What is the cross product of #[-3, 1, -1]# and #[0,1,2] #?

1 Answer
Jul 11, 2017

The vector is #=〈3,6,-3〉#

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,-1〉# and #vecb=〈0,1,2〉#

Therefore,

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

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

#=veci(1*2+1*1)-vecj(-3*2+0*1)+veck(-3*1-0*1)#

#=〈3,6,-3〉=vecc#

Verification by doing 2 dot products

#〈3,6,-3〉.〈-3,1,-1〉=-3*3+6*1+3*1=0#

#〈3,6,-3〉.〈0,1,2〉=3*0+6*1-3*2=0#

So,

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