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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.

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Perhaps $(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)$ is being used to denote the scalar triple product $\boldsymbol{a}_1 \cdot (\boldsymbol{a}_2 \times \boldsymbol{a}_3)$? It's hard to tell without more context, though. – Hans Lundmark Jun 17 '12 at 18:31
Yes it is correct $(\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)$ is only a notation for a scalar triple product. And this solves my problem. TY – 71GA Jun 17 '12 at 21:25

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:

$$ \boldsymbol{b}_1 = \frac{2 \pi \, \boldsymbol{\hat{i}} a^2}{a^3} = \frac{2\pi}{a} \, \boldsymbol{\hat{i}} = \frac{2\pi}{a} \, (1, 0, 0) $$

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I suspect that your question is ill posed. Division of a scalar by a vector is not a valid vector space operation.

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But it's not just a vector, that thing in the denominator. It's (something that looks like) a "vector of vectors"! – Hans Lundmark Jun 17 '12 at 18:32

$\def\va{{\bf a}} \def\vb{{\bf b}}$ 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.

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Thank you for pointing that out. Here is the link to the Slovenian Wikipedia site where they put it down wrong. – 71GA Jun 17 '12 at 22:39
@71GA: Sure thing. Cheers! – user26872 Jun 17 '12 at 22:58

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