How to divide by $(a_1,a_2,a_3)$ I have been searching for an explanation in Howard's Linear Algebra and couldn't find an identical example to the one below. 
The example tells me that vectors $\boldsymbol{a}_1$, $\boldsymbol{a}_2$ and $\boldsymbol{a}_3$ are:
$$\boldsymbol a_1 = (a,0,0)$$
$$\boldsymbol a_2 = (0,a,0)$$
$$\boldsymbol a_3 = (0,0,a)$$
And I have to calculate $\boldsymbol b_1$ using equation: 
$$\boldsymbol{b}_1 = \frac{2 \pi \, (\boldsymbol a_2 \times \boldsymbol a_3)}{(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)}$$
So far I've only managed to calculate the cross product $(\boldsymbol a_2 \times \boldsymbol a_3)$ using Sarrus' rule and what I get is:
$$\boldsymbol{b}_1 = \frac{2 \pi \, \hat{\boldsymbol{i}} \, a^2}{(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)}$$
But now I am stuck as I don't know how to calculate with a $(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)$, as this is first time I've come across something like this.  
Could you just point me to what to do next, or point me to a good html site as I still want to calculate this myself. 
Best regards.
 A: I have figured out that $(\boldsymbol a_2 \times \boldsymbol a_3)$ is a vector product which i can calculate like this: 
$$\boldsymbol a_2 \times \boldsymbol a_3 = 
\left|
\begin{array}{ccc}
\boldsymbol{\hat{i}}&\boldsymbol{\hat{j}}&\boldsymbol{\hat{k}}\\
0&a&0\\
0&0&a
\end{array}
\right|
=\boldsymbol{\hat{i}} a a + \boldsymbol{\hat{j}} 0 0 + \boldsymbol{\hat{k}} 0 0 - \boldsymbol{\hat{i}} 0 0 - \boldsymbol{\hat{j}} 0 a - \boldsymbol{\hat{k}} a 0 = \boldsymbol{\hat{i}} a^2$$
I also figured out (thanks to Hans Lundmark) that $(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)$ is actually only a notation for scalar tripple product  $\boldsymbol{a}_1 \cdot  (\boldsymbol{a}_2 \times \boldsymbol{a}_3)$, which i can calculate like this:
$$\boldsymbol a_1 \cdot (\boldsymbol a_2 \times \boldsymbol a_3) = 
\left|
\begin{array}{ccc}
a&0&0\\
0&a&0\\
0&0&a
\end{array}
\right|
= aaa + 000 + 000 - a00 - 00a - 0a0 = a^3$$
If i put all together in an equation for $\boldsymbol b_1$ i get a solution:
$$
\boldsymbol{b}_1 = \frac{2 \pi \, \boldsymbol{\hat{i}} a^2}{a^3} = \frac{2\pi}{a} \, \boldsymbol{\hat{i}} = \frac{2\pi}{a} \, (1, 0, 0)
$$
A: I suspect that your question is ill posed. Division of a scalar by a vector is not a valid vector space operation. 
A: $\def\va{{\bf a}}
\def\vb{{\bf b}}$
This is basically a fleshing out of the comments by @HansLundmark.
I suspect that what is written in the text (or, what was intended to be written) is
$$\vb_1 = \frac{2\pi(\va_2\times\va_3)}{[\va_1,\va_2,\va_3]}.$$
Note that $[\va_1,\va_2,\va_3]$ 
is a standard notation for the scalar triple product,
$$\begin{eqnarray*}
[\va_1,\va_2,\va_3] &=& \va_1\cdot(\va_2\times\va_3) \\
&=& \textrm{det}\langle\va_1,\va_2,\va_3\rangle \\
&=& |\langle\va_1,\va_2,\va_3\rangle|.
\end{eqnarray*}$$
We denote by $\langle\va_1,\va_2,\va_3\rangle$ the matrix whose columns are the vectors $\va_i$. 
(It is common to see this matrix written as $(\va_1,\va_2,\va_3)$, but we use angled brackets to avoid confusion with the notation in the question statement.)
It is possible, though unlikely, that what is intended is 
$$\vb_1 = 2\pi\langle\va_1,\va_2,\va_3\rangle^{-1} (\va_2\times\va_3).$$
This would be an abuse of notation, but is the most natural way for the multiplication to work if $(\va_1,\va_2,\va_3)$ is a matrix and the vectors are column vectors.
