Chapter 9.7, Problem 62AYU

Prove property (9).

[Hint: Use the result of Problem 57 and the fact that if $\theta$ the angle between $u\text{ and }v$, then $u\cdot v\text{ = }\Vert u\Vert \Vert v\Vert \cos \theta$.]

To determine

To prove: $\Vert \text{u} \times \text{v}\Vert = \Vert u\Vert \Vert v\Vert \sin \theta$, where $\theta$ is the angle between $u$ and $v$

Answer to Problem 62AYU

If $u$ and $v$ are vectors,

To prove that, $\Vert \text{u} \times \text{v}\Vert = \Vert u\Vert \Vert v\Vert \sin \theta$, where $\theta$ is the angle between $$ u$$ and $v$

We know that, ${\Vert \text{u} \times \text{v}\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(\text{u}\cdot \text{v}\right)}^2$

Put $\text{u}\cdot \text{v} = \Vert u\Vert \Vert v\Vert \cos \theta$

We get, $\begin{array}{lll}{\Vert \text{u} \times \text{v}\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\Vert u\Vert }^2{\Vert v\Vert }^2{\cos }^2\theta \hfill \\ \hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2\left(1 - {\cos }^2\theta \right)\hfill \\ \hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2{\sin }^2\theta \hfill \end{array}$

$\begin{array}{lll}{\Vert \text{u} \times \text{v}\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2{\sin }^2\theta \hfill \\ \Vert \text{u} \times \text{v}\Vert \hfill & = \hfill & \Vert u\Vert \Vert v\Vert \sin \theta ,\text{where}\theta \text{is}\text{the}\text{angle}\text{between}\text{u}\text{and}\text{v}\hfill \end{array}$

Hence proved

Explanation of Solution

Given:

If $u$ and $v$ are vectors, and $\theta$ is the angle between $u$ and $v$

Formula used:

If $u$ and $v$ are vectors, then ${\Vert \text{u} \times \text{v}\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(\text{u}\cdot \text{v}\right)}^2$

$\text{u}\cdot \text{v} = \Vert u\Vert \Vert v\Vert \cos \theta$, where $\theta$ is the angle between $u$ and $v$

Proof:

If $u$ and $v$ are vectors,

To prove that, $\Vert \text{u} \times \text{v}\Vert = \Vert u\Vert \Vert v\Vert \sin \theta$, where $\theta$ is the angle between $$ u$$ and $v$

We know that, ${\Vert \text{u} \times \text{v}\Vert }^2 = {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\left(\text{u}\cdot \text{v}\right)}^2$

Put $\text{u}\cdot \text{v} = \Vert u\Vert \Vert v\Vert \cos \theta$

We get, $\begin{array}{lll}{\Vert \text{u} \times \text{v}\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2 - {\Vert u\Vert }^2{\Vert v\Vert }^2{\cos }^2\theta \hfill \\ \hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2\left(1 - {\cos }^2\theta \right)\hfill \\ \hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2{\sin }^2\theta \hfill \end{array}$

$\begin{array}{lll}{\Vert \text{u} \times \text{v}\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^2{\Vert v\Vert }^2{\sin }^2\theta \hfill \\ \Vert \text{u} \times \text{v}\Vert \hfill & = \hfill & \Vert u\Vert \Vert v\Vert \sin \theta ,\text{where}\theta \text{is}\text{the}\text{angle}\text{between}\text{u}\text{and}\text{v}\hfill \end{array}$

Hence proved