What is the cross product of #<-3,5,8 ># and #<6, -2, 7 >#?

2 Answers
Apr 22, 2017

The vector is #=〈51,69,-24〉=#

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=〈-3,5,8〉# and #vecb=〈6,-2,7〉#

Therefore,

#| (veci,vecj,veck), (-3,5,8), (6,-2,7) | #

#=veci| (5,8), (-2,7) | -vecj| (-3,8), (6,7) | +veck| (-3,5), (6,-2) | #

#=veci(5*7+2*8)-vecj(-3*7-6*8)+veck(3*2-5*6)#

#=〈51,69,-24〉=vecc#

Verification by doing 2 dot products

#〈51,69,-24〉.〈-3,5,8〉=-51*3+69*5-24*8=0#

#〈51,69,-24〉.〈6,-2,7〉=51*6-69*2-24*7=0#

So,

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

Apr 22, 2017

#=(51,69,-24).#

Explanation:

Let, #vecx=(-3,5,8)=-3i+5j+8k,# and,

#vecy=(6,-2,7)=6i-2j+7k.#

#:. vecx xx vecy=|(i,j,k),(-3,5,8),(6,-2,7)|#

#=(35-(-16))i-(-21-48)j+(6-30)k#

#=51i+69j-24k#

#=(51,69,-24).#