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Use in Physics

In mechanics, the cross product is used in calculating torque and angular momentum. The cross product is also used in introductory electricity and magnetism, where calculations involving the production and effects of magnetic fields generally require the cross product.

Calculating Cross Products

Unit Vector Cross Products

By definition:

{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:

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

Odd

...

permutations

...

reverse

...

the

...

sign:

}\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 

\begin{large}$\hat{x}$\end{large}

\begin{large}$\hat{x}$\end{large}

and

...

the

...

cross

...

product

...

of

...

any

...

vector

...

with

...

itself

...

is

...

zero:

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

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

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

Using

...

the

...

relationships

...

given

...

above

...

for

...

the

...

cross

...

product

...

of

...

unit

...

vectors,

...

we

...

have:

}\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}

Shortcut Using Matrix Determinant

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

{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}

Geometric Methods

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 (

\begin{large}$\hat{x}\perp \hat{y}, \hat{z}$\end{large}

) or parallel (

\begin{large}$\hat{x} \parallel \hat{x}$\end{large}

) 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

\begin{large}$\vec{A} = \frac{1}{\sqrt{2}}\hat{x} + \frac{1}{\sqrt{2}}\hat{y}$\end{large}

. Two arbitrary vectors will usually not be perfectly parallel or perpendicular. Instead, they will form some angle θ as shown in the figures below.

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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:

\begin{large}\[ |\vec{A}\times \vec{B}| = |A||B|\sin\theta \]\end{large}

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 vector parallelogram.

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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 unambiguous. The diagonal should point "along with" the arrows of the sides.

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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

\begin{large}$\vec{A}\times\vec{B} =\vec{C}$\end{large}

, then our mathematical definition tells us that

\begin{large}$\vec{B}\times\vec{A} = - \vec{C}$\end{large}

. 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.

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 incorporated in the mathematical definition we gave above. The definition we stated makes the product

\begin{large}$\hat{x}\times\hat{y}$\end{large}

equal to plus

\begin{large}$\hat{z}$\end{large}

rather than minus

\begin{large}$\hat{z}$\end{large}

. 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

\begin{large}$\vec{A}\times\vec{B}$\end{large}

, 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.

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