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

We know that #vecA xx vecB = ||vecA|| * ||vecB|| * sin(theta) hatn#, where #hatn# is a unit vector given by the right hand rule.

So for of the unit vectors #hati#, #hatj# and #hatk# in the direction of #x#, #y# and #z# respectively, we can arrive at the following results.

#color(white)( (color(black){hati xx hati = vec0}, color(black){qquad hati xx hatj = hatk}, color(black){qquad hati xx hatk = -hatj}), (color(black){hatj xx hati = -hatk}, color(black){qquad hatj xx hatj = vec0}, color(black){qquad hatj xx hatk = hati}), (color(black){hatk xx hati = hatj}, color(black){qquad hatk xx hatj = -hati}, color(black){qquad hatk xx hatk = vec0}))#

Another thing that you should know is that cross product is distributive, which means

#vecA xx (vecB + vecC) = vecA xx vecB + vecA xx vecC#.

We are going to need all of these results for this question.

#<5,-3,8> xx <-2,7,3>#

#= (5hati - 3hatj + 8hatk) xx (-2hati + 7hatj + 3hatk)#

#= color(white)( (color(black){qquad 5hati xx (-2hati) + 5hati xx 7hatj + 5hati xx 3hatk}), (color(black){-3hatj xx (-2hati) - 3hatj xx 7hatj - 3hatj xx 3hatk}), (color(black){+8hatk xx (-2hati) + 8hatk xx 7hatj + 8hatk xx 3hatk}) )#

#= color(white)( (color(black){-10(vec0) + 35hatk qquad - 15hatj}), (color(black){-6hatk qquad - 21(vec0) - 9hati}), (color(black){qquad -16hatj qquad - 56hati qquad + 24(vec0)}) )#

#= -65hati - 31hatj + 29hatk#