What is the cross product of #[3, -4, 2]# and #[1, 1, 18] #?

1 Answer
Mar 12, 2016

Answer:

#[-74,-52,7]#

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,-4,2] xx [1,1,18] #

#= (3hati - 4hatj + 2hatk) xx (hati + hatj + 18hatk)#

#= color(white)( (color(black){qquad 3hati xx hati + 3hati xx hatj + 3hati xx 18hatk}), (color(black){-4hatj xx hati - 4hatj xx hatj - 4hatj xx 18hatk}), (color(black){+2hatk xx hati + 2hatk xx hatj + 2hatk xx 18hatk}) )#

#= color(white)( (color(black){quad 3(vec0) + 3hatk qquad - 54hatj}), (color(black){+4hatk qquad - 4(vec0) - 72hati}), (color(black){qquad +2hatj qquad - 2hati qquad + 36(vec0)}) )#

#= -74hati - 52hatj + 7hatk#

#= [-74,-52,7]#