. I prefer to make $a,b,c$ the columns of the matrix rather than the rows, but I wrote it this way in my answer since the formula above is how I've always seen the cross product mnemonic written. – 2017-02-09