# How do you find a unit vector that is orthogonal to both u= - 6i + 4j + k and v= 3i + j + 5k?

May 17, 2017

The unit vector is =1/1774〈19,33,-18〉

#### Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(d , e , f\right) , \left(g , h , i\right) |$

where 〈d,e,f〉 and 〈g,h,i〉 are the 2 vectors

Here, we have veca=〈-6,4,1〉 and vecb=〈3,1,5〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(- 6 , 4 , 1\right) , \left(3 , 1 , 5\right) |$

$= \vec{i} | \left(4 , 1\right) , \left(1 , 5\right) | - \vec{j} | \left(- 6 , 1\right) , \left(3 , 5\right) | + \vec{k} | \left(- 6 , 4\right) , \left(3 , 1\right) |$

$= \vec{i} \left(4 \cdot 5 - 1 \cdot 1\right) - \vec{j} \left(- 6 \cdot 5 - 1 \cdot 3\right) + \vec{k} \left(- 6 \cdot 1 - 4 \cdot 3\right)$

=〈19,33,-18〉=vecc

Verification by doing 2 dot products

〈19,33,-18〉.〈-6,4,1〉=-6*19+4*33-1*18=0

〈19,33,-18〉.〈3,1,5〉=19*3+1*33-5*18=0

So,

$\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$

The unit vector is

hatc=vecc/(||vecc||)=1/sqrt(19^2+33^2+18^2)*〈19,33,-18〉

=1/1774〈19,33,-18〉