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What is the cross product of #<6,-2,8 ># and #<1,3,-4 >#?

1 Answer
Jan 16, 2018

Answer:

The vector is #=〈-16,32,20〉#

Explanation:

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=〈6,-2,8〉# and #vecb=〈1,3,-4〉#

Therefore,

#| (veci,vecj,veck), (6,-2,8), (1,3,-4) | #

#=veci| (-2,8), (3,-4) | -vecj| (6,8), (1,-4) | +veck| (6,-2), (1,3) | #

#=veci((-2)*(-4)-(3)*(8))-vecj((6)*(-4)-(8)*(1))+veck((6)*(3)-(-2)*(1))#

#=〈-16,32,20〉=vecc#

Verification by doing 2 dot products

#〈-16,32,20〉.〈6,-2,8〉=(-16*6)+(32*-2)+(20*8)=0#

#〈-16,32,20〉.〈1,3,-4〉=(-16*1)+(32*3)+(20*-4)=0#

So,

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