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

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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.

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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

The formalism above has a simple geometric interpretation.  The cross product measures the "perpendicularity" of two vectors.  Since Cartesian unit vectors are always either perpendicular ({latex}\begin{large}$\hat{x}\perp \hat{y}, \hat{z}$\end{large}{latex}) or parallel ({latex}\begin{large}$\hat{x} \parallel \hat{x}$\end{large}{latex}) we get a cross product with either magnitude zero (for parallel unit vectors) or one (for perpendicular unit vectors).  The mathematical definitions given above, however, will let you construct cross products with vectors that are combinations of the unit vectors, such as {latex}\begin{large}$\vec{A} = \frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}$\end{large}{latex}.  Two arbitrary vectors will usually not be perfectly parallel or perpendicular.  Instead, they will form some angle θ.