Multiplication with vectors.

Question

Well, I'm not quite sure if I chose right terms for my problem but I will give it a chance. Here, I have some tasks and examples ( http://www.mif.vu.lt/matinf/asm/gr/p12.pdf ). On the bottom of the document there is an example and I don't know how calculations are made here. There is something like this (...)(i)(...). The first brackets are multiplied by (i) and I don't understand how to get that result. What calculations are made step by step. Later the new result is multiplied by second brackets and the answer is made, which I don't understand too. Can anybody explain or give some other examples how calculations like this are made ir a little clearer way? Thanks.

Answer 1

As Wouter already indicated in a comment, your formula uses quaternions. In a certain way, quaternions are a generalization of complex numbers where you have three imaginary units instead of one. Conventionally these are denoted as $\mathbf i, \mathbf j, \mathbf k$. The square of each of the three is $-1$. Every quaternion can be expressed using four real numbers:

$$a+b\mathbf i+c\mathbf j+d\mathbf k\qquad a,b,c,d\in\mathbb R$$

For their multiplication, the following multiplication table can be used:

$$\begin{array}{r|rrrr} & 1 & \mathbf i & \mathbf j & \mathbf k \\\hline 1 & 1 & \mathbf i & \mathbf j & \mathbf k \\ \mathbf i & \mathbf i & -1 & \mathbf k & -\mathbf j \\ \mathbf j & \mathbf j & -\mathbf k & -1 & \mathbf i \\ \mathbf k & \mathbf k & \mathbf j & -\mathbf i & -1 \\ \end{array}$$

Note that multiplication of quaternions is not commutative, so order matters.

So for example one step in your computation is

$$ (\sqrt2+\mathbf j+\mathbf k)\mathbf i = \sqrt2\mathbf i+(\mathbf{ji})+(\mathbf{ki}) = \sqrt2\mathbf i-\mathbf k+\mathbf j $$

The other steps all work pretty much the same: Use the distributive law to turn the product of two sums into the sum of a number of products, multiply real coefficients and use the above table to multiply the imaginary units.

Your text uses three-element vectors $(b;c;d)$ to describe purely imaginary quaternions (i.e. where the $a$ component is zero). One such is $\mathbf u=(0;1;1)=\mathbf j+\mathbf k$. It also uses the conjugate quaternion $\overline{\mathbf\alpha}$ to denote the inverse rotation. You get the conjugate by negating the three vector-part coefficients, i.e.

$$\overline{a+b\mathbf i+c\mathbf j+d\mathbf k}= a-b\mathbf i-c\mathbf j-d\mathbf k$$