SciCode / 38.1 / Given two vectors, return the cross-product of these two vectors. The input…

Background
Given the two input vectors
$$
\begin{aligned}
& \mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k} \\
& \mathbf{b}=b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k}
\end{aligned}
$$
their cross product a × b can be expanded using distributivity:
$$
\begin{aligned}
\mathbf{a} \times \mathbf{b}= & \left(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right) \times\left(b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k}\right) \\
= & a_1 b_1(\mathbf{i} \times \mathbf{i})+a_1 b_2(\mathbf{i} \times \mathbf{j})+a_1 b_3(\mathbf{i} \times \mathbf{k})+ \\
& a_2 b_1(\mathbf{j} \times \mathbf{i})+a_2 b_2(\mathbf{j} \times \mathbf{j})+a_2 b_3(\mathbf{j} \times \mathbf{k})+ \\
& a_3 b_1(\mathbf{k} \times \mathbf{i})+a_3 b_2(\mathbf{k} \times \mathbf{j})+a_3 b_3(\mathbf{k} \times \mathbf{k})
\end{aligned}
$$

From this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain:

$$
\begin{aligned}
\mathbf{a} \times \mathbf{b}= & a_1 b_1 \mathbf{0}+a_1 b_2 \mathbf{k}-a_1 b_3 \mathbf{j} \\
& -a_2 b_1 \mathbf{k}+a_2 b_2 \mathbf{0}+a_2 b_3 \mathbf{i} \\
& +a_3 b_1 \mathbf{j}-a_3 b_2 \mathbf{i}+a_3 b_3 \mathbf{0} \\
= & \left(a_2 b_3-a_3 b_2\right) \mathbf{i}+\left(a_3 b_1-a_1 b_3\right) \mathbf{j}+\left(a_1 b_2-a_2 b_1\right) \mathbf{k}
\end{aligned}
$$

For column vectors, we can represent the same result as follows:

$$
\mathbf{a} \times \mathbf{b}=\left[\begin{array}{l}
a_2 b_3-a_3 b_2 \\
a_3 b_1-a_1 b_3 \\
a_1 b_2-a_2 b_1
\end{array}\right]
$$

Background
Given the two input vectors
$$
\begin{aligned}
& \mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k} \\
& \mathbf{b}=b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k}
\end{aligned}
$$
their cross product a × b can be expanded using distributivity:
$$
\begin{aligned}
\mathbf{a} \times \mathbf{b}= & \left(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right) \times\left(b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k}\right) \\
= & a_1 b_1(\mathbf{i} \times \mathbf{i})+a_1 b_2(\mathbf{i} \times \mathbf{j})+a_1 b_3(\mathbf{i} \times \mathbf{k})+ \\
& a_2 b_1(\mathbf{j} \times \mathbf{i})+a_2 b_2(\mathbf{j} \times \mathbf{j})+a_2 b_3(\mathbf{j} \times \mathbf{k})+ \\
& a_3 b_1(\mathbf{k} \times \mathbf{i})+a_3 b_2(\mathbf{k} \times \mathbf{j})+a_3 b_3(\mathbf{k} \times \mathbf{k})
\end{aligned}
$$

From this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain:

$$
\begin{aligned}
\mathbf{a} \times \mathbf{b}= & a_1 b_1 \mathbf{0}+a_1 b_2 \mathbf{k}-a_1 b_3 \mathbf{j} \\
& -a_2 b_1 \mathbf{k}+a_2 b_2 \mathbf{0}+a_2 b_3 \mathbf{i} \\
& +a_3 b_1 \mathbf{j}-a_3 b_2 \mathbf{i}+a_3 b_3 \mathbf{0} \\
= & \left(a_2 b_3-a_3 b_2\right) \mathbf{i}+\left(a_3 b_1-a_1 b_3\right) \mathbf{j}+\left(a_1 b_2-a_2 b_1\right) \mathbf{k}
\end{aligned}
$$

For column vectors, we can represent the same result as follows:

$$
\mathbf{a} \times \mathbf{b}=\left[\begin{array}{l}
a_2 b_3-a_3 b_2 \\
a_3 b_1-a_1 b_3 \\
a_1 b_2-a_2 b_1
\end{array}\right]
$$