# What is the cross product of [1, 3, 4] and [2, -5, 8] ?

Mar 8, 2017

The vector is =〈44,0,-11〉

#### 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=〈1,3,4〉 and vecb=〈2,-5,8〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(1 , 3 , 4\right) , \left(2 , - 5 , 8\right) |$

$= \vec{i} | \left(3 , 4\right) , \left(- 5 , 8\right) | - \vec{j} | \left(1 , 4\right) , \left(2 , 8\right) | + \vec{k} | \left(1 , 3\right) , \left(2 , - 5\right) |$

$= \vec{i} \left(44\right) - \vec{j} \left(0\right) + \vec{k} \left(- 11\right)$

=〈44,0,-11〉=vecc

Verification by doing 2 dot products

$\vec{a} . \vec{c}$

=〈1,3,4>.〈44,0,-11〉=44-44=0

$\vec{b} . \vec{c}$

=〈2,-5,8〉.〈44,0,-11〉=88-88=0

So,

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