What is the cross product of #[-1,-1,2]# and #[1,-2,3] #?

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.

#[-1,-1,2] xx [1,-2,3] #

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

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

#= color(white)( (color(black){- 1(vec0) + 2hatk qquad + 3hatj}), (color(black){+hatk qquad + 2(vec0) - 3hati}), (color(black){qquad +2hatj qquad + 4hati qquad + 6(vec0)}) )#

#= hati + 5hatj + 3hatk#

#= [1,5,3]#