Let vectors \(\textbf{a}\) and \(\textbf{b}\) be defined as follows:
$$ \textbf{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \quad \textbf{b} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} $$
The cross product of \(\textbf{a} \times \textbf{b}\) is:
$$\textbf{a} \times \textbf{b} = \langle a_2b_3 - a_3b_2, -(a_1b_3 - a_3b_1), a_1b_2 - a_2b_1 \rangle$$
The magnitude of this product is:
$$\begin{gather}\sqrt {(a_2b_3 - a_3b_2)^2 + (a_1b_3 - a_3b_1)^2 + (a_1b_2 - a_2b_1)^2} \\ \sqrt {(a_2^2b_3^2 -2a_2a_3b_2b_3 + a_3^2b_2^2) + (a_1^2b_3^2 -2a_1a_3b_1b_3 + a_3^2b_1^2) + (a_1^2b_2^2 -2a_1a_2b_1b_2 + a_2^2b_1^2)} \end{gather}$$
The square of the magnitude is:
$$\begin{gather} a_2^2b_3^2+ a_3^2b_2^2 + a_1^2b_3^2 + a_3^2b_1^2 + a_1^2b_2^2 + a_2^2b_1^2 \\ -2a_2a_3b_2b_3 -2a_1a_3b_1b_3 -2a_1a_2b_1b_2 \end{gather}$$
The dot product \(\textbf{a} \cdot \textbf{b}\) is:
$$\begin{align} \Vert\textbf{a}\Vert \ \Vert\textbf{b}\Vert \ \cos(\theta) &= a_1b_1 + a_2b_2 + a_3b_3\end{align}$$
The square of this is:
$$\begin{align} \Vert\textbf{a}\Vert^2 \ \Vert\textbf{b}\Vert^2 \ \cos^2(\theta) &= (a_1b_1 + a_2b_2 + a_3b_3)^2 \\ &= (a_1^2b_1^2 + a_1b_1a_2b_2 + a_1b_1a_3b_3) + \\ & \quad (a_2b_2a_1b_1 + a_2^2b_2^2 + a_2b_2a_3b_3) + (a_3b_3a_1b_1 + a_3b_3a_2b_2 + a_3^2b_3^2) \end{align}$$
Rearranging:
$$ \Vert\textbf{a}\Vert^2 \ \Vert\textbf{b}\Vert^2 \ \cos^2(\theta) = a_1^2b_1^2 + a_2^2b_2^2 + a_3^2b_3^2 + 2a_1b_1a_2b_2 + 2a_1b_1a_3b_3 + 2a_3b_3a_2b_2$$
Notice how there are common terms in \(\Vert \textbf{a} \times \textbf{b} \Vert^2 \) and \((\textbf{a} \cdot \textbf{b})^2\). Which means if we add them:
$$\begin{align} \Vert \textbf{a} \times \textbf{b} \Vert^2 + (\textbf{a} \cdot \textbf{b})^2 &= a_2^2b_3^2+ a_3^2b_2^2 + a_1^2b_3^2 + a_3^2b_1^2 + a_1^2b_2^2 + a_2^2b_1^2 + a_1^2b_1^2 + a_2^2b_2^2 + a_3^2b_3^2 \\ &= a_1^2(b_1^2 + b_2^2+b_3^2)+a_2^2(b_1^2 + b_2^2+b_3^2)+a_3^2(b_1^2 + b_2^2+b_3^2) \\ &= (a_1^2 +a_2^2+a_3^2)(b_1^2 + b_2^2+b_3^2) \end{align}$$
Rearranging:
$$\begin{align} \Vert \textbf{a} \times \textbf{b} \Vert^2 &= (a_1^2 +a_2^2+a_3^2)(b_1^2 + b_2^2+b_3^2) - (\textbf{a} \cdot \textbf{b})^2 \\ &= (\Vert\textbf{a}\Vert^2) (\Vert\textbf{b}\Vert^2) - (\Vert\textbf{a}\Vert^2 \ \Vert\textbf{b}\Vert^2 \ \cos^2(\theta)) \\ &= (\Vert\textbf{a}\Vert^2 \ \Vert\textbf{b}\Vert^2) (1- \cos^2(\theta)) \end{align}$$
Since \(\sin^2(\theta) + \cos^2(\theta) = 1\):
$$\Vert \textbf{a} \times \textbf{b} \Vert^2 = (\Vert\textbf{a}\Vert^2 \ \Vert\textbf{b}\Vert^2) (\sin^2(\theta))$$
Since \(\sin^2(\theta) = \sin(\theta)\) when \(0 \le \theta \le 180\):
$$\Vert \textbf{a} \times \textbf{b} \Vert = \Vert\textbf{a}\Vert \ \Vert\textbf{b}\Vert \ \sin(\theta) $$