Question

Prove Property$\left(9\right).$

Short answer

We had proved$\Vert \mathit{u}\times \mathit{v}\Vert =\Vert \mathit{u}\Vert \Vert \mathit{v}\Vert \sin \theta .$

Step-by-step solution

Step 1. The value of u×v

We know that,

$\mathbf{(}\mathit{u}\times \mathit{v}{\mathbf{)}}^2=\Vert \mathit{u}{\Vert }^2\Vert v{\Vert }^2-(\mathit{u}·\mathit{v}{)}^2$

Substituting$u·v=\Vert u\Vert \Vert v\Vert \cos \theta$

$=\Vert u{\Vert }^2\Vert v{\Vert }^2-(\Vert u\Vert \Vert v\Vert \cos \theta {)}^2$

$=\Vert \mathit{u}{\Vert }^2\Vert v{\Vert }^2-\Vert \mathit{u}{\Vert }^2\Vert v{\Vert }^2{\cos }^2\theta$

Since, $\left({\sin }^2\theta +{\cos }^2\theta =1\right)$, we can write the above as,

$=\Vert \mathit{u}{\Vert }^2\Vert \mathit{v}{\Vert }^2\left(1-{\cos }^2\theta \right)$

$=\Vert u{\Vert }^2\Vert v{\Vert }^2{\sin }^2\theta$

Step 2. Deduce the equation

The equation obtained is :

$$\begin{gathered} \Vert u\times v{\Vert }^2=\Vert \mathit{u}{\Vert }^2\Vert \mathit{v}{\Vert }^2{\sin }^2\theta \\ \Vert u\times v\Vert =\Vert u\Vert \Vert v\Vert \sin \theta \end{gathered}$$

Hence Proved.