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Assume the direction vectors are nonzero. The cross product equation is equivalent toso $\mathbf{x}-\mathbf{a}$ is parallel to $\mathbf{u}$ and the locus is the line $\mathbf{x}=\mathbf{a}+\lambda \mathbf{u}$.

For nonparallel directions, ${\mathbf{u}}_1\times {\mathbf{u}}_2$ is normal to both lines. Projecting their displacement onto the corresponding unit vector gives the distance between skew linesIf the directions are parallel, the corresponding formula is

Let $\mathbf{\sigma }$ be the stated Maxwell stress tensor. Taking its divergence and usinggivesThe Maxwell equations in vacuum with sources areSubstitution, together with antisymmetry of the cross product, yieldsTherefore the required vector is the electromagnetic momentum densityandThis is local conservation of electromagnetic momentum. The tensor component ${\sigma }_{ij}$ is the flux of $i$-momentum in the $j$ direction in the convention of the question, $\mathbf{g}$ is field momentum per unit volume, and $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ is the Lorentz force density transferred to matter.

For $A=(A_1,A_2,A_3)$ and $B=(B_1,B_2,B_3)$, the scalar product is the inner productThe cross product isIt is perpendicular to both vectors and has magnitude $|A||B|\sin \theta$.

Put $S(\mathbf{x})=\mathbf{x}\cdot \mathbf{A}\cdot \mathbf{x}$. Since $\mathbf{u}=-S{r}^{-5}\mathbf{x}$ and $\nabla \times \mathbf{x}=0$, the product rule givesFor the symmetric stresslet tensor $\mathbf{A}$,The final term is parallel to $\mathbf{x}$ and drops out of the cross product, leaving

Parametrize the loop by $r^{\prime }=R{\hat{e}}_r(\varphi )$ and $d{\ell }^{\prime }=R{\hat{e}}_{\varphi }(\varphi )d\varphi$. At the observation point $r=z{\hat{e}}_z$, the Biot-Savart law for a line current givesThe components perpendicular to the axis cancel by rotational symmetry, while the axial component of the cross product is $R^{2}d\varphi$. ThereforeThe direction is fixed by the right-hand rule and reverses when the electric current reverses. At the centre,This is the on-axis magnetic field of a circular current loop.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated Body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

Write $k=\mu (1-\beta )$. Taking the cross product of the equation of motion with $\mathbf{r}$ givesSince $\mathbf{r}\cdot \mathbf{h}=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum $\mathbf{h}$. Taking the inner product with $\dot{\mathbf{r}}$ gives the conserved specific orbital energyFor $\beta >1$, the effective inverse-square force is repulsive and the potential term is positive.

The right-hand rule fixes the orientation of a cross product and relates the circulation direction of an electric current to the direction of its magnetic field.

Differentiating $R^{T}R=I$ gives $\mathfrak{so}(3)=\{A\in M_3(\mathbb{R}):A^T=-A\}$. The generators $L_{a}v=e_a\times v$ satisfy $[L_a,L_b]={ϵ}_{abc}{L}_c$ by the cross product identity. This identifies it with the SU(2) Lie algebra by matching a skew-Hermitian Pauli basis.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

In the SO(3) vector Higgs model, a real scalar field $\Phi \in {\mathbb{R}}^3$ transforms as $\Phi \mapsto R\Phi$. Its gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$ for generators acting by the cross product.

The graph $z=g(x,y)$ has upward-oriented vector area element $d\mathbf{S}=(-g_x,-g_y,1)dxdy$ and scalar area element $dS=\sqrt{1+g_x^2+g_y^2}dxdy$. These follow from the cross product of the parametrization derivatives $(1,0,g_x)$ and $(0,1,g_y)$. The vector element is used in flux surface integrals and Stokes theorem; reversing orientation changes its sign.