# What is the cross product of #<< -1, -1, 2 >># and #<< 4,3,6 >> #?

##### 1 Answer

Well, you have at least two ways to do it.

The first way:

Let

#color(blue)(vecu xx vecv) = << u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 >>#

#= << -1*6 - 2*3, 2*4 - (-1*6), -1*3 - (-1*4) >>#

#= color(blue)(<< -12, 14, 1 >>)#

Assuming you didn't know that formula, the second way (which is a little more foolproof) is recognizing that:

#hati xx hatj = hatk#

#hatj xx hatk = hati#

#hatk xx hati = hatj#

#hatA xx hatA = vec0#

#hatA xx hatB = -hatB xx hatA# where

#hati = << 1,0,0 >># ,#hatj = << 0,1,0 >># , and#hatk = << 0,0,1 >># .

Thus, rewriting the vectors in unit vector form:

#(-hati - hatj + 2hatk)xx(4hati + 3hatj + 6hatk)#

#= cancel(-4(hati xx hati))^(0) - 3(hati xx hatj) - 6(hati xx hatk) - 4(hatj xx hati) - cancel(3(hatj xx hatj))^(0) - 6(hatj xx hatk) + 8(hatk xx hati) + 6(hatk xx hatj) + cancel(12(hatk xx hatk))^(0)#

#= -3hatk + 6hatj + 4hatk - 6hati + 8hatj - 6hati#

#= - 12hati + 14hatj + hatk#

#= color(blue)(<< -12, 14, 1 >>)#

as expected.