# Cross product

The cross product, also known as the "vector product", is a vector associated with a pair of vectors in 3-dimensional space.

## Geometric Definition

The cross product is written with a "times sign" between the vectors. Letting ${\vec {C}}$ be the cross product of ${\vec {A}}$ "times" ${\vec {B}}$, we have:
       ${\vec {C}}={\vec {A}}\times {\vec {B}}$
The magnitude of ${\vec {C}}$ is product of the lengths of the two vectors times the sine of the angle between them:
       $||{\vec {A}}\times {\vec {B}}||=AB\sin \theta$,
where $A=||{\vec {A}}||$ and $B=||{\vec {B}}||$, and pointing in the direction perpendicular to the plane formed by the two vectors:
This leaves an ambiguity—it could point in either of two opposite directions. The ambiguity is resolved by the "right hand rule"—if the fingers of the right hand (except the thumb) are curled as if to depict the rotation of the first vector into the second (following the angular direction of the smallest angle between them), then the thumb points in the direction of the cross product.
A few properties can be deduced from this definition:
* It is homogeneous in each argument: $(\lambda {\vec {A}})\times {\vec {B}}={\vec {A}}\times (\lambda {\vec {B}})=\lambda ({\vec {A}}\times {\vec {B}})\,$.
* If the two vectors point in the same direction, or opposite directions, their cross product is zero. This is because the sine of 0° or 180° is zero. It is fortunate that this is so, because the right hand rule couldn't be applied in that case.
* The cross product of two vectors is perpendicular to each of those vectors.
* It is anticommutative: ${\vec {A}}\times {\vec {B}}=-{\vec {B}}\times {\vec {A}}$.
* The cross product has an intrinsic "handedness" or chirality, due to the use of the right hand rule. If one looks in a mirror at two vectors and their cross product, the cross product will appear to point in the wrong direction.

## Algebraic Definition

The cross product has a remarkable, convenient, and elegant algebraic formulation. If the components of a vector in 3-dimensional Cartesian space are given with subscripts: $A_{x}$, $A_{y}$, and $A_{z}$, then the cross product of vectors ${\vec {A}}$ and ${\vec {B}}$ is given by:
       $({\vec {A}}\times {\vec {B}})_{x}=(A_{y}B_{z}-A_{z}B_{y})$
       $({\vec {A}}\times {\vec {B}})_{y}=(A_{z}B_{x}-A_{x}B_{z})$
       $({\vec {A}}\times {\vec {B}})_{z}=(A_{x}B_{y}-A_{y}B_{x})$
       It needs to be emphasized that this formula only works if the components of the vectors are described in Cartesian space. In other coordinate systems, it may be necessary to use other, more complicated formulas.
This can be put into a particularly convenient and easy-to-remember form through the use of determinants. Letting ${\hat {x}}$, ${\hat {y}}$, and ${\hat {z}}$ be the Cartesian unit basis vectors, and taking a slight liberty in the meaning of multiplication, we have:
       ${\vec {A}}\times {\vec {B}}={\begin{vmatrix}{\hat {x}}&{\hat {y}}&{\hat {z}}\\A_{x}&A_{y}&A_{z}\\B_{x}&B_{y}&B_{z}\end{vmatrix}}$
The theory of determinants tells us that the determinant changes sign if any two rows of the matrix are exchanged, so swapping the two vectors changes the sign—the cross product is anticommutative. The theory of determinants also tells us that it is homogeneous. One can also work out that it is distributive:
       $({\vec {A}}+{\vec {B}})\times {\vec {C}}=({\vec {A}}\times {\vec {C}})+({\vec {B}}\times {\vec {C}})$

## Equivalence of the two Definitions

It is important to prove that the geometric and algebraic definitions of the cross product are the same. The proof is somewhat difficult. First, we need a few lemmas, that is, preliminary theorems.
Notation: We will use a fixed vector ${\vec {C}}$ throughout. Consider the plane $P\,$ perpendicular to ${\vec {C}}$. The projection of any other vector (say ${\vec {A}}$) onto $P\,$ will be denoted ${\vec {A}}^{*}$. See Figure 1.
    What if ${\vec {C}}$ is zero? Its perpendicular plane is undefined. But in that case the lemma we are trying to prove (distributivity) will be trivially true.
Lemma 1: Using the geometric definition of the cross product:
       ${\vec {A}}\times {\vec {C}}={\vec {A}}^{*}\times {\vec {C}}$
Proof: Let $\theta \,$ be the angle between ${\vec {A}}$ and ${\vec {C}}$. We have:
       $\|{\vec {A}}^{*}\|=\|{\vec {A}}\|\sin \theta$
Since ${\vec {A}}^{*}$ is perpendicular to ${\vec {C}}$, we have:
       $\|{\vec {A}}\times {\vec {C}}\|=\|{\vec {A}}\|\|{\vec {C}}\|\sin \theta =\|{\vec {A}}^{*}\|\|{\vec {C}}\|=\|{\vec {A}}^{*}\|\|{\vec {C}}\|\sin 90^{\circ }=\|{\vec {A}}^{*}\times {\vec {C}}\|$
So they have the same length. Also, ${\vec {A}}\times {\vec {C}}$ and ${\vec {A}}^{*}\times {\vec {C}}$ point in the same direction, the direction perpendicular to the plane formed by ${\vec {C}}$ and ${\vec {A}}$. See Figure 2.
Q.E.D.
Lemma 2: Letting ${\vec {B}}$ be another vector:
       $({\vec {A}}+{\vec {B}})^{*}={\vec {A}}^{*}+{\vec {B}}^{*}$
Proof: We use the "parallelogram rule" for vector addition. In perspective, the vectors might look like Figures 3 and 4.
But if we look directly down on the plane, ${\vec {A}}$ and ${\vec {A}}^{*}$ are the same, as are ${\vec {B}}$ and ${\vec {B}}^{*}$, and $({\vec {A}}+{\vec {B}})$ and $({\vec {A}}+{\vec {B}})^{*}$.
Q.E.D.
Lemma 3: The cross product, using the geometric definition, obeys the distributive law:
       $({\vec {A}}+{\vec {B}})\times {\vec {C}}=({\vec {A}}\times {\vec {C}})+({\vec {B}}\times {\vec {C}})$
Proof:
If we look directly down on the plane perpendicular to C, we can once again see the addition of ${\vec {A}}^{*}$ and ${\vec {B}}^{*}$ as shown in Figure 5. If we rotate that diagram clockwise by 90° and scale by the length of ${\vec {C}}$, we get the cross products of all those vectors with C, as shown in Figure 6. Those cross products all lie in the plane perpendicular to C, and the addition of ${\vec {A}}^{*}+{\vec {B}}^{*}$, by the parallelogram rule, is faithfully carried over into the addition of ${\vec {A}}^{*}\times {\vec {C}}+{\vec {B}}^{*}\times {\vec {C}}$.
This shows that
       ${\vec {A}}^{*}\times {\vec {C}}+{\vec {B}}^{*}\times {\vec {C}}=({\vec {A}}^{*}+{\vec {B}}^{*})\times {\vec {C}}$
By lemma 2, this becomes
       ${\vec {A}}^{*}\times {\vec {C}}+{\vec {B}}^{*}\times {\vec {C}}=({\vec {A}}+{\vec {B}})^{*}\times {\vec {C}}$
And by lemma 1, this becomes
       ${\vec {A}}\times {\vec {C}}+{\vec {B}}\times {\vec {C}}=({\vec {A}}+{\vec {B}})\times {\vec {C}}$
Q.E.D.
By the antisymmetry of the cross product, it is also distributive in the second argument:
       ${\vec {C}}\times ({\vec {A}}+{\vec {B}})=({\vec {C}}\times {\vec {A}})+({\vec {C}}\times {\vec {B}})$
Theorem: The geometric and algebraic definitions of the cross product are the same.
Proof: Let the vectors ${\vec {A}}$ and ${\vec {B}}$ be expressed as linear combinations of the Cartesian unit basis vectors ${\hat {x}}$, ${\hat {y}}$, and ${\hat {z}}$:
       ${\vec {A}}=A_{x}{\hat {x}}+A_{y}{\hat {y}}+A_{z}{\hat {z}}$
       ${\vec {B}}=B_{x}{\hat {x}}+B_{y}{\hat {y}}+B_{z}{\hat {z}}$
(The numbers $A_{x}\dots B_{z}$ are the components of the two vectors.)
By lemma 3, and linearity, we can expand the cross product:
       ${\vec {A}}\times {\vec {B}}=A_{x}\ ({\hat {x}}\times {\vec {B}})+A_{y}\ ({\hat {y}}\times {\vec {B}})+A_{z}\ ({\hat {z}}\times {\vec {B}})$
       $=A_{x}\ ({\hat {x}}\times (B_{x}{\hat {x}}+B_{y}{\hat {y}}+B_{z}{\hat {z}}))$
       $+\ A_{y}\ ({\hat {y}}\times (B_{x}{\hat {x}}+B_{y}{\hat {y}}+B_{z}{\hat {z}}))$
       $+\ A_{z}\ ({\hat {z}}\times (B_{x}{\hat {x}}+B_{y}{\hat {y}}+B_{z}{\hat {z}}))$
       $=A_{x}B_{x}\ ({\hat {x}}\times {\hat {x}})+A_{x}B_{y}\ ({\hat {x}}\times {\hat {y}})+A_{x}B_{z}\ ({\hat {x}}\times {\hat {z}})$
       $+\ A_{y}B_{x}\ ({\hat {y}}\times {\hat {x}})+A_{y}B_{y}\ ({\hat {y}}\times {\hat {y}})+A_{y}B_{z}\ ({\hat {y}}\times {\hat {z}})$
       $+\ A_{z}B_{x}\ ({\hat {z}}\times {\hat {x}})+A_{z}B_{y}\ ({\hat {z}}\times {\hat {y}})+A_{z}B_{z}\ ({\hat {z}}\times {\hat {z}})$
Now inspection of the cross products of the basis vectors, using the geometric definition and the right hand rule, shows that:
       ${\hat {x}}\times {\hat {x}}=0\ \ \ \ \ \ \ \ {\hat {x}}\times {\hat {y}}={\hat {z}}\ \ \ \ \ \ \ \ {\hat {x}}\times {\hat {z}}=-{\hat {y}}$
       ${\hat {y}}\times {\hat {x}}=-{\hat {z}}\ \ \ \ \ \ {\hat {y}}\times {\hat {y}}=0\ \ \ \ \ \ \ \ {\hat {y}}\times {\hat {z}}={\hat {x}}$
       ${\hat {z}}\times {\hat {x}}={\hat {y}}\ \ \ \ \ \ \ \ {\hat {z}}\times {\hat {y}}=-{\hat {x}}\ \ \ \ \ \ {\hat {z}}\times {\hat {z}}=0$
So
       ${\vec {A}}\times {\vec {B}}=(A_{y}B_{z}-A_{z}B_{y})\ {\hat {x}}+(A_{z}B_{x}-A_{x}B_{z})\ {\hat {y}}+(A_{x}B_{y}-A_{y}B_{x})\ {\hat {z}}$
which is the same as the algebraic definition.
Q.E.D.

## Footnotes and References

* ↑ A real, old-fashioned, times sign. Not an asterisk.