Question

What is the cross product of `<-3,0,1>` and `<1,2,-4>`?

Answer 1

The cross product of (-3,0,1) and (1,2,-4) is (-2,11,-6).

Explanation:

We have two vectors `A=(A_x,A_y,A_z)` and `B=(B_x,B_y,B_z)`.

The cross product can be calculated nicely using matrices.
It is `det((hat(x),hat(y),hat(z)),(A_x,A_y,A_z),(B_x,B_y,B_z))=`

In the first row we have unit vectors (length 1, along each of the axes) next are coordinates of first vector, and then second vector.

#=hat(x)A_yB_z+hat(y)A_zB_x+hat(z)A_xB_y -hat(x)A_zB_y-hat(y)A_xB_z-hat(z)A_yB_x =hat(x)(A_yB_z-A_zB_y)+hat(y)(A_zB_x-A_xB_z) +hat(z)(A_xB_y-A_yB_x) =(A_yB_z-A_zB_y, A_zB_x-A_xB_z, A_xB_y-A_yB_x)#

This expression is resulting vector. The formula could be memorized by the people, who don't know matrices yet. Notice how Neither `A_x` nor `B_x` has any impact on x coordinate of the result. The same for y and z. That's the sign of perpendicularity of the result to both vectors. Notice how result disappears, when `A~~B`.

Plugging values in:
`A=(-3,0,1)` and `B=(1,2,-4)`
`A times B`
#=(0*(-4)-1*2, 1*1-(-3)(-4), (-3)*2-0*1) =(0-2, 1-12, -6-0)=(-2,11,-6)#