Feeds:
Posts

The vector (or cross) product

In this previous blog, I mentioned that there are two ways in which to multiply vectors, either the dot (scalar) product of the cross (vector) product. The dot product is pretty easy to do. If we have two vectors $\vec{a} \text{ and } \vec{b}$, the dot product is just given by

$\boxed{ \vec{a} \cdot \vec{b} = \lvert a \rvert \; \lvert b \rvert \; \cos(\theta) }$ where $\theta$ is the angle between them.

The vector (cross) product is a little more complicated.

Let us suppose our vectors are $\vec{a} \text{ and } \vec{b}$ and that $\vec{a}$ can bet written as $\vec{a} = a_{x} \hat{x} + a_{y} \hat{y} + a_{z} \hat{z}$ and $\vec{b} = b_{x} \hat{x} + b_{y} \hat{y} + b_{z} \hat{z}$, to find the vector product it is easiest to use matrices. This is the way I have always done it, and the way I teach it to my students, but if anyone reading this has a different method they wish to share that would be great.

So, we write $\vec{a}$ again as $\left( a_{x} \hat{x} + a_{y} \hat{y} + a_{z} \hat{z} \right)$ and $\vec{b}$ as $\left( b_{x} \hat{x} + b_{y} \hat{y} + b_{z} \hat{z} \right)$ where the brackets indicate that these are now matrices.

Using the determinant matrix to calculate the vector product

$\vec{c} = \vec{a} \times \vec{b} = \left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ \end{array} \right|$ where the expression on the right is the determinant matrix.

To calculate each component we work out the determinant of three separate $2 \times 2$ matrices, as shown below. For the $\hat{x}$ component, we cross out the top line and first column of the $3 \times 3$ matrix, and compute the determinant of the remaining $2 \times 2$ matrix.

To calculate the $\hat{x}$ component we work out the determinant of the $2 \times 2$ matrix as shown.

For our example, this will be $(a_{y}b_{z} - a_{z}b_{y}))\hat{x}$. Similarly, for the $\hat{y}$ component

To calculate the $\hat{y}$ component we work out the determinant of the $2 \times 2$ matrix as shown.

which will be $(a_{x}b_{z} - a_{z}b_{x})\hat{y}$, but not that we take the negative of this. Finally for the $\hat{z}$ component we have

To calculate the $\hat{z}$ component we work out the determinant of the $2 \times 2$ matrix as shown.

which will be $(a_{x}b_{y} - a{y}b_{x})\hat{z}$.

Summarising, we can write $\boxed{ \vec{c} = \vec{a} \times \vec{b} = (a_{y}b_{z} - a_{z}b_{y})\hat{x} - (a_{x}b_{z} - a_{z}b_{x})\hat{y} + (a_{x}b_{y} - a_{y}b_{x}) \hat{z} }$. This is the vector product of the two vectors.

In the simple case where $\vec{a} = a_{x}\hat{x}$ and $\vec{b} = b_{y}\hat{y}$ then $\vec{c} = \vec{a} \times \vec{b}$ will simply be

$\vec{c} = \vec{a} \times \vec{b} = \left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ a_{x} & 0 & 0 \\ 0 & b_{y} & 0 \\ \end{array} \right|$

which comes to be $(0)\hat{x} - (0)\hat{y} + (a_{x}b_{y} - 0)\hat{z} = (a_{x}b_{y})\hat{z}$.

If we want $\vec{b} \times \vec{a}$ then we must write

$\vec{d} = \vec{b} \times \vec{a} = \left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ 0 & b_{y} & 0 \\ a_{x} & 0 & 0 \\ \end{array} \right|$

which comes to be $\vec{d} = (0)\hat{x} - (0)\hat{y} + (0 - a_{x}b_{y})\hat{z} = -(a_{x}b_{y})\hat{z} = -\vec{c}$. This is shown in the figure below.

The cross (vector) product is not commutative, so $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{a} = -\vec{c}$.

One Response

1. […] Writing this in terms of vectors, and remembering that division of vectors is not defined, we instead write that where is the vector product (or cross-product), as I discussed in this blog here. […]