What is the cross product of <6,-2,8 > and <1,3,-4 >?

Jan 16, 2018

The vector is =〈-16,32,20〉

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=〈6,-2,8〉 and vecb=〈1,3,-4〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(6 , - 2 , 8\right) , \left(1 , 3 , - 4\right) |$

$= \vec{i} | \left(- 2 , 8\right) , \left(3 , - 4\right) | - \vec{j} | \left(6 , 8\right) , \left(1 , - 4\right) | + \vec{k} | \left(6 , - 2\right) , \left(1 , 3\right) |$

$= \vec{i} \left(\left(- 2\right) \cdot \left(- 4\right) - \left(3\right) \cdot \left(8\right)\right) - \vec{j} \left(\left(6\right) \cdot \left(- 4\right) - \left(8\right) \cdot \left(1\right)\right) + \vec{k} \left(\left(6\right) \cdot \left(3\right) - \left(- 2\right) \cdot \left(1\right)\right)$

=〈-16,32,20〉=vecc

Verification by doing 2 dot products

〈-16,32,20〉.〈6,-2,8〉=(-16*6)+(32*-2)+(20*8)=0

〈-16,32,20〉.〈1,3,-4〉=(-16*1)+(32*3)+(20*-4)=0

So,

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