Question

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

Answer 1

The vector is `=〈-16,32,20〉`

Explanation:

The cross product of 2 vectors is calculated with the determinant

`| (veci,vecj,veck), (d,e,f), (g,h,i) |`

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,

`| (veci,vecj,veck), (6,-2,8), (1,3,-4) |`

`=veci| (-2,8), (3,-4) | -vecj| (6,8), (1,-4) | +veck| (6,-2), (1,3) |`

`=veci((-2)*(-4)-(3)*(8))-vecj((6)*(-4)-(8)*(1))+veck((6)*(3)-(-2)*(1))`

`=〈-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,

`vecc` is perpendicular to `veca` and `vecb`