Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors

Theorem

Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space of $3$ dimensions:

Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.

Then $\norm {\mathbf a \times \mathbf b}$ equals the area of the parallelogram two of whose sides are $\mathbf a$ and $\mathbf b$.

$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$

where:

$\norm {\mathbf a}$ denotes the length of $\mathbf a$

$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction

$\mathbf {\hat n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.

As $\mathbf {\hat n}$ is the unit vector:

$\norm {\paren {\norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \hat {\mathbf n} } } = \norm {\mathbf a} \norm {\mathbf b} \sin \theta$

By Area of Parallelogram, the area of the parallelogram equals the product of one of its bases and the associated altitude.

Let $\mathbf a$ denote the base of the parallelogram.

Then its altitude is $\norm {\mathbf b} \sin \theta$.

The result follows.

$\blacksquare$