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

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Answer 1 · 2017-01-13T07:54:21

The answer is $=〈20,6,11〉$

The cross product 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 $veca=〈3,1,-6〉$ and $vecb=〈-2,3,2〉$

Therefore,

$| (veci,vecj,veck), (3,1,-6), (-2,3,2) |$

$=veci| (1,-6), (3,2) | -vecj| (3,-6), (-2,2) | +veck| (3,1), (-2,3) |$

$=veci(20)-vecj(-6)+veck(11)$

$=〈20,6,11〉=vecc$

Verification by doing 2 dot products

$〈20,6,11〉.〈3,1,-6〉=60+6-66=0$

$〈20,6,11〉.〈-2,3,2〉=-40+18+22=0$

So,

$vecc$ is perpendicular to $veca$ and $vecb$

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