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

1 Answer
Oct 1, 2017

Answer:

The vector is #=〈48,-6,-36〉#

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,-6,-3〉# and #vecb=〈-5,2,-7〉#

Therefore,

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

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

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

#=〈48,-6,-36〉=vecc#

Verification by doing 2 dot products (inner products)

#〈48,-6,-36〉.〈-3,-6,-3〉=-48*3+6*6+36*3=0#

#〈48,-6,-36〉.〈-5,2,-7〉=-48*5-6*2+36*7=0#

So,

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