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For a solenoidal vector field $J$ with compact support and suitable integrability, apply the divergence theorem to $x_i{x}_{j}J$. It gives zero boundary term and hence the displayed antisymmetry. Equivalently, for any constant vector $C$, the vector triple product identity yields $\int (C\cdot r)JdV=\frac{1}{2}C\times \int (J\times r)dV$. A piecewise smooth field must have no normal jump at its support boundary; otherwise its weak divergence contains a surface contribution and the argument fails.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.

With all electric dipole moments and time derivatives evaluated at $t_0=t-r/c$, the retarded potentials of a point electric dipole give, for $r>0$,For a short verification, apply $\nabla t_0=-\mathbf{n}/c$ and ${\partial }_i{n}_j=({\delta }_{ij}-n_i{n}_j)/r$ when computing $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. This produces the displayed $r^{-3}$, $r^{-2}$ and $r^{-1}$ terms. The radiation electric field is transverse and obeys ${\mathbf{B}}_{\mathrm{rad}}=\mathbf{n}\times {\mathbf{E}}_{\mathrm{rad}}/c$. The vector triple product then gives the leading Poynting vector $\mathbf{n}[|{\mathbf{p}}^{\prime \prime }{|}^2-(\mathbf{n}\cdot {\mathbf{p}}^{\prime \prime }{)}^2]/(16{\pi }^2{ϵ}_0{c}^3{r}^2)$, displaying the angular distribution of electric dipole radiation.

For a unit vector $\mathbf{n}$, real $\alpha \ne 0$, and real $\beta$, the displayed linear equation has the unique vector solutionTo prove it, split vectors into components parallel and orthogonal to $\mathbf{n}$. On the parallel line, the linear map is multiplication by $\alpha$. On the perpendicular plane, the operator $N\mathbf{v}=\mathbf{n}\times \mathbf{v}$ obeys $N^2=-I$, by the vector triple product. Therefore $(\alpha I-\beta N)(\alpha I+\beta N)=({\alpha }^2+{\beta }^2)I$ on that plane. Combining its inverse with the parallel solution gives the formula. This avoids component calculations in equations with an axial cross product term.

The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$ be the scalar triple product. The reciprocal basis satisfies $\mathbf{a}\cdot {\mathbf{a}}^{\prime }=1$ and all mismatched dot products vanish, with cyclic analogues. Apply the given vector triple product identity to obtainDotting with ${\mathbf{a}}^{\prime }$ givesThe sign is retained: a negatively oriented original basis has a negatively oriented reciprocal basis. Equivalently, if the original vectors are columns of $B$, their reciprocals are columns of $B^{-T}$, whose determinant is $1/\det B$.

Write both cross products with the Levi-Civita symbol. Its contraction identity givesTherefore the vector triple product identity isThe order matters: the cross product is not associative, and exchanging the two nested arrangements changes the expression.

The vector triple product with $|u|=1$ givesThis is the orthogonal projection onto a plane perpendicular to $u$, so it is a linear map. Its image is the plane $u^{\perp }$ through the origin, while its kernel is the normal line $\mathrm{span}\{u\}$. In components,The equation $Ar=r$ is equivalent to $u(u\cdot r)=0$, hence to $u\cdot r=0$. Its solution set is precisely the two-dimensional plane $u^{\perp }$, with infinitely many nonzero vectors, not merely the normal line. These are the eigenvectors with eigenvalue one; the normal direction has eigenvalue zero.

Use the Levi-Civita symbol and Einstein summation convention. The tensor contraction ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$, expressed through the Kronecker delta, givesThe last two contractions are dot products, proving the vector triple product identity $\boxed{a\times (b\times c)=(a\cdot c)b-(a\cdot b)c}$.

Choose the linear relationThen the displayed tensor contraction is $(a_i{c}_i)b_k-(a_i{b}_i)c_k=(a\cdot c)b_k-(a\cdot b)c_k$, precisely the $k$th vector component of $a\times (b\times c)$ by the vector triple product identity. The matrix $M$ need not be a symmetric matrix.

The three vectors are coplanar precisely when their scalar triple product vanishes:This includes linearly dependent or zero vectors. It expresses that the determinant with those vectors as columns is zero.

For the two moving particles, distinguish closest approach at a common time from the geometric distance between their supporting lines. For fixed reference times, putTheir relative position is ${\mathbf{d}}_0+t\mathbf{v}$. If $\mathbf{v}\ne 0$, the closest approach of two uniformly moving particles occurs atIf $\mathbf{v}=0$, the distance is constant. If only $t\ge 0$ is permitted, replace $t_{\ast }$ by $\max (0,t_{\ast })$ before evaluating the distance.

Now suppose the velocities are nonparallel, and allow the relative starting-time offset to be chosen so that the particles reach the closest points of the two lines simultaneously. Write $\Delta ={\mathbf{a}}_2-{\mathbf{a}}_1$, $A=|{\mathbf{b}}_1{|}^2$, $B={\mathbf{b}}_1\cdot {\mathbf{b}}_2$, $C=|{\mathbf{b}}_2{|}^2$ and $D=AC-B^2>0$. The independent line parameters $s_1,s_2$ at the closest points satisfyThese equations make the connecting vector perpendicular to both velocities. Solving the two-by-two linear system givesThe corresponding distance between skew lines is $|\Delta \cdot ({\mathbf{b}}_1\times {\mathbf{b}}_2)|/|{\mathbf{b}}_1\times {\mathbf{b}}_2|$. Simultaneous arrival requires $t_1+s_1=t_2+s_2$, so $t_1-t_2=s_2-s_1$. Since the vector triple product givesthis condition is preciselyThus the printed formula has a valid interpretation as a synchronization condition for attaining the line-distance lower bound. It is not an identity that must hold for arbitrary fixed reference times at the time of closest approach.

For a counterexample to that stronger reading, take ${\mathbf{a}}_1=0$, ${\mathbf{a}}_2=(1,0,1)$, ${\mathbf{b}}_1=(1,0,0)$, ${\mathbf{b}}_2=(0,1,0)$ and $t_1=t_2=0$. The actual squared distance is $2(t-1/2{)}^2+3/2$, minimized at $t=1/2$. The printed right-hand side is $-1$, whereas the fixed difference $t_1-t_2$ is zero. The line-distance bound is one, but it is not attained by that synchronized motion. Parallel velocities also require separate treatment because the displayed denominator then vanishes.

Assume $\mathbf{a}\ne 0$ and allow all $\mathbf{R}\ne 0$. Put $\hat{\mathbf{a}}=\mathbf{a}/|\mathbf{a}|$ and $\mathbf{u}=\mathbf{R}/|\mathbf{R}|$. Every resulting vector is perpendicular to $\mathbf{a}$ and has $|\mathbf{r}|=|c|\sin \theta \le |c|$. Conversely, for $c\ne 0$ and any $\mathbf{q}=\mathbf{r}/c$ perpendicular to $\hat{\mathbf{a}}$ with $|\mathbf{q}|\le 1$, chooseThe vector triple product gives $\mathbf{u}\times \hat{\mathbf{a}}=\mathbf{q}$, and $|\mathbf{u}|=1$. Thus every point of the closed disk centered at the origin, perpendicular to $\mathbf{a}$, of radius $|c|$ is attained. The boundary circle alone would omit allowed directions of $\mathbf{R}$. For $c=0$ the locus is the origin; for $\mathbf{a}=0$ the original expression is undefined.

Use the vector triple product identity:For the $i$th component of the second integral, the antisymmetric first moment of a localized solenoidal field with $C^{\prime }=e_i$ gives ${\int }_V{r}_i(C\cdot J)dV=-{\int }_V(C\cdot r)J_{i}dV$. Thus the two terms on the right have opposite integrals. Consequently

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.

Neglecting gravity, the equation of motion is $m\ddot{\mathbf{x}}=a\dot{\mathbf{x}}\times \mathbf{x}/r^3$, where $r=|\mathbf{x}|>0$. Dotting with the velocity gives $d(m|\dot{\mathbf{x}}{|}^2/2)/dt=0$, because a cross product is perpendicular to its factors. Thus the speed is constant. The vector triple product givesHence the generalized angular momentum in a radial monopole field isDotting with $\mathbf{x}$ gives $\mathbf{L}\cdot \mathbf{x}=-ar$, or $\hat{\mathbf{L}}\cdot \hat{\mathbf{x}}=-a/|\mathbf{L}|$. For $\mathbf{L}\ne 0$, the direction of $\mathbf{x}$ therefore makes the fixed angle $\boxed{\cos \beta =-a/|\mathbf{L}|}$ with the axis $\mathbf{L}$, describing a right circular conical surface with vertex at the North Pole. Orthogonality gives $|\mathbf{L}{|}^2=m^2|\mathbf{x}\times \dot{\mathbf{x}}{|}^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary angular momentum is nonzero the cone is the usual orbital plane. If $\mathbf{L}=0$, the norm identity forces $a=0$ and $\mathbf{x}\times \dot{\mathbf{x}}=0$, giving a radial straight line.

For the unheaded continuation, write the vector triple product asThe Levi-Civita symbol contraction is ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$. One way to check its sign is to fix distinct $i,j$: the only nonzero term has $k$ the remaining index, and the pairs $(l,m)=(i,j),(j,i)$ contribute $+1,-1$ respectively. When $i=j$, both sides vanish. Contracting with the vectors now gives both requested notations:If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a}\cdot \mathbf{b}=0$. With the specified $\mathbf{c}$, the perpendicularity of the cross product gives $\mathbf{a}\cdot \mathbf{c}=|\mathbf{a}{|}^2$. ThereforeThe same answer covers the degenerate case in which one of these two vectors is zero.

Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.

The vector triple product removes the nested cross products:Since $\mathbf{\omega }$ is constant, the divergence isThe cube has volume $2^3=8$. With the outward unit normal, the divergence theorem therefore gives

For $\psi =\mu /r$, the radial gradient givesThe force $-\nabla \varphi$ is radial and has no torque about the origin. For the remaining force, use the vector triple product:Meanwhile the radial unit vector obeysSince $\psi \mathbf{r}=\mu \hat{\mathbf{r}}$, these derivatives cancel. Thus the generalized angular momentum in a radial monopole field isThe argument applies away from $r=0$, where the field is defined, for any radial $\varphi$. In contrast to ordinary angular momentum, this conserved vector has $\mathbf{L}\cdot \hat{\mathbf{r}}=-\mu$; it fixes a cone rather than generally fixing a plane.