What is the cross product of #[2, 6, -1]# and #[3, -4, 2] #?

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Answer 1 · 2017-05-10T05:44:38

The andwer is #=〈8,-7,-26〉#

The cross product of 2 vectors 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=〈2,6,-1〉# and #vecb=〈3,-4,2〉#

Therefore,

#| (veci,vecj,veck), (2,6,-1), (3,-4,2) | #

#=veci| (6,-1), (-4,2) | -vecj| (2,-1), (3,2) | +veck| (2,6), (3,-4) | #

#=veci(6*2-1*4)-vecj(2*2+1*3)+veck(-2*4-6*3)#

#=〈8,-7,-26〉=vecc#

Verification by doing 2 dot products

#〈8,-7,-26〉.〈2,6,-1〉=8*2-6*7+26*1=0#

#〈8,-7,-26〉.〈3,-4,2〉=8*3+7*4-26*2=0#

So,

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