# How do you find a unit vector perpendicular to both vector u(1, -1,-1)and vector v(2, -2, 3)?

##### 1 Answer
Jul 11, 2016

$\pm \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 0\right)$

#### Explanation:

The vector perpendicular to both $u \mathmr{and} v$ is $\pm \frac{u X v}{|} u X v |$

$= \pm \frac{\left(1. - 1 , - 1\right) X \left(2 , - 2 , 3\right)}{|} \left(1. - 1 , - 1\right) X \left(2 , - 2 , 3\right) |$

Now, the numerator vector is

$\left(\left(- 1\right) \left(3\right) - \left(- 1\right) \left(- 2\right) , \left(- 1\right) \left(2\right) - \left(1\right) \left(3\right) , \left(1\right) \left(- 2\right) - \left(- 1\right) \left(2\right)\right)$

$= \pm \left(- 5 , - 5 , 0\right)$ and $| \left(- 5 , - 5 , 0\right) | = 5 \sqrt{2}$'

So, the answer is $\pm$(-5, -5, 0)$/ \left(5 \sqrt{2}\right)$

$= \pm \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 0\right)$.