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

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

#### Explanation:

We have two vectors $A = \left({A}_{x} , {A}_{y} , {A}_{z}\right)$ and $B = \left({B}_{x} , {B}_{y} , {B}_{z}\right)$.

The cross product can be calculated nicely using matrices.
It is $\det \left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ {A}_{x} & {A}_{y} & {A}_{z} \\ {B}_{x} & {B}_{y} & {B}_{z}\end{matrix}\right) =$

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 $A \approx B$.

Plugging values in:
$A = \left(- 3 , 0 , 1\right)$ and $B = \left(1 , 2 , - 4\right)$
$A \times B$
=(0*(-4)-1*2, 1*1-(-3)(-4), (-3)*2-0*1) =(0-2, 1-12, -6-0)=(-2,11,-6)