Cross product

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The cross product is one way of taking the product of two vectors (the other being the dot product). This method yields a third vector perpendicular to both. Unlike the dot product, it is only defined in $\mathbb R^{3}$ (that is, three dimensions). It is commonly used in physics, engineering, vector calculus, and linear algebra. It is defined by the formula

$\mathbf a\times\mathbf b=\|\mathbf a\|\|\mathbf b\|\sin(\theta)\,\mathbf n$

where $\mathbf{n}$ is the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ . It can be computed other ways as well:

${\begin{aligned}\mathbf {a} \times \mathbf {b} &={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{x}&a_{y}&a_{z}\\b_{x}&b_{y}&b_{z}\\\end{vmatrix}}={\begin{vmatrix}a_{y}&a_{z}\\b_{y}&b_{z}\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}a_{x}&a_{z}\\b_{x}&b_{z}\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}a_{x}&a_{y}\\a_{x}&a_{y}\end{vmatrix}}\mathbf {k} \\&=(a_{y}b_{z}-a_{z}b_{y})\mathbf {i} +(a_{z}b_{x}-a_{x}b_{z})\mathbf {j} +(a_{x}b_{y}-a_{y}b_{x})\mathbf {k} \end{aligned}}$

The space $\mathbb R^{3}$ together with the cross product is an algebra over the real numbers, which is neither commutative nor associative, but is a Lie algebra with the cross product being the Lie bracket.

Properties

• ${\vec {v}}\times {\vec {0}}={\vec {0}}$
• ${\vec {v}}\times {\vec {v}}={\vec {0}}$
• ${\vec {v}}\times {\vec {w}}=-{\vec {w}}\times {\vec {v}}$
• $a{\vec {v}}\times {\vec {w}}=a({\vec {v}}\times {\vec {w}})={\vec {v}}\times a{\vec {w}}$
• ${\vec {u}}\times ({\vec {v}}+{\vec {w}})=({\vec {u}}\times {\vec {v}})+({\vec {u}}\times {\vec {w}})$

Conversion to matrix multiplication

The vector cross product also can be expressed as the product of a skew-symmetric matrix and a vector:

$\mathbf {a} \times \mathbf {b} =[\mathbf {a} ]_{\times }\mathbf {b} ={\begin{bmatrix}\,0&\!-a_{3}&\,\,a_{2}\\\,\,a_{3}&0&\!-a_{1}\\-a_{2}&\,\,a_{1}&\,0\end{bmatrix}}{\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}$

$\mathbf {a} \times \mathbf {b} =[\mathbf {b} ]_{\times }^{\mathrm {T} }\mathbf {a} ={\begin{bmatrix}\,0&\,\,b_{3}&\!-b_{2}\\-b_{3}&0&\,\,b_{1}\\\,\,b_{2}&\!-b_{1}&\,0\end{bmatrix}}{\begin{bmatrix}a_{1}\\a_{2}\\a_{3}\end{bmatrix}},$

where superscript ^T refers to the transpose operation, and [a]_× is defined by:

$[\mathbf {a} ]_{\times }{\stackrel {\rm {def}}{=}}{\begin{bmatrix}\,\,0&\!-a_{3}&\,\,\,a_{2}\\\,\,\,a_{3}&0&\!-a_{1}\\\!-a_{2}&\,\,a_{1}&\,\,0\end{bmatrix}}.$

The columns [a]_×,i of the skew-symmetric matrix for a vector a can be also obtained by calculating the cross-product with unit vectors, i.e.:

$[\mathbf {a} ]_{\times ,i}=\mathbf {a} \times \mathbf {{\hat {e}}_{i}} ,\;i\in \{1,2,3\}$

Also, if a is itself expressed as a cross product:

$\mathbf {a} =\mathbf {c} \times \mathbf {d}$

then

$[\mathbf {a} ]_{\times }=\mathbf {d} \mathbf {c} ^{\mathrm {T} }-\mathbf {c} \mathbf {d} ^{\mathrm {T} }.$

Proof by substitution

Evaluation of the cross product gives $\mathbf {a} =\mathbf {c} \times \mathbf {d} ={\begin{pmatrix}c_{2}d_{3}-c_{3}d_{2}\\c_{3}d_{1}-c_{1}d_{3}\\c_{1}d_{2}-c_{2}d_{1}\end{pmatrix}}$ Hence, the left hand side equals $[\mathbf {a} ]_{\times }={\begin{bmatrix}0&c_{2}d_{1}-c_{1}d_{2}&c_{3}d_{1}-c_{1}d_{3}\\c_{1}d_{2}-c_{2}d_{1}&0&c_{3}d_{2}-c_{2}d_{3}\\c_{1}d_{3}-c_{3}d_{1}&c_{2}d_{3}-c_{3}d_{2}&0\end{bmatrix}}$ Now, for the right hand side, $\mathbf {c} \mathbf {d} ^{\mathrm {T} }={\begin{bmatrix}c_{1}d_{1}&c_{1}d_{2}&c_{1}d_{3}\\c_{2}d_{1}&c_{2}d_{2}&c_{2}d_{3}\\c_{3}d_{1}&c_{3}d_{2}&c_{3}d_{3}\end{bmatrix}}$ And its transpose is $\mathbf {d} \mathbf {c} ^{\mathrm {T} }={\begin{bmatrix}c_{1}d_{1}&c_{2}d_{1}&c_{3}d_{1}\\c_{1}d_{2}&c_{2}d_{2}&c_{3}d_{2}\\c_{1}d_{3}&c_{2}d_{3}&c_{3}d_{3}\end{bmatrix}}$ Evaluation of the right hand side gives $\mathbf {d} \mathbf {c} ^{\mathrm {T} }-\mathbf {c} \mathbf {d} ^{\mathrm {T} }={\begin{bmatrix}0&c_{2}d_{1}-c_{1}d_{2}&c_{3}d_{1}-c_{1}d_{3}\\c_{1}d_{2}-c_{2}d_{1}&0&c_{3}d_{2}-c_{2}d_{3}\\c_{1}d_{3}-c_{3}d_{1}&c_{2}d_{3}-c_{3}d_{2}&0\end{bmatrix}}$ Comparison shows that the left hand side equals the right hand side.

This result can be generalized to higher dimensions using geometric algebra. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector. In three dimensions bivectors are dual to vectors so the product is equivalent to the cross product, with the bivector instead of its vector dual. In higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors.

This notation is also often much easier to work with, for example, in epipolar geometry.

From the general properties of the cross product follows immediately that

$[\mathbf {a} ]_{\times }\,\mathbf {a} =\mathbf {0}$   and   $\mathbf {a} ^{\mathrm {T} }\,[\mathbf {a} ]_{\times }=\mathbf {0}$

and from fact that [a]_× is skew-symmetric it follows that

$\mathbf {b} ^{\mathrm {T} }\,[\mathbf {a} ]_{\times }\,\mathbf {b} =0.$

The triple product expansion (bac–cab rule) can be easily proven using this notation.

The Lie algebra R³ with cross product (three-dimensional Euclidean space R³ with the Lie bracket given by the cross product?) is isomorphic to the Lie algebra so(3), whose elements can be identified with the 3×3 skew-symmetric matrices. The map a → [a]_× provides an isomorphism between R³ and so(3). Under this map, the cross product of 3-vectors corresponds to the commutator of 3x3 skew-symmetric matrices.

Matrix conversion for cross product with canonical base vectors

Denoting with $\mathbf {e} _{i}\in \mathbf {R} ^{3\times 1}$ the $i$-th canonical base vector, the cross product of a generic vector $\mathbf {v} \in \mathbf {R} ^{3\times 1}$ with $\mathbf{e}_i$ is given by: $\mathbf {v} \times \mathbf {e} _{i}=\mathbf {C} _{i}\mathbf {v}$, where $\mathbf {C} _{1}={\begin{bmatrix}0&0&0\\0&0&1\\0&-1&0\end{bmatrix}},\quad \mathbf {C} _{2}={\begin{bmatrix}0&0&-1\\0&0&0\\1&0&0\end{bmatrix}},\quad \mathbf {C} _{3}={\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix}}$ These matrices share the following properties: $\mathbf {C} _{i}^{T}=-\mathbf {C} _{i}$ (skew-symmetric); Both trace and determinant are zero; ${\text{rank}}(\mathbf {C} _{i})=2$; $\mathbf {C} _{i}\mathbf {C} _{i}^{T}=\mathbf {P} _{\mathbf {e} _{i}}^{^{\perp }}$ (see below); The orthogonal projection matrix of a vector $\mathbf {v} \neq \mathbf {0}$ is given by $\mathbf {P} _{\mathbf {v} }=\mathbf {v} (\mathbf {v} ^{T}\mathbf {v} )^{-1}\mathbf {v} ^{T}$. The projection matrix onto the orthogonal complement is given by $\mathbf {P} _{\mathbf {v} }^{^{\perp }}=\mathbf {I} -\mathbf {P} _{\mathbf {v} }$, where $\mathbf{I}$ is the identity matrix. For the special case of $\mathbf {v} =\mathbf {e} _{i}$, it can be verified that $\mathbf {P} _{\mathbf {e} _{1}}^{^{\perp }}={\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}},\quad \mathbf {P} _{\mathbf {e} _{2}}^{^{\perp }}={\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix}},\quad \mathbf {P} _{\mathbf {e} _{3}}^{^{\perp }}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}}$ For other properties of orthogonal projection matrices, see projection (linear algebra).

Infinitesimal generators of rotations

The cross product conveniently describes the infinitesimal generators of rotations in R³. Specifically, if n is a unit vector in R³ and R(φ, n) denotes a rotation about the axis through the origin specified by n, with angle φ (measured in radians, counterclockwise when viewed from the tip of n), then

$\left.{d \over d\phi }\right|_{\phi =0}R(\phi ,{\boldsymbol {n}}){\boldsymbol {x}}={\boldsymbol {n}}\times {\boldsymbol {x}}$

for every vector x in R³. The cross product with n therefore describes the infinitesimal generator of the rotations about n. These infinitesimal generators form the Lie algebra so(3) of the rotation group SO(3), and we obtain the result that the Lie algebra R³ with cross product is isomorphic to the Lie algebra so(3).

See also

• Dot product
• Scalar triple product