Page History

{excerpt}Also known as the vector product, the cross product is a way of multiplying two vectors to yield another vector.{excerpt}

||Page Contents||
|{toc:indent=10px|style=none}|

----
h2. Use in Physics

In mechanics, the cross product is used in calculating [torque|torque (one-dimensional)] and [angular momentum|angular momentum (one-dimensional)].  The cross product is also used in introductory electricity and magnetism.  Calculations involving the production and effects of magnetic fields generally involve the cross product.

----
h2. Calculating Cross Products

h4. Unit Vector Cross Products

By definition:

{latex}\begin{large}\[\hat{x}\times \hat{y}= \hat{z}\]\end{large}{latex}

and the same holds for even permutations of the order of the unit vectors, thus:

{latex}\begin{large}\[ \hat{y} \times \hat{z} = \hat{x} \]
\[ \hat{z}\times \hat{x} = \hat{y}\]\end{large}{latex}

Odd permutations reverse the sign:

{latex}\begin{large}\[ \hat{y}\times\hat{x} = -\hat{z}\]
\[\hat{z}\times\hat{y} = -\hat{x}\]
\[\hat{x}\times\hat{z} = -\hat{y}\]\end{large}{latex}

{info}For three dimensions, the sign of the cross product of two unit vectors can be easily remembered by checking if the unit vectors are in a special version of alphabetical order.  Start with the position of the {latex}\begin{large}$\hat{x}$\end{large}{latex} vector and read to the right.  When you get to the end of the equation, wrap to the beginning and keep reading until you return to {latex}\begin{large}$\hat{x}$\end{large}{latex}.  If you get x, y, z, then the sign of on the right hand side is positive.  If you get x, z, y then the sign is negative.{info}

and the cross product of any vector with itself is zero:

{latex}\begin{large}\[ \hat{x}\times\hat{x} = 0\]
\[\hat{y}\times\hat{y} = 0\]
\[\hat{z}\times\hat{z} = 0\]\end{large}{latex}

{note}Note that reversing the order of the two vectors being multiplied switches the sign of the result.{note}

Using this definition, it is possible to find the componentwise cross product of two vectors:

{latex}\begin{large}\[\vec{A}\times\vec{B}=(A_{x}\hat{x}+A_{y}\hat{y}+A_{z}\hat{z})\times(B_{x}\hat{x}+B_{y}\hat{y}+B_{z}\hat{z}) = A_{x}B_{x}\hat{x}\times\hat{x} + A_{x}B_{y}\hat{x}\times\hat{y} + A_{x}B_{z}\hat{x}\times\hat{z} + A_{y}B_{x}\hat{y}\times\hat{x} +A_{y}B_{y}\hat{y}\times\hat{y}+A_{y}B_{z}\hat{y}\times\hat{z}+A_{z}B_{x}\hat{z}\times\hat{x}+A_{z}B_{y}\hat{z}\times\hat{y} + A_{z}B_{z}\hat{z}\times\hat{z}\]\end{large}{latex}

Using the relationships given above for the cross product of unit vectors, we have:

{latex}\begin{large}\[ A_{x}B_{y}\hat{z} - A_{x}B_{z}\hat{y}-A_{y}B_{x}\hat{z}+A_{y}B_{z}\hat{x} + A_{z}B_{x}\hat{y}-A_{z}B_{y}\hat{x} = (A_{y}B_{z}-A_{z}B_{y})\hat{x} + (A_{z}B_{x} - A_{x}B_{z})\hat{y} +(A_{x}B_{y}-A_{y}B_{x})\hat{z}\]\end{large}{latex}

h4. Shortcut Using Matrix Determinant

One way to remember the formula derived in the section above is to use a matrix determinant:

{latex}\begin{large}\[ \vec{A}\times\vec{B} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{vmatrix} = (A_{y}B_{z}-A_{z}B_{y})\hat{x} + (A_{z}B_{x} - A_{x}B_{z})\hat{y} +(A_{x}B_{y}-A_{y}B_{x})\hat{z}\]\end{large}{latex}

h4. Magnitudes from Trigonometry