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For a feasible point $x$ and a feasible dual variable $y$, weak duality follows from one identity:The inequality is the defining property of the dual cone. Every feasible pair therefore encloses the optimal objective function value:Both optima are finite when such a pair exists. Consequently,These are absolute cost bounds, rather than bounds on distance to an optimizing vector. Zero gap certifies both optima immediately. A small positive gap bounds the remaining improvement even if no strong duality theorem has been invoked.

For the nonnegative-orthant optimality certificate example, $x=(2.9,0.2{)}^T$ is feasible and costs $3.3$. Pairing it with $y=(0,1,1{)}^T$ gives $g=0.3$, so its cost exceeds the optimum by at most $0.3$. Here the optimum is known to be $3$, and the bound is exact. Pairing the same $x$ with the weaker $r=1/2$ certificate gives $g=1.8$: a valid but less informative guarantee.

The identity assumes exact primal and dual feasibility. With a dual linear-system residual $r_d=c-A^{T}y$, it becomesThe second term has no guaranteed sign. Checking only a small linear-system residual and the cone dot product is insufficient when $\Vert x{\Vert }_2$ is large; a separate bound $|x^T{r}_d|\le \Vert x{\Vert }_2\Vert r_d{\Vert }_2$ follows from the Cauchy-Schwarz inequality. Approximate cone membership likewise needs a quantified correction before claiming a certificate.

For an exact arithmetic example, minimize $x$ subject to $x\ge 10^9$. The point $x=10^9$ is optimal. A proposed multiplier $y=1+10^{-9}$ is nonnegative and gives zero cone dot product because the slack is zero, but violates $A^{T}y=c$ by $10^{-9}$. Its purported lower bound is $10^9+1$, exceeding the optimum. The missing term $x^T{r}_d=-1$ explains the failure.

For the cubic polynomial, an oriented, normalized tangent vector isIts Euclidean norm is one and $DF(z_k)t_k=0$. The sign here follows increasing $x$; for a general solution branch, choose the sign of each new tangent vector so its dot product with the previous one is positive. First predict $z_{\mathrm{p}}=z_k+h{t}_k$, then apply Newton method starting at $z_{\mathrm{p}}$ toThe second equation keeps the correction on the transverse line in this two-dimensional example; in higher dimensions it is a hyperplane. Solving the linear system $DH(z^{(j)})\delta z=-H(z^{(j)})$ and updating $z^{(j+1)}=z^{(j)}+\delta z$ corrects both coordinates together.

This step gives $z_{\mathrm{p}}\approx (0.617556,0.424641)$ and $z_{k+1}\approx (0.635532,0.378840)$. Check both equations: the solution must satisfy $x^3-x+\lambda =0$ and $t_k^{\mathsf{T}}(z-z_{\mathrm{p}})=0$. Checking only the first could accept any point on the solution branch. The new $x$ exceeds $1/\sqrt{3}$, so the step crosses the continuation fold, even though the predictor's $\lambda$ exceeds $2/(3\sqrt{3})$.

At the seed, set $a=3x_k^2-1$. The determinant of the correction Jacobian matrix with the exact unit vector above isAt either continuation fold, $a=0$, $t_k=(1,0)$ and $DH(z_k)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: the correction remains invertible. More generally, if $DF$ has full rank and a one-dimensional kernel spanned by $t_k$, adding $t_k^{\mathsf{T}}$ gives an invertible matrix, since a vector annihilated by both rows must be zero. This is a local argument; the old transverse hyperplane need not work arbitrarily far along the curve. Reduce $h$ and retry when correction fails, and check the equations and orientation after successful steps. Scaling state and parameter coordinates also matters, because it changes the Euclidean norm used to normalize the tangent vector.

// Target: analysis.bigb

The convex conjugate records a function using slopes rather than input positions. Let $f:{\mathbb{R}}^n\to \mathbb{R}\cup \{+\infty \}$ be a proper extended-real function: it is finite somewhere and never takes $-\infty$. Its convex conjugate iswhere $p\cdot x$ is the dot product and $\mathrm{dom}f=\{x:f(x)<+\infty \}$ is the effective domain. An infinite value of $f$ excludes that input from the supremum. The value $f^{\ast }(p)$ can be $+\infty$, but cannot be $-\infty$ because an input with finite $f(x)$ supplies a finite lower bound.

The convex conjugate is automatically a convex function, even when $f$ is not convex: each fixed input gives an affine function $p\mapsto p\cdot x-f(x)$, and their supremum is convex. It is also lower semicontinuous. It need not be proper: for $f(x)=-x^2$ on $\mathbb{R}$, $px-f(x)=px+x^2$ is unbounded above for every $p$. The definition and the automatic convexity are discussed in Boyd and Vandenberghe, Convex Optimization (2004), §3.3.1.

The best lower bound with a prescribed slope. Whenever $f^{\ast }(p)$ is finite, the affine functionis an affine minorant of $f$. Indeed, an affine function $p\cdot x+b$ lies below $f$ everywhere exactly when $b\le {\inf }_x(f(x)-p\cdot x)=-f^{\ast }(p)$. Thus $-f^{\ast }(p)$ is the largest possible intercept at slope $p$. The convex conjugate trades positions for slopes: $x$ specifies where to inspect $f$, whereas $p$ specifies which global affine minorant to use. Taking the supremum over all these affine minorants produces the biconjugate; the Fenchel-Moreau theorem says this recovers a proper convex function when it is lower semicontinuous.

A quadratic, computed by completing the square. For the convex quadratic function $f(x)=x^2/2$,Consequently $f^{\ast }(p)=p^2/2$, and the unique maximizing input is $x=p$. The affine minorant ${\ell }_p(x)=px-p^2/2$ is the tangent line at $(p,p^2/2)$: its slope is $p$ and its intercept is $-f^{\ast }(p)=-p^2/2$. For example, $p=2$ gives $f^{\ast }(2)=2$ and the tangent line $2x-2$, touching at $(2,2)$. This is the same curve described either by its point values or by the intercepts of its tangent lines.

A corner turns into a constraint on slopes. For the absolute value $f(x)=|x|$, if $|p|\le 1$ then $px-|x|\le 0$, with value zero at $x=0$. If $p>1$, taking $x\to +\infty$ makes $(p-1)x\to +\infty$; if $p<-1$, taking $x\to -\infty$ does the same. Therefore its convex conjugate is zero on $[-1,1]$ and $+\infty$ outside: it is the indicator functional of a constraint set for this interval. For $|p|<1$ the unique maximizing input is zero; for $p=1$ every $x\ge 0$ maximizes, and for $p=-1$ every $x\le 0$ maximizes. The corner's many supporting slopes are exactly the subdifferential $\partial f(0)=[-1,1]$.

A supremum need not be attained. For a differentiable convex function finite on all of ${\mathbb{R}}^n$, an input maximizes $p\cdot x-f(x)$ exactly when $p=\nabla f(x)$. If $f$ is strictly convex, there is at most one such input; existence is a separate issue. For the exponential function $f(x)=e^x$, the derivative is positive. At $p>0$, solving $p=e^x$ gives $x=\log p$ and $f^{\ast }(p)=p\log p-p$. At $p=0$, the supremum of $-e^x$ is zero as $x\to -\infty$, with no maximizing input. At $p<0$, $px-e^x\to +\infty$ as $x\to -\infty$. Thus a strictly convex differentiable function can have both unattained finite conjugate values and infinite conjugate values.

A coordinate of a vector in a chosen basis is one entry of its coordinate vector. The corresponding coordinate functional in the dual basis extracts it. For an orthonormal basis, it can instead be read as an inner product with the basis vector, as in orthonormal eigenbasis. For a nonorthogonal basis, ordinary dot products need not give the coordinates.

The instantaneous power of the Coriolis force, evaluated with the velocity in the same rotating reference frame, vanishes:This follows from the perpendicularity of the cross product and the dot product definition of power. The Coriolis force alone does no work and cannot change the rotating-frame kinetic energy $m|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2$. It can change the direction of the relative velocity. Actual forces and the centrifugal force can still change that kinetic energy, as the free particle in rotating coordinates illustrates. Conservation of an appropriate total energy has the additional conditions stated under energy balance in a uniformly rotating frame.

For a source-free plane electromagnetic wave in vacuum, let $n^{\mu }=(1,\mathbf{s})$ be its null propagation Lorentz four-vector. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$, and define the dual electromagnetic field tensor by ${\tilde{F}}^{0i}=-B_i$, ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. Lower both indices with the Minkowski metric. Direct contraction givesBoth contractions vanish exactly when the transverse plane electromagnetic wave Maxwell equations hold. Being zero covectors, they remain zero in every inertial frame. Thus the transformed spatial wavevector is perpendicular to both transformed fields, even though the separate three-dimensional dot products are not themselves Lorentz scalars for arbitrary fields.

One contraction alone is insufficient without the other Maxwell equations. In particular, adding a magnetic field parallel to $\mathbf{s}$ leaves $n^{\mu }{F}_{\mu \nu }$ unchanged but violates $\mathbf{s}\cdot \mathbf{B}=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$ between field strengths and directions.

The vector space ${\mathbb{R}}^3$ with the cross product as Lie bracket, $[x,y]=x\times y$. Bilinearity and $x\times x=0$ give an Alternating Lie bracket; the vector triple product identity proves the Jacobi identity because the three expressions $x\times (y\times z)=y(x\cdot z)-z(x\cdot y)$ cancel cyclically, using the symmetric dot product.

Its relation to the SO(3) Lie algebra is explicit. For a vector $x=(x_1,x_2,x_3)$, define the skew-symmetric matrixThen $[\hat{x},\hat{y}]=\widehat{x\times y}$: acting on $v$, the left side is $x\times (y\times v)-y\times (x\times v)=(x\times y)\times v$. Every real $3\times 3$ skew-symmetric matrix occurs uniquely, so this hat map is a Lie algebra isomorphism. With the standard basis vectors $e_1,e_2,e_3$,and their matrix commutators are $[X,Y]=Z$, $[Y,Z]=X$, $[Z,X]=Y$. With active rotation matrices acting on column vectors, $\exp (t\hat{x})$ rotates through angle $t\Vert x\Vert$ about the oriented axis $x$ when $x\ne 0$. In particular $\exp (tZ)e_1=(\cos t,\sin t,0)$, fixing the positive-angle convention used in the small rotation commutator.

The map $x\mapsto -(i/2)x\cdot \sigma$ is also a Lie algebra isomorphism to the compact real SU(2) Lie algebra, using the Pauli matrix commutator identity. These two identifications concern Lie algebras; they do not identify the corresponding Lie groups globally.

For any real three-dimensional vectors $a,b$, the displayed linear equation has the unique solutionTaking a dot product with $b$ first yields $(1+|b{|}^2)(b\cdot x)=a\cdot b$. Taking a cross product with $b$ and using the vector triple product identity givesSubstitution into the original equation cancels the longitudinal terms and proves the formula. Conversely direct substitution verifies it for all $b$, including $b=0$, since the denominator is positive. If $C_{b}x=b\times x$, the underlying identity for linear maps is $(I-C_b)(I+C_b+b{b}^T)=(1+|b{|}^2)I$, using $C_b^2=b{b}^T-|b{|}^{2}I$.

The current density $\mathbf{J}(\mathbf{r},t)$ describes the amount and direction of local electric charge transport per area per time. Its SI unit is $A/{\mathrm{m}}^2$. Through a small area element with unit normal $\mathbf{n}$, the contribution to electric current is $dI=\mathbf{J}\cdot \mathbf{n}dA$: the dot product selects transport through the surface. A current density tangent to the surface contributes zero. If current density is uniform and perpendicular to a wire's transverse slice, its magnitude $j=|\mathbf{J}|$ satisfiesHere $A$ is the wire cross-sectional area. For nonuniform current density, $I/A$ is only the average normal component, and the surface integral is needed. ^[ref].

Consider two separate uniform-flow cases, each carrying electric current $I=2\mathrm{A}$. A wire cross-sectional area of $1\mathrm{m}{\mathrm{m}}^2=10^{-6}{\mathrm{m}}^2$ gives $j=2\times 10^{6}A/{\mathrm{m}}^2$; doubling the wire cross-sectional area to $2\mathrm{m}{\mathrm{m}}^2$ gives $j=10^{6}A/{\mathrm{m}}^2$. Equal electric current can therefore mean different local transport intensity. This distinction matters for Joule heating, whose local power per volume depends on current density and electric field.

For a toy Electron transport model, stipulate mobile electron number density $n=8\times 10^{28}{\mathrm{m}}^{-3}$ and drift velocity ${\mathbf{v}}_d=-10^{-4}m/s\hat{\mathbf{x}}$, with the positive unit normal $\hat{\mathbf{x}}$. Each Electron has electric charge $-e$, where the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$ is exact in SI. These are defined model inputs, rather than measured parameters of a particular material. ^[ref] The current density and its magnitude areFor wire cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$ and constant transport over time $10\mathrm{s}$, this givesThe two minus signs make conventional current point opposite to the Electron drift velocity. A small drift velocity can carry an appreciable electric current because many charge carriers contribute; it does not specify the propagation speed of an electrical disturbance.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

If $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable at the point $\gamma (t)$ of a differentiable parametric curve, the multivariable chain rule givesThe gradient supplies the local spatial rates; the velocity vector ${\gamma }^{\prime }(t)$ supplies the direction and rate of movement. Their dot product is the change in the sampled scalar field per unit parameter. If the parameter is time, this is a rate per unit time; a directional derivative per unit distance instead uses a unit vector.

Take $F(x,y)=x^2+2y^2$ and the unit circle path $\gamma (t)=(\cos t,\sin t)$. Along the path,At $t=\pi /4$, the gradient is $(\sqrt{2},2\sqrt{2})$, the velocity vector is $(-1/\sqrt{2},1/\sqrt{2})$, and their dot product is $-1+2=1$. The sampled value is $3/2$ and increases at rate one, even though one coordinate is decreasing. This is why a moving observation must combine every coordinate's contribution.

For a level curve, $F(\gamma (t))$ is constant, so the derivative along a curve is zero and its velocity vector is orthogonal to the gradient. Along other paths, the same formula combines a field's geometry with the observer's motion. ^[ref].

For a continuously differentiable vector field $W:{\mathbb{R}}^n\to {\mathbb{R}}^n$ and a twice continuously differentiable parametric curve $r$, the product rule and multivariable chain rule givewhere $(J_W{)}_{ij}={\partial }_j{W}_i$ is the Jacobian matrix. Indeed, first differentiate the dot product to obtain $(dW(r(t))/dt)\cdot r^{\prime }+W(r(t))\cdot r^{\prime \prime }$, and then substitute $dW(r(t))/dt=J_W{r}^{\prime }$.

The first term describes variation in the sampled vector field; the second records the changing velocity vector of the parametric curve. In three dimensions one may take $W=\nabla \times F$ for a sufficiently smooth vector field $F$. The formula then involves first and second spatial partial derivatives of $F$ and the velocity and acceleration of the parametric curve. Omitting the second term would be valid only when its dot product vanishes, for example for a straight path with constant velocity.

In the same uniform stationary sound model, let $\tau$ be emission time, $t$ reception time, and ${\mathbf{x}}_s(\tau )$ and ${\mathbf{x}}_o(t)$ source and receiver positions. The direct-path propagation condition and its direction areThe unit vector $\mathbf{n}$ points from the source's old emission position to the receiver's reception position. With velocities ${\mathbf{v}}_s=d{\mathbf{x}}_s/d\tau$ and ${\mathbf{v}}_o=d{\mathbf{x}}_o/dt$, taking a derivative of the propagation condition givesHere the dot products project onto the actual propagation direction. The last equality counts source wave phase cycles at reception for constant emitted frequency $f_s$. It assumes both speeds are below $c$, a unique direct arrival and smooth motion; interpreting a measured short-window frequency spectrum additionally requires little frequency change within that window. The source velocity must be evaluated at emission, not at reception. This delayed geometry matters when the direction changes.

For an original pass-by model, a source follows ${\mathbf{x}}_s(\tau )=(v\tau ,0)$ and a stationary receiver is at $(0,b)$ with $b>0$. The source passes its closest position at emission time $\tau =0$. ThenThese parametric equations give received frequency against reception time. For $\tau <0$ the shift is positive, for $\tau >0$ negative, and the unshifted emission at $\tau =0$ reaches the receiver at $t=b/c$. Thus zero shift occurs after the source's geometric closest passage in the material's reference frame. Increasing $b$ spreads the transition over a longer interval even though $v$ is unchanged. Far before and after passage, the frequency approaches $f_{s}c/(c-v)$ and $f_{s}c/(c+v)$ respectively.

A dot product measures how much two Euclidean vectors point along each other. This article uses real Euclidean vectors in ${\mathbb{R}}^n$, with the usual Cartesian basis: multiply matching coordinates and add,The result is a scalar, while $\Vert \mathbf{a}\Vert$ is the Euclidean norm. This real convention has no complex conjugation; for vectors over the complex numbers, use the convention specified in inner product.

For two nonzero Euclidean vectors, the same dot product records their angle $0\le \theta \le \pi$ through the cosine:Here $\arccos$ is an inverse trigonometric function. The Cauchy-Schwarz inequality guarantees that the normalized ratio lies in $[-1,1]$. The zero vector has dot product zero with every vector, but its angle is undefined: the ratio would divide by zero. To outline the geometric identity, expand $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}$ and compare with the law of cosines.^[ref]

Take $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$. Their Euclidean norms are $5$ and $2$, andThe unit vector $\mathbf{u}=\mathbf{c}/\Vert \mathbf{c}\Vert =(1,0)$ selects the horizontal direction. The scalar projection $\mathbf{a}\cdot \mathbf{u}=3$ says how far $\mathbf{a}$ reaches along that direction; its vector projection is $3\mathbf{u}=(3,0)$. Thus $\mathbf{a}=(3,0)+(0,4)$ splits into an aligned part and an orthogonal projection residual.

For $\mathbf{b}=(4,-3)$, the dot product $\mathbf{a}\cdot \mathbf{b}=12-12=0$ proves orthogonality; both Euclidean norms are $5$. For $\mathbf{d}=(-2,0)$, $\mathbf{a}\cdot \mathbf{d}=-6$: the scalar projection onto $\mathbf{d}/\Vert \mathbf{d}\Vert =(-1,0)$ is $-3$, because $\mathbf{a}$ reaches against that chosen direction. Its angle with $\mathbf{d}$ is $126.8699^{\circ }$.

Keeping $\mathbf{a}=(3,4)$ fixed and choosing $\mathbf{b}$ of Euclidean norm $2$ gives $\mathbf{a}\cdot \mathbf{b}=10\cos \theta$. At angles $0^{\circ },60^{\circ },90^{\circ },120^{\circ },180^{\circ }$, the dot products are $10,5,0,-5,-10$. Positive values mean partial alignment, zero means orthogonality, and negative values mean a component against the other direction.

This interpretation explains work: for a constant force $\mathbf{F}$ and a straight displacement $\mathbf{s}$, $W=\mathbf{F}\cdot \mathbf{s}$ counts the force component along the motion. A sideways force contributes zero work, and an opposing force contributes negative work.^[ref]