# What is the cross product of <4 , 5 ,0 > and <4, 1 ,-2 >?

May 24, 2017

$< - 10 , - 8 , - 16 >$

#### Explanation:

We'll call the vector $< 4 , 5 , 0 > \vec{A}$, and the vector $< 4 , 1 , - 2 > \vec{B}$

The cross product of $\vec{A}$ and $\vec{B}$ is a vector $\vec{C}$ with components

${C}_{x} = {A}_{y} {B}_{z} - {A}_{z} {B}_{y}$

${C}_{y} = {A}_{x} {B}_{z} - {A}_{z} {B}_{x}$

${C}_{z} = {A}_{x} {B}_{y} - {A}_{y} {B}_{x}$

We have our components for vectors $\vec{A}$ and $\vec{B}$ expressed in their position vectors, and we'll use these values to calculate the components of $\vec{C}$:

${C}_{x} = \left(5\right) \left(- 2\right) - \left(0\right) \left(1\right) = - 10$

${C}_{y} = \left(4\right) \left(- 2\right) - \left(0\right) \left(4\right) = - 8$

${C}_{z} = \left(4\right) \left(1\right) - \left(5\right) \left(4\right) = - 16$

Using unit vectors, this vector product $\vec{C}$ is

$\left(- 10\right) \vec{i} + \left(- 8\right) \vec{j} + \left(- 16\right) \vec{k}$

Or, expressed as a position vector,

$< - 10 , - 8 , - 16 >$