What is the cross product of #<-1, 2 ,7 ># and #<-3 ,1 ,-1 >#?

1 Answer
Jul 12, 2016

Answer:

#(-1,2,7)xx(-3,1,-1) = (-9,-22,5)#

Explanation:

Cross products can be evaluated using the aptly named determinant rule. If you are trying to determine #vecA xx vecB# with components:
#vecA = (a_1,a_2,a_3)#
#vecB = (b_1,b_2,b_3)#

Then you can construct a matrix of the form:
#M = ((hati,hatj,hatk),(a_1,a_2,a_3),(b_1,b_2,b_3))#

Where #hati#, #hatj#, and #hatk# are your orthonormal basis vectors. (This will work in higher dimensions, but quickly becomes cumbersome). Then you can write the cross product as:
#vecA xx vecB = det(M) = |(hati,hatj,hatk),(a_1,a_2,a_3),(b_1,b_2,b_3)|#
#= hati |(a_2,a_3),(b_2,b_3)| - hatj |(a_1,a_3),(b_1,b_3)| + hatk |(a_1,a_2),(b_1,b_2)|#
#= (a_2b_3 - a_3b_2)hati - (a_1b_3 - a_3b_1)hatj + (a_1b_2 - a_2b_1)hatk#

You can see that the result is going to be a vector. Plugging in the initial conditions you gave us, we end up with:
#(-1,2,7)xx(-3,1,-1) = |(hati,hatj,hatk),(-1,2,7),(-3,1,-1)|#
#= hati |(2,7),(1,-1)| - hatj |(-1,7),(-3,-1)| + hatk |(-1,2),(-3,1)|#
#= (-2 - 7)hati - (1 + 21) hatj + (-1 +6) hatk#
#= -9hati -22hatj + 5hatk#
#= (-9,-22,5)#