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The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

Every traceless skew-Hermitian matrix of size two has the formso it equals $\mathcal{M}(c)$ for $c=(2p,2\mathrm{Re}w,2\mathrm{Im}w)$. If $c=0$, take $A=B=0$. Otherwise choose a unit vector $u$ perpendicular to $c$ and set $a=u$, $b=c\times u$. The vector triple product gives $a\times b=c$.

By part (iii), choosing $A=\mathcal{M}(a)$ and $B=\mathcal{M}(b)$ gives the required representationIn fact both factors can themselves be chosen as traceless skew-Hermitian matrices. The cross-product model of su(2) explains this geometric commutator construction.

Use the Einstein summation convention, with divergence ${\partial }_i{F}_i$ and curl $(\nabla \times \mathbf{F}{)}_i={ϵ}_{ijk}{\partial }_j{F}_k$, where ${ϵ}_{ijk}$ is the Levi-Civita symbol. For the first vector field,The derivative matrix is symmetric, so its contraction with the Levi-Civita symbol vanishes. For the second vector field,For the third vector field, $H_i={ϵ}_{ikl}{a}_k{x}_l/r$. Its divergence is zero because the Levi-Civita symbol contracts with symmetric products. For its curl, either contract two Levi-Civita symbols or use the vector triple product:ThusAll formulas hold on $r>0$, the domain where the radial unit vector is defined.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

Since $\mathbf{E}\cdot {\mathbf{e}}_z=0$, the vector triple product yields $\mathbf{E}\times ({\mathbf{e}}_z\times \mathbf{E})=|\mathbf{E}{|}^2{\mathbf{e}}_z$. Consequently the Poynting vector isIts average over a full period uses $⟨{\cos }^2⟩=1/2$:This is the mean electromagnetic energy flux per unit area. For positive frequency it points in the propagation direction and has magnitude $E_0^2/(2{\mu }_{0}c)={ϵ}_{0}c{E}_0^2/2$.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

The vector triple product identity givesApplying the same identity after reversing the outer cross product givesThe formulas differ because the cross product is not associative.

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

Because $\tilde{\mathbf{n}}\cdot (\tilde{\mathbf{n}}\times {\partial }_t\tilde{\mathbf{n}})=0$, the component of $\mathbf{q}$ parallel to $\tilde{\mathbf{n}}$ is $-\mathbf{B}\cdot \tilde{\mathbf{n}}$. The Gaussian integral over $\lambda$ removes precisely this longitudinal component. ThusUsing the vector triple-product identity and $|\tilde{\mathbf{n}}|=1$, this becomes the $O(3)$ nonlinear sigma model in a background field:The magnetic field acts as the temporal component of an $O(3)$ background gauge connection.

Suppose ${\epsilon }_{ij}{e}_j={\epsilon }_e{e}_i$. Taking the scalar product of the equation in part (d) with $\mathbf{k}$ gives $\mathbf{k}\cdot \mathbf{e}=0$. The vector triple product then reduces the equation toso $k^2={\omega }^2\mu {\epsilon }_e$. From part (d),ConsequentlyThus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction $\mathbf{k}$.

In components, part (i) givesTherefore the vector triple product is

PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.