# dot and cross product

#### topsquark

Math Team
Phyiscally speaking this is mimics a Crystallography topic. Given three vectors (spanning 3D space) $$\displaystyle \vec{a_i}$$ (typically with a unit volume $$\displaystyle V_a \equiv \vec{a_1} \cdot ( \vec{a_2} \times \vec{a_3} ) = 1$$, though you don't need this condition) you are given to find the basis vectors of the reciprocal lattice. I don't believe that your given solution for the $$\displaystyle \vec{b_i}$$'s in terms of the $$\displaystyle \vec{a_i}$$ is, in general, unique, but they do have the same relations $$\displaystyle b_j a_i = 2 \pi \delta _{ji}$$.

The reason the word "reciprocal" is used here (it's also called a "Bravais lattice") is that we have the inverse relations:
$$\displaystyle \vec{a_1} = 2 \pi \dfrac{( \vec{b_2} \times \vec{b_3} )}{ \vec{b_1} \cdot ( \vec{b_2} \times \vec{b_3} )}$$
etc for the other components.

Let's define a few things:
1) $$\displaystyle V_a = \vec{a_1} \cdot ( \vec{a_2} \times \vec{a_3} ) = \vec{a_2} \cdot ( \vec{a_3} \times \vec{a_1} ) = \vec{a_3} \cdot ( \vec{a_1} \times \vec{a_2} )$$ is the volume of the cell in ordinary 3D space.
(Note that I am using $$\displaystyle \vec{a_2} \cdot ( \vec{a_3} \times \vec{a_1} )$$ instead of your given $$\displaystyle \vec{a_2} \cdot ( \vec{a_1} \times \vec{a_3} )$$. Both give the same answers but my form is more in line with the box product of three vectors.)

2) $$\displaystyle \vec{b_1} = \left ( \dfrac{ 2 \pi }{V_a} \right ) ( \vec{a_2} \times \vec{a_3} )$$.
etc.

Okay, so there are two cases I'm going to calculate. I'll leave the rest to you.

Let's take j = i = 1. Then we expect that $$\displaystyle b_1 a_1 = 2 \pi$$.

$$\displaystyle b_1 a_1 = \left ( \dfrac{ 2 \pi}{V_a} \right ) ~ | \vec{a_2} \times \vec{a_3} | ~ | \vec{a_1} | = \left ( \dfrac{ 2 \pi}{V_a} \right ) ~ ( \vec{a_2} \times \vec{a_3} ) \cdot \vec{a_1} = \left ( \dfrac{2 \pi }{V_a} \right ) \vec{a_1} \cdot ( \vec{a_2} \times \vec{a_3} )$$ (the change from the absolute value to the vector form is true in terms of scalars so we can get away with this.)

$$\displaystyle b_1 a_1 = \left ( \dfrac{2 \pi}{V_a} \right ) V_a = 2 \pi$$
So this checks.

Next case: let j = 1, i = 2. Then we expect $$\displaystyle b_1 a_2 = 0$$.

$$\displaystyle b_1 a_2 = \left ( \dfrac{2 \pi}{V_a} \right ) ~ | \vec{a_2} \times \vec{a_3} | | \vec{a_2} | = = \left ( \dfrac{ 2 \pi}{V_a} \right ) ~ \vec{a_2} \cdot (\vec{a_2} \times \vec{a_3})$$

This is just the familiar form for $$\displaystyle \vec{A} \cdot ( \vec{B} \times \vec{C} ) = \left | \begin{matrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{matrix} \right |$$.

Here we have that $$\displaystyle \vec{A} = \vec{a_2}$$, $$\displaystyle \vec{B} = \vec{a_2}$$, and $$\displaystyle \vec{C} = \vec{a_3}$$.

So
$$\displaystyle b_1 a_2 = \left | \begin{matrix} a_{2x} & a_{2y} & a_{2x} \\ a_{2x} & a_{2y} & a_{2z} \\ a_{3x} & a_{3y} & a_{3z} \end{matrix} \right | = 0$$

Give the rest a try and see what you get.

-Dan