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