# What are cross products?

Mar 15, 2018

See explanation...

#### Explanation:

When you encounter vectors in $3$ dimensions then you meet two ways of multiplying two vectors together:

Dot product

Written $\vec{u} \cdot \vec{v}$, this takes two vectors and produces a scalar result.

If $\vec{u} = < {u}_{1} , {u}_{2} , {u}_{3} >$ and $\vec{v} = < {v}_{1} , {v}_{2} , {v}_{3} >$ then:

$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2} + {u}_{3} {v}_{3}$

Cross product

Written $\vec{u} \times \vec{v}$, this takes two vectors and produces a vector perpendicular to both of them, or the zero vector if $\vec{u}$ and $\vec{v}$ are parallel.

If $\vec{u} = < {u}_{1} , {u}_{2} , {u}_{3} >$ and $\vec{v} = < {v}_{1} , {v}_{2} , {v}_{3} >$ then:

$\vec{u} \times \vec{v} = < {u}_{2} {v}_{3} - {u}_{3} {v}_{2} , \textcolor{w h i t e}{.} {u}_{3} {v}_{1} - {u}_{1} {v}_{3} , \textcolor{w h i t e}{.} {u}_{1} {v}_{2} - {u}_{2} {v}_{1} >$

This is sometimes described in terms of a determinant of a $3 \times 3$ matrix and the three unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$:

$\vec{u} \times \vec{v} = \left\mid \begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ {u}_{1} & {u}_{2} & {u}_{3} \\ {v}_{1} & {v}_{2} & {v}_{3}\end{matrix} \right\mid$

Neither dot product nor cross product allow division of vectors. To find how to divide vectors you can look at the quaternions. The quaternions form a $4$ dimensional vector space over the real numbers and have arithmetic with non-commutative multiplication that can be expressed as a combination of dot product and cross product. Actually that's the wrong way around, since quaternion arithmetic predates the modern presentation of vectors, dot and cross products.
$\left({r}_{1} , \vec{{v}_{1}}\right) + \left({r}_{2} , \vec{{v}_{2}}\right) = \left({r}_{1} + {r}_{2} , \vec{{v}_{1}} + \vec{{v}_{2}}\right)$
$\left({r}_{1} , \vec{{v}_{1}}\right) \cdot \left({r}_{2} , \vec{{v}_{2}}\right) = \left({r}_{1} {r}_{2} - \vec{{v}_{1}} \cdot \vec{{v}_{2}} , {r}_{1} \vec{{v}_{2}} + {r}_{2} \vec{{v}_{1}} + \vec{{v}_{1}} \times \vec{{v}_{2}}\right)$