# Cross Products

The cross product is an operation that finds a third vector, $\vec{w}$, that is orthogonal to two vectors, $\vec{v}$ and $\vec{u}$. The problem of finding a vector orthogonal to another two comes up frequently in math and physics, which is the motivation for the cross product.

Suppose we have two vectors $\vec{v}$ and $\vec{u}$ that are in $\mathbb{R}^3$. We can conclude that $\vec{w}$ is orthogonal to both vectors if:

$\vec{u} * \vec{w} = u_1w_1+u_2w_2+u_3w_3 = 0$

$\vec{v} * \vec{w} = v_1w_1 + v_2w_2 + v_3w_3 = 0$

If we were to solve this linear system, we would end up with the following vector for $\vec{w}$.

$\vec{w} = \left[\begin{array}{cc} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{array}\right]$

This tells us that in $\mathbb{R}^3$, the cross product operation can be defined as:

$\vec{u} \times \vec{v} = \left[\begin{array}{cc} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{array}\right]$

**Example:
**Determine $\left[\begin{array}{cc} 2 \\ 3 \\ 5 \end{array}\right] \times
\left[\begin{array}{cc} -1 \\ 1 \\ 2 \end{array}\right]$

$\left[\begin{array}{cc} 2 \\ 3 \\ 5 \end{array}\right] \times \left[\begin{array}{cc} -1 \\ 1 \\ 2 \end{array}\right] = \left[\begin{array}{cc} (3)(2) - (5)(1) \\ (5)(-1) - (2)(2) \\ (2)(1)-(3)(-1) \end{array}\right] = \left[\begin{array}{cc} 1 \\ -9 \\ 5 \end{array}\right]$

For now, we will only worry about cross products in $\mathbb{R}^3$. It is possible to do a cross product in $\mathbb{R}^n$, however it is much more complex and out of the scope of introduction level Linear Algebra.

There are a set of properties that apply to the cross product operation. Let $\vec{v}, \vec{u}, \vec{w} \in \mathbb{R}^3$, and $t \in \mathbb{R}$. The following properties hold:

- $\vec{v} \times \vec{u} = -\vec{u} \times \vec{v}$
- $\vec{v} \times \vec{v} = \vec{0}$
- $\vec{v} \times (\vec{u} \times \vec{w}) = \vec{v} \times \vec{u} + \vec{v} \times \vec{w}$
- $(t\vec{v}) \times \vec{u} = t (\vec{v} \times \vec{u})$

One example of a cross product application is using it to find the normal of a plane. A plane has an equation in the form $\vec{x} = \vec{p} + s\vec{u} + t\vec{v}$, where $\vec{v}$ and $\vec{u}$ are linearly independent. A normal vector is vector $\vec{n}$ that is perpendicular to both $\vec{u}$ and $\vec{v}$. Therefore, we can conclude that $\vec{n} = \vec{u} \times \vec{v}$