# What is the cross product of <9,2,8 > and <1,3,-4 >?

Apr 3, 2017

The vector is =〈-32,44,25〉

#### Explanation:

The cross product is a vector perpendiculat to 2 other vectors

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

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

Here, we have veca=〈9,2,8〉 and vecb=〈1,3,-4〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(9 , 2 , 8\right) , \left(1 , 3 , - 4\right) |$

$= \vec{i} | \left(2 , 8\right) , \left(3 , - 4\right) | - \vec{j} | \left(9 , 8\right) , \left(1 , - 4\right) | + \vec{k} | \left(9 , 2\right) , \left(1 , 3\right) |$

$= \vec{i} \left(- 2 \cdot 4 - 3 \cdot 8\right) - \vec{j} \left(- 9 \cdot 4 - 8 \cdot 1\right) + \vec{k} \left(9 \cdot 3 - 2 \cdot 1\right)$

=〈-32,44,25〉=vecc

Verification by doing 2 dot products

〈-32,44,25〉.〈9,2,8〉=-32*9+44*2+25*8=0

〈-32,44,25〉.〈1,3,-4〉=-32*1+44*3-4*25=0

So,

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