# How do you find a unit vector that is orthogonal to both u = (1, 0, 1) v = (0, 1, 1)?

Jul 23, 2016

$\frac{u \times v}{| | u \times v | |} = \left(- \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$

#### Explanation:

The cross product of $u = \left({u}_{1} , {u}_{2} , {u}_{3}\right)$ and $v = \left({v}_{1} , {v}_{2} , {v}_{3}\right)$ is given by:

$\left({u}_{1} , {u}_{2} , {u}_{3}\right) \times \left({v}_{1} , {v}_{2} , {v}_{3}\right) = \left(\left\mid \begin{matrix}{u}_{2} & {u}_{3} \\ {v}_{2} & {v}_{3}\end{matrix} \right\mid , \left\mid \begin{matrix}{u}_{3} & {u}_{1} \\ {v}_{3} & {v}_{1}\end{matrix} \right\mid , \left\mid \begin{matrix}{u}_{1} & {u}_{2} \\ {v}_{1} & {v}_{2}\end{matrix} \right\mid\right)$

This will be orthogonal to both $u$ and $v$, but will need scaling to make it unit length.

So we find:

$u \times v = \left(1 , 0 , 1\right) \times \left(0 , 1 , 1\right)$

$= \left(\left\mid \begin{matrix}0 & 1 \\ 1 & 1\end{matrix} \right\mid , \left\mid \begin{matrix}1 & 1 \\ 1 & 0\end{matrix} \right\mid , \left\mid \begin{matrix}1 & 0 \\ 0 & 1\end{matrix} \right\mid\right)$

$= \left(- 1 , - 1 , 1\right)$

Then:
||""(-1, -1, 1) || = sqrt((-1)^2+(-1)^2+1^2) = sqrt(1+1+1) = sqrt(3)

So to make $\left(- 1 , - 1 , 1\right)$ into a unit vector, divide it by $\sqrt{3}$:

$\frac{1}{\sqrt{3}} \left(- 1 , - 1 , 1\right) = \frac{\sqrt{3}}{3} \left(- 1 , - 1 , 1\right) = \left(- \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$