What is the cross product of $<-1, 2 ,0 >$ and $<-3 ,1 ,9 >$ ?

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Answer 1 · 2017-02-20T16:29:27

The answer is $=〈18,9,5〉$

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

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

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

Here, we have $veca=〈-1,2,0〉$ and $vecb=〈-3,1,9〉$

Therefore,

$| (veci,vecj,veck), (-1,2,0), (-3,1,9) |$

$=veci| (2,0), (1,9) | -vecj| (-1,0), (-3,9) | +veck| (-1,2), (-3,1) |$

$=veci(18)-vecj(-9)+veck(-1+6)$

$=〈18,9,5〉=vecc$

Verification by doing 2 dot products

$〈-1,2,0>.〈18,9,5〉=-18+18=0$

$〈-3,1,9〉.〈18,9,5〉=-54+9+45=0$

So,

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

What is the cross product of <-1, 2 ,0 ><-1, 2 ,0 > and <-3 ,1 ,9 ><-3 ,1 ,9 >?