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

The cross product between two vectors #vecA# and #vecB# is defined to be

#vecA xx vecB = ||vecA|| * ||vecB|| * sin(theta) * hatn#,

where #hatn# is a unit vector given by the right hand rule, and #theta# is the angle between #vecA# and #vecB# and must satisfy #0<=theta<=pi#.

For of the unit vectors #hati#, #hatj# and #hatk# in the direction of #x#, #y# and #z# respectively, using the above definition of cross product gives the following set of 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}))#

Also, note that cross product is distributive.

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

So for this question.

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

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

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

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

#= -6hati - 3hatk#

#= [-6,0,-3]#