# What is the cross product of [2, 6, -1] and [3, -4, 2] ?

May 10, 2017

The andwer is =〈8,-7,-26〉

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

Here, we have veca=〈2,6,-1〉 and vecb=〈3,-4,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , 6 , - 1\right) , \left(3 , - 4 , 2\right) |$

$= \vec{i} | \left(6 , - 1\right) , \left(- 4 , 2\right) | - \vec{j} | \left(2 , - 1\right) , \left(3 , 2\right) | + \vec{k} | \left(2 , 6\right) , \left(3 , - 4\right) |$

$= \vec{i} \left(6 \cdot 2 - 1 \cdot 4\right) - \vec{j} \left(2 \cdot 2 + 1 \cdot 3\right) + \vec{k} \left(- 2 \cdot 4 - 6 \cdot 3\right)$

=〈8,-7,-26〉=vecc

Verification by doing 2 dot products

〈8,-7,-26〉.〈2,6,-1〉=8*2-6*7+26*1=0

〈8,-7,-26〉.〈3,-4,2〉=8*3+7*4-26*2=0

So,

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