# What is the cross product of [2, 5, 4] and [1, -4, 0] ?

Mar 4, 2017

$\left[16 , 4 , - 13\right] .$

#### Explanation:

$\left[2 , 5 , 4\right] \times \left[1 , - 4 , 0\right] = | \left(i , j , k\right) , \left(2 , 5 , 4\right) , \left(1 , - 4 , 0\right) | ,$

$= 16 i + 4 j - 13 k ,$

$= \left[16 , 4 , - 13\right] .$

Mar 4, 2017

The vector is =〈16,4,-13〉

#### 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=〈2,5,4〉 and vecb=〈1,-4,0〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , 5 , 4\right) , \left(1 , - 4 , 0\right) |$

$= \vec{i} | \left(5 , 4\right) , \left(- 4 , 0\right) | - \vec{j} | \left(2 , 4\right) , \left(1 , 0\right) | + \vec{k} | \left(2 , 5\right) , \left(1 , - 4\right) |$

$= \vec{i} \left(16\right) - \vec{j} \left(- 4\right) + \vec{k} \left(- 13\right)$

=〈16,4,-13〉=vecc

Verification by doing 2 dot products

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

=〈2,5,4>.〈16,4,-13〉=32+20-52=0

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

=〈1,-4,0〉.〈16,4,-13〉=16-16+0=0

So,

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