What is the cross product of #[3,-1,2]# and #[5,1,-3] #?

1 Answer
Mar 16, 2016

Answer:

#[1,19,8]#

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.

#[3,-1,2] xx [5,1,-3]#

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

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

#= color(white)( (color(black){15(vec0) + 3hatk + 9hatj}), (color(black){+5hatk qquad - vec0 quad + 3hati}), (color(black){quad +10hatj quad - 2hati - 6(vec0)}) )#

#= hati + 19hatj + 8hatk#

#= [1,19,8]#