How do I prove that $\Vert u \times v\rVert = \lVert u\rVert\lVert v\rVert\sin(\theta)$?

Question

I already know the following properties

$$\langle u, u\rangle = \lVert u\rVert^2$$

and $$\langle u\times v, w\rangle = \langle u ,v\times w\rangle$$

However, I am unable to use these properties in order to prove the above property in the question.

Answer 1

• If $\vec u$ or $\vec v$ is the zero vector, then $\left\lVert\vec u\right\rVert=0$, or $\left\lVert\vec v\right\rVert=0$ by the positive definiteness norm axiom.

Clearly $\vec u\times\vec v$ is also the zero vector, so $\left\lVert\vec u\times \vec v \right\rVert=0$.

Hence, equality $\left\lVert \vec u \times \vec v\right\rVert = \lVert \vec u\rVert\lVert \vec v\rVert\sin \theta$ holds.

• If both $\vec u$ or $\vec v$ are non-zero vectors, then we have

$$ \begin{align} \left\lVert \vec u \times \vec v\right\rVert^2 &=\left\lVert\vec u\right\rVert^2\left\lVert\vec v\right\rVert^2−(\vec u\cdot\vec v)^2\\ &=\left\lVert\vec u\right\rVert^2\left\lVert\vec v\right\rVert^2−\left\lVert\vec u\right\rVert^2\left\lVert\vec v\right\rVert^2 \cos^2\theta\\ &=\left\lVert\vec u\right\rVert^2\left\lVert\vec v\right\rVert^2(1−\cos^2\theta)\\ &=\left\lVert\vec u\right\rVert^2\left\lVert\vec v\right\rVert^2\sin^2\theta \end{align} $$ $$\implies\left\lVert \vec u \times \vec v\right\rVert = \left\lVert \vec u\right\rVert\left\lVert \vec v\right\rVert\sin \theta$$ since square of real number is non-negative.