For a $2\times 2$ matrix $$\begin{pmatrix}a&b\\c&d \end{pmatrix}
$$
The determinant is given by $ad-bc$. And the cross product of $$\begin{pmatrix} a\\b\\0\end{pmatrix}\times \begin{pmatrix} c\\d\\0\end{pmatrix} =\begin{pmatrix} 0\\0\\ad-bc\end{pmatrix}$$
We can also note that $|a\times b|=|a|b|\sin\theta$ where $\theta$ is the angle between $a$ and $b$. Hence is there any way to relate the determinant to the equation with sine? I recently saw that $$\text{Re}(a)\text{Im}(b)-\text{Im}(a)\text{Re}(b)=|ab|\sin{\text{arg}(a/b)}$$
How would one verify/prove this?
Best Answer
You can actually define the cross product of two vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$ to the be unique vector $\mathbf{a} \times \mathbf{b} \in \mathbb{R}^3$ such that $$ \forall \mathbf{c} \in \mathbb{R}^3, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}), $$ where $(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c})$ denotes the $3 \times 3$ matrix whose columns are $\mathbf{a},\mathbf{b},\mathbf{c}$ in that order. In particular, you can recover $\mathbf{a} \times \mathbf{b}$ as $$ \mathbf{a} \times \mathbf{b} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{i})\mathbf{i} + \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{j})\mathbf{j} + \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{k})\mathbf{k}, $$ which can be massaged using determinant identities to give you the usual ghastly explicit formula; in the special case that $\mathbf{a}$ and $\mathbf{b}$ lie in the $xy$-plane, you immediately recover your observation above. Moreover, it immediately follows that $\mathbf{a} \times \mathbf{b}$ is perpendicular to $\mathbf{a}$, $\mathbf{b}$, and any linear combination of $\mathbf{a}$ and $\mathbf{b}$, since by basic determinant identities, including the fact that a square matrix with repeated columns has a vanishing determinant, $$ (\mathbf{a} \times \mathbf{b}) \cdot (s\mathbf{a}+t\mathbf{b}) = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,s\mathbf{a}+t\mathbf{b}) = s\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{a}) + t\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{b}) = 0. $$ Anyhow, the point of all this you're comfortable with basic linear algebra, especially with how the determinant behaves under elementary row and column operations, then you can derive the identity $$ \|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}} $$ from the identity $$ \forall \mathbf{c} \in \mathbb{R}^3, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}), $$ without too much trouble.
For simplicity, let's assume that $\mathbf{a} \neq \mathbf{0}$ and $\mathbf{b} \neq \mathbf{0}$; otherwise the claim is trivial. Actually, let's show that for any $\mathbf{c} \in \operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$, i.e., for any $\mathbf{c}$ perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, that $$ \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{c}\|. $$ If $\mathbf{a} \times \mathbf{b} \neq \mathbf{0}$, then we can plug in $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ to get $$ \|\mathbf{a} \times \mathbf{b}\|^2 = \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b})\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{a}\times\mathbf{b}\|, $$ and hence $$ \|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}} $$ If $\mathbf{a} \times \mathbf{b} = \mathbf{0}$, then since $\operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$ is at least $1$-dimensional, take any non-zero vector $\mathbf{c} \in \operatorname{Span}\{\mathbf{a},\mathbf{b}\}^\perp$ to get $$ 0 = \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \left(\|\mathbf{a}\|\,\|\mathbf{b}\|\sin(\theta_{\mathbf{a},\mathbf{b}})\right)\|\mathbf{c}\|, $$ which yields $\sin(\theta_{\mathbf{a},\mathbf{b}}) = 0$ and hence $$ \|\mathbf{a} \times \mathbf{b}\| = 0 = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta_{\mathbf{a},\mathbf{b}}. $$
First, by the defining identity for cross products, $$ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}). $$
Next, since determinants are preserved under column additions (e.g., $\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}) = \det(\mathbf{a}\,\vert\,\mathbf{b}+s\mathbf{a}\,\vert\,\mathbf{c})$), we have that $$ \det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c}) = \det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c}), $$ where $$ \mathbf{b}^\prime := \mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a} $$ is the orthogonal projection of $\mathbf{b}$ onto $\operatorname{Span}\{\mathbf{a}\}^\perp$, i.e., onto the plane through the origin with normal vector $\mathbf{a}$; geometrically, if you believe that $\det(\mathbf{a}\,\vert\,\mathbf{b}\,\vert\,\mathbf{c})$ is the signed volume of the parallelepiped spanned by $\mathbf{a},\mathbf{b},\mathbf{c}$, then we're essentially saying that the parallelpiped spanned by $\mathbf{a},\mathbf{b},\mathbf{c}$ has the same volume as the paralleliped spanned by spanned by $\mathbf{a},\mathbf{b}^\prime,\mathbf{c}$ by Cavalieri's principle. Observe, in particular, that $\mathbf{a}$, $\mathbf{b}^\prime$, and $\mathbf{c}$ are pairwise orthogonal by construction.
Next, since $\mathbf{a}$, $\mathbf{b}^\prime$, and $\mathbf{a} \times \mathbf{b}$ are pairwise orthogonal, $$ \lvert\det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})\rvert = \sqrt{\det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^2}\\ = \sqrt{\det\left((\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^T\right) \det(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})}\\ = \sqrt{\det\left((\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c})^T(\mathbf{a}\,\vert\,\mathbf{b}^\prime\,\vert\,\mathbf{c}) \right)}\\ = \begin{vmatrix}\|\mathbf{a}\|^2&0&0\\0&\|\mathbf{b}^\prime\|^2&0\\0&0&\|\mathbf{c}\|^2\end{vmatrix}^{1/2}\\ = \|\mathbf{a}\|\,\|\mathbf{b}^\prime\|\,\|\mathbf{c}\|. $$
At last, since the angle $\theta_{\mathbf{a},\mathbf{b}} \in [0,\pi]$ between the non-zero vectors $\mathbf{a},\mathbf{b}$ is given by the formula $$ \cos \theta_{\mathbf{a},\mathbf{b}} = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|} $$ it follows that $$ \|\mathbf{b}^\prime\|^2 = \left(\mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right) \cdot \left(\mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right)\\ =\|\mathbf{b}\|^2 - 2 \mathbf{b} \cdot \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a} + \left\|\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2}\mathbf{a}\right\|^2\\ =\|\mathbf{b}\|^2 - \frac{(\mathbf{a} \cdot \mathbf{b})^2}{\|\mathbf{a}\|^2}\\ =\|\mathbf{b}\|^2\left(1 - \left(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}\right)^2\right)\\ =\|\mathbf{b}\|^2(1-\cos^2\theta_{\mathbf{a},\mathbf{b}})\\ =\|\mathbf{b}\|^2\sin^2\theta_{\mathbf{a},\mathbf{b}}, $$ and hence that $$ \left\lvert(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\right\rvert = \|\mathbf{a}\|\,\|\mathbf{b}^\prime\|\,\|\mathbf{c}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\left(\theta_{\mathbf{a},\mathbf{b}}\right)\|\mathbf{c}\|, $$ as was claimed.