Norm of Vector Cross Product

Theorem
Let $\mathbf a$ and $\mathbf b$ be vectors in the Euclidean space $\R^3$.

Let $\times$ denote the vector cross product.

Then:


 * $(1): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2 = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2$


 * $(2): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert = \left\Vert{\mathbf a}\right\Vert \left\Vert{\mathbf b}\right\Vert \left\vert{\sin \theta}\right\vert$

where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$, or an arbitrary number if $\mathbf a$ or $\mathbf b$ is the zero vector.

Proof
Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.

Then:

This proves $(1)$.

If $\mathbf a$ or $\mathbf b$ is the zero vector, then $\left\Vert{\mathbf a}\right\Vert = 0$, or $\left\Vert{\mathbf b}\right\Vert = 0$ by the positive definiteness norm axiom.

By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector, so $\left\Vert{\mathbf a \times \mathbf b}\right\Vert = 0$.

Hence, equality $(2)$ holds.

If both $\mathbf a$ or $\mathbf b$ are non-zero vectors, we continue the calculations from the first section:

Equality $(2)$ now follows after taking the square root of both sides of the equality.

This is possible as Square of Real Number is Non-Negative.