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The vector projection multiplies the scalar projection by the chosen unit vector, giving the orthogonal projection onto the linear span of a nonzero real Euclidean vector $\mathbf{c}$:For $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$, this gives $\mathbf{p}=(6/4)(2,0)=(3,0)$ and $\mathbf{r}=(0,4)$; the dot product $\mathbf{r}\cdot \mathbf{c}=0$ checks the orthogonal projection residual. Replacing $\mathbf{c}$ by $(-2,0)$ gives the same $\mathbf{p}=(-6/4)(-2,0)=(3,0)$: reversing direction changes the scalar projection, but preserves the vector projection onto the line.

For a short proof of the nearest-point property, any vector $\mathbf{z}$ on this line satisfies $\Vert \mathbf{a}-\mathbf{z}{\Vert }^2=\Vert \mathbf{r}{\Vert }^2+\Vert \mathbf{p}-\mathbf{z}{\Vert }^2$ by the Pythagorean identity. The second term is uniquely minimized at $\mathbf{z}=\mathbf{p}$. This connects a directional measurement to the orthogonal projection used in approximation.^[ref]

A vector area element combines the scalar surface area element with a chosen unit normal vector. Reversing the order of the two parameter directions reverses its sign. Flux through an oriented surface is the integral of the dot product of a vector field with this vector element. For a graph $z=f(x,y)$, the upward choice is $(-f_x,-f_y,1)dxdy$.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]

A wavenumber measures how rapidly a wave phase changes with distance. Two conventions coexist: angular wavenumber counts radians per unit length, while ordinary spatial frequency counts cycles per unit length. For a monochromatic wave with wavelength $\lambda >0$,Here $K\ge 0$ is the magnitude of the angular wavenumber. The symbol $k$ can instead denote a signed one-dimensional angular wavenumber: a factor $e^{ikx}$ has wavelength $2\pi /|k|$ for real $k\ne 0$. Both conventions have dimensions of inverse length; writing $\mathrm{rad}{\mathrm{m}}^{-1}$ or “cycles per metre” exposes the convention. A quoted value in ${\mathrm{m}}^{-1}$ alone does not resolve the factor of $2\pi$. Spectroscopic wavenumber usually uses the cycles-per-length convention.^[ref]

For a real monochromatic plane wave, fix the conventionThe wavevector $\mathbf{k}$ supplies both the spatial wave phase slope and the normal to the wavefronts; its dot product with position makes the wave phase constant on parallel planes. The angular frequency is $\omega =2\pi f$, with $f>0$ the ordinary frequency. Moving a distance $\lambda$ along $\mathbf{k}$ changes the wave phase by $2\pi$. Holding a wave crest's wave phase fixed gives its phase velocity $\omega \mathbf{k}/K^2$ and phase speedIn one dimension the signed phase velocity is $\omega /k$. This identifies a moving pattern; it does not identify the motion of the medium's particles or a wave packet's group velocity.

For the synthetic plane wave below, choose $\lambda =0.40\mathrm{m}$ and $f=5\mathrm{Hz}$. Then $K=5\pi \mathrm{rad}{\mathrm{m}}^{-1}$, ${\nu }_x=2.5$ cycles per metre, $\omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}$, and $v_p=2\mathrm{m}{\mathrm{s}}^{-1}$. The wave crests are $0.40\mathrm{m}$ apart at one instant; after a quarter oscillation period, each has advanced $0.10\mathrm{m}$. The lower panel retains whole turns of wave phase, making its slope visible instead of reducing it modulo $2\pi$.

A lattice direction $[uvw]$ lies parallel to a plane $(hkl)$ exactly when $hu+kv+lw=0$. Indeed, their direct and reciprocal lattice vectors have dot product proportional to this expression. This statement holds for noncubic cells as well; angles between directions still require the direct-lattice metric tensor.

Work measures the energy transferred by a force as its point of application moves. For a point mass following a path $C$, add the contribution of the force along each small displacement:This line integral uses the actual path, and the dot product selects the component of force along the motion. A large force perpendicular to every small displacement does zero work; holding something still also does zero mechanical work on it, even if holding it feels tiring. The SI unit is $\mathrm{J}=\mathrm{N}\mathrm{m}$. ^[ref]

The usefulness of work is that it connects a description in terms of forces and distances to one in terms of energy. The work-energy theorem gives the change in kinetic energy from the total net work, often without solving for the entire motion. The work of one selected force equals that change only when the other forces contribute zero total work.

For a fixed force $\mathbf{F}$ and endpoint displacement $\Delta \mathbf{r}$, the line integral reduces to a dot product:Here $\theta$ is the angle between the force and the displacement. For straight motion, $d$ is also the distance traveled. A $3\mathrm{N}$ force acting along a $4\mathrm{m}$ displacement does $12\mathrm{J}$ of work. Turning the same force perpendicular to the displacement gives $\cos (\pi /2)=0$ and zero work; reversing it gives $\cos \pi =-1$ and $-12\mathrm{J}$. At $\theta =\pi /3$, the work is $6\mathrm{J}$. These signs describe energy transferred to or from the moving body. ^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

For a wrench with nonzero resultant force $\mathbf{F}$ and moment ${\mathbf{\tau }}_O$ about a stated point $O$, its wrench pitch isIt has units of length: torque divided by force. It measures the signed moment remaining parallel to the resultant, per unit force. The force/moment representation and its frame transformations are developed in the primary robotics text’s wrench supplement.^[ref]

The wrench pitch does not depend on the moment origin. Moving from $O$ to $P$, with displacement $\mathbf{r}=\overrightarrow{OP}$ expressed in the same axes, gives ${\mathbf{\tau }}_P={\mathbf{\tau }}_O-\mathbf{r}\times \mathbf{F}$. Taking the dot product with $\mathbf{F}$ removes the added term because $\mathbf{F}\cdot (\mathbf{r}\times \mathbf{F})=0$. A common rotation of both vectors also preserves the ratio. Scaling every applied load by the same nonzero scalar leaves $h$ unchanged.

The central axis of a wrench makes the meaning visible. DefineThe vector triple-product identity givesEvery point on the line ${\mathbf{r}}_0+s\hat{\mathbf{F}}$ has that same parallel residual moment. This line also minimizes the moment magnitude over all choices of origin: the perpendicular part can be removed, while the invariant parallel part cannot. The minimum is $|h|\Vert \mathbf{F}\Vert$.

For an original example, let $\mathbf{F}=(0,0,20)\mathrm{N}$ and ${\mathbf{\tau }}_O=(4,-2,6)\mathrm{N}\mathrm{m}$. Then $h=0.30\mathrm{m}$ and ${\mathbf{r}}_0=(0.10,0.20,0)\mathrm{m}$. Translating the moment origin to that point leaves $(0,0,6)\mathrm{N}\mathrm{m}$: a $20\mathrm{N}$ axial force together with a $6\mathrm{N}\mathrm{m}$ axial force couple. No relocation of a single force can eliminate this remaining couple.

If $h=0$ with $\mathbf{F}\ne 0$, the central-axis moment vanishes and one force on that line reproduces the external load system. If $\mathbf{F}=0$ and the moment is nonzero, the load is a pure force couple; the formula for $h$ is undefined. An infinite-pitch description is a limiting screw-theory convention, not a finite value supplied by this quotient. Both zero force and zero moment give the zero wrench and no distinguished axis.

A proposed load-sensor check combines a known force on an offset line with a known parallel couple. Recompute moments about several origins and compare the resulting $h$ values. Errors in origin translation can corrupt the moment components while the exact invariant remains fixed; near zero resultant force, the division by $\Vert \mathbf{F}{\Vert }^2$ makes pitch sensitive to measurement error. This geometric load parameter is not automatically a physical screw’s thread spacing.