What is the cross product of #<0,8,5># and #<-1,-1,2>#?

1 Answer
Mar 16, 2016

Answer:

#<21,-5,8>#

Explanation:

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.

#<0,8,5> xx <-1,-1,2>#

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

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

#= color(white)( (color(black){8hatk - 8(vec0) + 16hati}), (color(black){-5hatj + 5hati qquad + 10(vec0)}) )#

#= 21hati - 5hatj + 8hatk#

#= <21,-5,8>#