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

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Answer 1 · 2018-01-16T06:48:04

The vector is $=〈-16,32,20〉$

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 $vecb$

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