# How do you find a Unit vector perpendicular to vector a=(4,-3,1) and vector b=(2,3,-1)?

Feb 21, 2017

$\hat{\vec{n}} = \frac{\sqrt{10}}{10} \left(- \vec{j} + 3 \vec{k}\right)$

#### Explanation:

$\text{ A vector perpendicular to: " veca "& "vecb" can be found by finding the cross product} \vec{a} \times \vec{b}$

$\text{the corresponding unit vector is found by } \hat{\vec{u}} = \frac{\vec{u}}{|} \vec{u} |$

$\vec{a} = < 4 , - 3 , 1 > , \vec{b} = < 2 , 3 , - 1 >$

$\vec{a} \times \vec{b} = | \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(4 , - 3 , 1\right) , \left(2 , 3 , - 1\right) |$

expanding by Row 1

$\vec{a} \times \vec{b} = \vec{i} | \left(- 3 , \textcolor{w h i t e}{-} 1\right) , \left(\textcolor{w h i t e}{-} 3 , - 1\right) | - \vec{j} | \left(4 , 1\right) , \left(2 , - 1\right) | + \vec{k} | \left(4 , - 3\right) , \left(2 , 3\right) |$

$\vec{a} \times \vec{b} = \vec{i} \left(3 - 3\right) + \vec{j} \left(- 4 - 2\right) + \vec{k} \left(12 - - 6\right)$

$\vec{a} \times \vec{b} = - 6 \vec{j} + 18 \vec{k}$

since we are only wanting the direction we can simplify teh vector by removing any common factors.

so a perpendicular vector in this direction can be written

$\vec{n} = - \vec{j} + 3 \vec{k} \implies | \vec{n} | = \sqrt{{1}^{2} + {3}^{2}} = \sqrt{10}$

so a unit vector will be:

$\hat{\vec{n}} = \frac{1}{\sqrt{10}} \left(- \vec{j} + 3 \vec{k}\right) = \frac{\sqrt{10}}{10} \left(- \vec{j} + 3 \vec{k}\right)$