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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.2012-06-17
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    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. TY2012-06-17

3 Answers 3