Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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:

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

share|cite|improve this answer

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

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.