# What is the unit vector that is normal to the plane containing <2i+7j-2k> and <8i-2j+3k>?

May 13, 2017

The unit vector is =1/sqrt(4373)*〈17,-22,-60〉

#### 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=〈2,7,-2〉 and vecb=〈8,-2,3〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , 7 , - 2\right) , \left(8 , - 2 , 3\right) |$

$= \vec{i} | \left(7 , - 2\right) , \left(- 2 , 3\right) | - \vec{j} | \left(2 , - 2\right) , \left(8 , 3\right) | + \vec{k} | \left(2 , 7\right) , \left(8 , - 2\right) |$

$= \vec{i} \left(7 \cdot 3 - 2 \cdot 2\right) - \vec{j} \left(2 \cdot 3 + 8 \cdot 2\right) + \vec{k} \left(2 \cdot - 2 - 7 \cdot 8\right)$

=〈17,-22,-60〉=vecc

Verification by doing 2 dot products

〈17,-22,-60〉.〈2,7,-2〉=17*2-22*7-60*-2=0

〈17,-22,-60〉.〈8,-2,3〉=17*8-22*-2-60*3=0

So,

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

The unit vector is

hatc=vecc/||c||=1/sqrt(4373)*〈17,-22,-60〉