What is the cross product of #[3, 2, 5]# and #[4,3,6] #?

1 Answer
Mar 8, 2017

Answer:

The vector is #=〈-3,2,1〉#

Explanation:

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=〈3,2,5〉# and #vecb=〈4,3,6〉#

Therefore,

#| (veci,vecj,veck), (3,2,5), (4,3,6) | #

#=veci| (2,5), (3,6) | -vecj| (3,5), (4,6) | +veck| (3,2), (4,3) | #

#=veci(-3)-vecj(-2)+veck(1)#

#=〈-3,2,1〉=vecc#

Verification by doing 2 dot products

#veca.vecc#

#=〈3,2,5>.〈-3,2,1〉=-9+4+5=0#

#vecb.vecc#

#=〈4,3,6〉.〈-3,2,1〉=-12+6+6=0#

So,

#vecc# is perpendicular to #veca# and #vecb#