# What is the cross product of <-2, 5 ,-2 > and <7 ,3 ,2 >?

Apr 11, 2018

The vector is =〈16,-10,-41〉

#### Explanation:

The cross product of 2 vectors is calculated with the determinant

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

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

Here, we have veca=〈-2,5,-2〉 and vecb=〈7,3,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(- 2 , 5 , - 2\right) , \left(7 , 3 , 2\right) |$

$= \vec{i} | \left(5 , - 2\right) , \left(3 , 2\right) | - \vec{j} | \left(- 2 , - 2\right) , \left(7 , 2\right) | + \vec{k} | \left(- 2 , 5\right) , \left(7 , 3\right) |$

$= \vec{i} \left(\left(5\right) \cdot \left(2\right) - \left(- 2\right) \cdot \left(3\right)\right) - \vec{j} \left(\left(- 2\right) \cdot \left(2\right) - \left(- 2\right) \cdot \left(7\right)\right) + \vec{k} \left(\left(- 2\right) \cdot \left(3\right) - \left(5\right) \cdot \left(7\right)\right)$

=〈16,-10,-41〉=vecc

Verification by doing 2 dot products

〈16,-10,-41〉.〈-2,5,-2〉=(16)*(-2)+(-10)*(5)+(-41)*(-2)=0

〈16,-10,-41〉.〈7,3,2〉=(16)*(7)+(-10)*(3)+(-41)*(2)=0

So,

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