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

Jul 22, 2016

Reqd. vector$= \left(\frac{3}{\sqrt{14}} , \frac{1}{\sqrt{14}} , - \frac{2}{\sqrt{14}}\right)$.

#### Explanation:

A well-known Property of the Vector Product will be useful in this case.

Given two vectors $\vec{x} \mathmr{and} \vec{y}$, we know that, $\vec{x}$ x $\vec{y}$

is a vector that is $\bot$ to both vecx & vecy

Therefore, taking $\vec{u} \times \vec{v} = \vec{w} ,$ say, we get,

$\vec{w} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(0 , 2 , 1\right) , \left(1 , - 1 , 1\right) |$

$= 3 \hat{i} + \hat{j} - 2 \hat{k} = \left(3 , 1 , - 2\right)$

Now reqd. unit vector, i.e., $\hat{w}$ is given by, $\frac{\vec{w}}{|} | \vec{w} | |$,

where, $| | \vec{w} | | = \sqrt{{3}^{2} + {1}^{2} + {\left(- 2\right)}^{2}} = \sqrt{14}$

Hence, reqd. vector$\hat{w} = \left(\frac{3}{\sqrt{14}} , \frac{1}{\sqrt{14}} , - \frac{2}{\sqrt{14}}\right)$.