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

1 Answer
Apr 11, 2018

Answer:

The vector is #=〈16,-10,-41〉#

Explanation:

The cross product of 2 vectors is calculated with the determinant

#| (veci,vecj,veck), (d,e,f), (g,h,i) | #

where #veca=〈d,e,f〉# and #vecb=〈g,h,i〉# are the 2 vectors

Here, we have #veca=〈-2,5,-2〉# and #vecb=〈7,3,2〉#

Therefore,

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

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

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

#=〈16,-10,-41〉=vecc#

Verification by doing 2 dot products

#〈16,-10,-41〉.〈-2,5,-2〉=(16)*(-2)+(-10)*(5)+(-41)*(-2)=0#

#〈16,-10,-41〉.〈7,3,2〉=(16)*(7)+(-10)*(3)+(-41)*(2)=0#

So,

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