Chapter 9.7, Problem 58AYU

To determine

To prove: If $u$ and $v$ are orthogonal, then $\Vert u \times v\Vert = \Vert u\Vert \Vert v\Vert$

Answer to Problem 58AYU

If $u$ and $v$ are orthogonal vectors,

To prove that $\Vert u \times v\Vert = \Vert u\Vert \Vert v\Vert$

Since, ${\Vert u \times v\Vert }^2 = {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2 - {\left(\text{u}\cdot \text{v}\right)}^2$

$\begin{array}{lll}{\Vert u \times v\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2 - {\mathit{\left(}\mathit{0}\mathit{\right)}}^2\hfill \end{array}$

Since If $u$ and $v$ are orthogonal, then $u.v=0$

$\begin{array}{llll}{\Vert u \times v\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2\hfill & \hfill \\ \Vert u \times v\Vert \hfill & = \hfill & \Vert u\Vert \Vert v\Vert \hfill & \hfill \end{array}$

Hence proved.

Explanation of Solution

Given:

If $u$ and $v$ are orthogonal vectors

Formula used:

If $u$ and $v$ are orthogonal, then $\text{u}\cdot \text{v} = \mathit{0}$

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

Proof:

If $u$ and $v$ are orthogonal vectors,

To prove that $\Vert u \times v\Vert = \Vert u\Vert \Vert v\Vert$

Since, ${\Vert u \times v\Vert }^2 = {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2 - {\left(\text{u}\cdot \text{v}\right)}^2$

$\begin{array}{lll}{\Vert u \times v\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2 - {\mathit{\left(}\mathit{0}\mathit{\right)}}^2\hfill \end{array}$

Since If $u$ and $v$ are orthogonal, then $u.v=0$

$\begin{array}{llll}{\Vert u \times v\Vert }^2\hfill & = \hfill & {\Vert u\Vert }^{\uparrow }2{\Vert v\Vert }^{\uparrow }2\hfill & \hfill \\ \Vert u \times v\Vert \hfill & = \hfill & \Vert u\Vert \Vert v\Vert \hfill & \hfill \end{array}$

Hence proved.