{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 the result 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}
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h2. Geometric Methods
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 one (for perpendicular unit vectors) or zero (for parallel 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 as shown in the figures below.
!VecAngle.png|height=150!
By using the mathematical definition, it is possible to show that for the case of two vectors *A* and *B* that meet at an angle θ, the magnitude of the cross product will be:
{latex}\begin{large}\[ |\vec{A}\times \vec{B}| = |A||B|\sin\theta \]\end{large}{latex}
{tip}Note that this definition gives a magnitude of one for the product of two perpendicular unit vectors and a magnitude of zero for two parallel unit vectors.{tip}
h4. Magnitudes from Vector Parallelograms
When manipulating vectors, it is sometimes useful to imagine the parallelogram constructed by adding the two vectors in both possible orders (e.g., *A* + *B* and *B* + *A*). The magnitude of the sum of the two vectors can then be interpreted as the length of the diagonal of the parallelogram. The cross product can be similarly interpreted. The magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by adding the vectors.
!VecParallelogram.png|height=150!
h4. Direction from Right Hand Rule
We have given two geometric interpretations of the size of the cross product. Unfortunately, the direction of the cross product is not similarly meaningful. Consider a comparison between vector addition and the cross product. Vector addition is commutative, which means that if *A* + *B* = *C*, then it is also true that *B* + *A* = *C*. For this reason, the preferred direction of the diagonal of the vector parallelogram is obvious. The diagonal should point "along with" the arrows of the sides.
!VecAddCross.png|height=150!
The preferred direction for the cross product is not obvious in the same way. One signal of the difficulty is that the cross product is not commutative. If {latex}\begin{large}$\vec{A}\times\vec{B} =\vec{C}$\end{large}{latex}, then our mathematical definition tells us that {latex}\begin{large}$\vec{B}\times\vec{A} = - \vec{C}$\end{large}{latex}. The _order_ of the vectors in the equation matters for determining the direction of the cross product. This difficulty shows up in the geometric interpretation of the cross product by noticing that if we define the direction of the cross product to be perpendicular to the surface of the parallelogram, there are _two equally good choices_. If we construct a parallelogram that lies in the x,y plane, for example, then either the +z or -z direction is perpendicular to the parallelogram. {info}The situation is even worse if we tried to define a direction for the area that wasn't perpendicular to the parallelogram. Instead of only two directions, we would have an infinite number to choose from!{info}
The fact is that the only way to define a direction for the cross product is to make an _arbitrary_ rule. The rule has already been set up in our mathematical definition. The rule we stated makes the product {latex}\begin{large}$\hat{x}\times\hat{y}$\end{large}{latex} equal to *plus* {latex}\begin{large}$\hat{z}$\end{large}{latex} rather than *minus* {latex}\begin{large}$\hat{z}$\end{large}{latex}. This was an arbitrary choice, based on the traditional ordering of those unit vectors. Once that choice has been made, all we need is a simple rule to remember the consequences. The most widely taught mnemonic is the "right hand rule". To find the direction of the cross product of two vectors, start by carefully reading the order of the vectors. For {latex}\begin{large}$\vec{A}\times\vec{B}$\end{large}{latex}, begin by laying the fingers of your _right hand_ along vector *A* (the first in the product). Then, curl your fingers toward *B*. Your thumb will indicate the direction of the product vector.
!VecCrossProduct.png|height=150!
{warning}You will get the *wrong answer* if you use your left hand.{warning}
{tip}Check that this definition reverses the direction of the product if you start with your fingers along *B* and curl toward *A*.{tip}
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