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

1 Answer
Jan 13, 2017

Answer:

The answer is #=〈20,6,11〉#

Explanation:

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#