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 $\left\lvert{\mathbf a \times \mathbf b}\right\rvert$ equals the area of the parallelogram two of whose sides are $\mathbf a$ and $\mathbf b$.

Proof
By definition of vector cross product:
 * $\mathbf a \times \mathbf b = \left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta \hat {\mathbf n}$

where:
 * $\left\vert{a}\right\vert$ denotes the length of $\mathbf a$
 * $\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction
 * $\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right hand rule.

As $\hat {\mathbf n}$ is the unit vector:
 * $\left\vert{\left({\left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta \hat {\mathbf n} }\right)}\right\vert = \left\vert{\mathbf a}\right\vert \, \left\vert{\mathbf b}\right\vert \sin \theta$

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 $\left\vert{\mathbf b}\right\vert \sin \theta$.

The result follows.