Equivalence of Definitions of Vector Cross Product

Proof
Let $\mathbf a \times \mathbf b$ be a vector cross product by definition $2$.

Then by definition:
 * $(1): \quad \mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$

By Angle Between Vectors in Terms of Dot Product:
 * $\cos \theta = \ds \frac {\mathbf a \cdot \mathbf b} {\norm {\mathbf a} \norm {\mathbf b} }$


 * Cross product equivalency circle.png

Note that $\theta$ is measured from $0$ to $\pi$.

If $\dfrac \pi 2 < \theta < \pi$, the dot product is negative and it exists in quadrant $2$.

The sine ratio is unaltered.

This gives:

Next we have:

Then:

Hence:

To simplify subsequent algebra, let us define $u$:

Let $\mathbf {\hat n} = \begin {bmatrix} n_i \\ n_j \\ n_k \end {bmatrix}$

Now, by definition $2$ of the cross product:

Hence by definition of dot product:

Then:

Substituting from $(12)$ into $(11)$:

and from $(12)$ into $(10)$:

Hence from $(12)$, $(13)$ and $(14)$ we have the components of $\mathbf {\hat n}$:

Substituting back into $(5)$:

Hence the result.