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} = \begin{bmatrix} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{bmatrix}

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

\vec{u} \times \vec{v} = \begin{bmatrix} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{bmatrix}

Example: Determine \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} \times \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix}

\begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} \times \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} (3)(2) - (5)(1) \\ (5)(-1) - (2)(2) \\ (2)(1)-(3)(-1) \end{bmatrix} = \begin{bmatrix} 1 \\ -9 \\ 5 \end{bmatrix}

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:

  1. \vec{v} \times \vec{u} = -\vec{u} \times \vec{v}
  2. \vec{v} \times \vec{v} = \vec{0}
  3. \vec{v} \times (\vec{u} \times \vec{w}) = \vec{v} \times \vec{u} + \vec{v} \times \vec{w}
  4. (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}               

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