RadfordMathematics.com

Properties of the Cross Product

(Properties of the Vector Product of Two Vectors)

In this section we learn about the properties of the cross product. In particular, we learn about each of the following:

  • anti-commutatibity of the cross product
  • distributivity
  • multiplication by a scalar
  • collinear vectors
  • magnitude of the cross product

Anti-Commutativity of the Cross Product

Given two vectors \(\vec{u}\) and \(\vec{v}\) \[\vec{u}\times \vec{v} = - \vec{v} \times \vec{u}\]

Proof

Given

Distributivity

Given three vectors \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\): \[\vec{u} \times \begin{pmatrix} \vec{v} + \vec{w}\end{pmatrix} = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} \]

Multiplication by a Scalar

Given two vectors \(\vec{u}\) and \(\vec{v}\) and a scalar \(k\in \mathbb{R}\): \[k\begin{pmatrix}\vec{u}\times \vec{v}\end{pmatrix} = \begin{pmatrix} k\vec{u}\end{pmatrix} \times \vec{v} = \vec{u} \times \begin{pmatrix} k\vec{v}\end{pmatrix}\]

Why is this useful?

The fact that \(k\begin{pmatrix}\vec{u}\times \vec{v}\end{pmatrix} = \begin{pmatrix} k\vec{u}\end{pmatrix} \times \vec{v} = \vec{u} \times \begin{pmatrix} k\vec{v}\end{pmatrix}\) can often be used to make calculation easier.

Example

For example, say we're given \(\vec{u} = 20\vec{i} + 60 \vec{j} + 40 \vec{k}\) and \(\vec{v} = \vec{i} + 5\vec{j} - 4\vec{k}\) and that we have to find \(\vec{u} \times \vec{v}\). Then we can use this property to make our calculations a little simpler (and therefore faster) by noticing that \(\vec{u} = 20 \begin{pmatrix} \vec{i} + 3 \vec{j} + 2 \vec{k} \end{pmatrix}\) and using this property to write: \[\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 20 & 60 & 40 \\ 1 & 5 & -4 \end{vmatrix} = 20 \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & 2 \\ 1 & 5 & -4 \end{vmatrix}\]

Example

Another example could be, to calculate \(\vec{u}\times \vec{v}\), where \(\vec{u} = 5 \vec{i} - 20 \vec{j} + 10 \vec{k}\) and \(\vec{v} = 9 \vec{i} +6 \vec{j} -3 \vec{k}\). Noticing that \(\vec{u} = 5\begin{pmatrix}1\vec{i} - 4\vec{j} + 2 \vec{k} \end{pmatrix}\) and \(\vec{v} = 3 \begin{pmatrix} 3\vec{i} + 2\vec{j} - \vec{k}\end{pmatrix}\) we can write: \[\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 5 & -20 & 10 \\ 9 & 6 & -3 \end{vmatrix} = 5\times 3\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -4 & 2 \\ 3 & 2 & -1\end{vmatrix} = 15\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -4 & 2 \\ 3 & 2 & -1\end{vmatrix}\]

Collinear Vectors (Parallel Vectors)

Given two vectors \(\vec{u}\) and \(\vec{v}\), then the cross product \(\vec{u}\times \vec{v}\) is such that: \[\vec{u} \times \vec{v} = \vec{0} \iff \vec{u}\text{ and } \vec{v} \text{ are collinear (parallel)} \] if and only of the vectors \(\vec{u}\) and \(\vec{v}\) are collinear (\(\vec{u}\) and \(\vec{v}\) are parallel).

Test for Collinearity

This property provides us with a useful test for collinearity. Indeed, to check if two vectors, \(\vec{u}\) and \(\vec{v}\), are collinear all we have to do is calculate the cross product \(\vec{u}\times \vec{v}\) then if:

  • \(\vec{u}\times \vec{v} = \vec{0}\) the two vectors are collinear
  • \(\vec{u}\times \vec{v} \neq \vec{0}\) the two vectors aren't collinear.
For instance, we can show that the vectors \(\vec{u} = \vec{i} - 3 \vec{j} + 5 \vec{k}\) and \(\vec{v} = -3 \vec{i} + 9 \vec{j} - 15 \vec{k}\) are collinear by calculating \(\vec{u}\times \vec{v}\): \[\begin{aligned} \vec{u}\times \vec{v} & = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -3 & 5 \\ -3 & 9 & -15 \end{vmatrix} \\ & = \vec{i} \begin{vmatrix} -3 & 5 \\ 9 & -15 \end{vmatrix} - \vec{j} \begin{vmatrix} 1 & 5 \\ -3 & -15 \end{vmatrix} + \vec{k} \begin{vmatrix} 1 & -3 \\ -3 & 9 \end{vmatrix} \\ & = \vec{i} \begin{pmatrix} - 3 \times (-15) - 9 \times 5 \end{pmatrix} - \vec{j} \begin{pmatrix}1 \times (-15) - (-3)\times 5 \end{pmatrix} + \vec{k} \begin{pmatrix} 1 \times 9 - (-3)\times (-3)\end{pmatrix} \\ & = \vec{i} \begin{pmatrix} 45 - 45 \end{pmatrix} - \vec{j} \begin{pmatrix} -15 + 15 \end{pmatrix} + \vec{k} \begin{pmatrix} 9 - 9 \end{pmatrix} \\ & = 0\vec{i} + 0\vec{j} + 0\vec{k} \\ \vec{u}\times \vec{v} & = \vec{0} \end{aligned}\] Since \(\vec{u}\times \vec{v} = \vec{0}\), we can state that the vectors \(\vec{u}\) and \(\vec{v}\) are collinear.

Exercise 1

  1. For two vectors \(\vec{a}\) and \(\vec{b}\), simplify: \[\begin{pmatrix} \vec{a}-\vec{b}\end{pmatrix} \times \begin{pmatrix} \vec{a} + \vec{b}\end{pmatrix}\]
  2. For two vectors \(\vec{a}\) and \(\vec{b}\), simplify: \[\begin{pmatrix} \vec{a}-\vec{b}\end{pmatrix} \times \begin{pmatrix} \vec{a} - \vec{b}\end{pmatrix}\]
  3. For two vectors \(\vec{a}\) and \(\vec{b}\), simplify: \[\begin{pmatrix} \vec{a} + \vec{b}\end{pmatrix} \times \begin{pmatrix} \vec{a} + \vec{b}\end{pmatrix}\]
  4. Find a vector normal to the plane containing the points \(A\begin{pmatrix}2,\ -1,\ 3\end{pmatrix}\), \(B\begin{pmatrix}5,\ 0,\ 2\end{pmatrix}\) and \(A\begin{pmatrix}-6,\ 3,\ 7\end{pmatrix}\).

Note: this exercise can be downloaded as a worksheet to practice with: Worksheet 1

Scan this QR-Code with your phone/tablet and view this page on your preferred device.

Subscribe to Our Channel

Subscribe Now and view all of our playlists & tutorials.