How do you find a unit vector that is perpendicular to both the vector u = 0,2,1 and v = 1, -1, 1?

Sep 15, 2016

$\pm \frac{1}{\sqrt{14}} \left(3 , 1 , - 2\right)$, for opposite directions.

Explanation:

If vector $u = \left({u}_{1} , {u}_{2} , {u}_{3}\right) \mathmr{and} v = \left({v}_{1} , {v}_{2} , {v}_{3}\right)$, then

vectors $\pm u X v = \pm \left({u}_{2} {v}_{3} - {u}_{3} {v}_{2} , {u}_{3} {v}_{1} - {u}_{1} {v}_{3} , {u}_{1} {v}_{2} - {u}_{2} {v}_{1}\right)$

are perpendicular to both $u \mathmr{and} v$, in the opposite directions.

Here, $u \left(0 , 2 , 1\right) \mathmr{and} v = \left(1 , - 1 , 1\right)$. So,

$\pm u X v$

=+-((2)(1)-(1)(-1), (1)(1)-(0)((1), (0)(-1)-((2)(1))

$= \pm \left(3 , 1 , - 2\right)$.

For unit vectors,

divide by the modulus $| \left(3 , 1 , - 2\right) | = \sqrt{{3}^{2} + {\left(1\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{14}$,