What is the cross product of #(14i - 7j - 7k)# and #(-5i + 12j + 2 k)#?

Question

Charles Aeschliman · Accepted Answer

#70hati + 7hatj + 133hatk#

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.

#(14hati - 7hatj - 7hatk) xx (-5hati + 12hatj + 2hatk)#

#= color(white)( (color(black){qquad 14hati xx (-5hati) + 14hati xx 12hatj + 14hati xx 2hatk}), (color(black){-7hatj xx (-5hati) - 7hatj xx 12hatj - 7hatj xx 2hatk}), (color(black){-7hatk xx (-5hati) - 7hatk xx 12hatj - 7hatk xx 2hatk}) )#

#= color(white)( (color(black){-70(vec0) + 168hatk qquad - 28hatj}), (color(black){-35hatk qquad - 84(vec0) - 14hati}), (color(black){qquad +35hatj qquad + 84hati qquad - 14(vec0)}) )#

#= 70hati + 7hatj + 133hatk#

Benjamin Asiedu · Answer

undefined The cross product of vectors $ (14i - 7j - 7k) $ and $ (-5i + 12j + 2k) $ is:

$$ (14 \times -5)i + (-7 \times 2)j + (-7 \times 12)k $$
$$ = -70i - 14j - 84k $$