# What is the cross product of <3 ,1 ,-6 > and <-2 ,3 ,2 >?

Jan 13, 2017

The answer is =〈20,6,11〉

#### Explanation:

The cross product is calculated with the determinant

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

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

Here, we have veca=〈3,1,-6〉 and vecb=〈-2,3,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(3 , 1 , - 6\right) , \left(- 2 , 3 , 2\right) |$

$= \vec{i} | \left(1 , - 6\right) , \left(3 , 2\right) | - \vec{j} | \left(3 , - 6\right) , \left(- 2 , 2\right) | + \vec{k} | \left(3 , 1\right) , \left(- 2 , 3\right) |$

$= \vec{i} \left(20\right) - \vec{j} \left(- 6\right) + \vec{k} \left(11\right)$

=〈20,6,11〉=vecc

Verification by doing 2 dot products

〈20,6,11〉.〈3,1,-6〉=60+6-66=0

〈20,6,11〉.〈-2,3,2〉=-40+18+22=0

So,

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