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Since $A{A}^T=I$ and $\det A=1$, $A$ is an orthogonal matrix in the special orthogonal group. It also satisfies $A^{T}A=I$ and thus preserves the dot product. Normalize the fixed eigenvector to $e_3=v/\Vert v\Vert$. For $u\cdot e_3=0$,so the orthogonal complement of the fixed axis is invariant. Choose an oriented orthonormal basis $e_1,e_2,e_3$. In this basis the matrix has form $\mathrm{diag}(B,1)$, where $B$ is a real $2\times 2$ orthogonal matrix of determinant one.

Write the first column of $B$ as $(\cos \theta ,\sin \theta {)}^T$. Its unit second column must be perpendicular to the first; positivity of the determinant selects $(-\sin \theta ,\cos \theta {)}^T$. Thus $B=R(\theta )$, and $A$ fixes the axis through $v$ while rotating its perpendicular plane by $\theta$. Since trace is invariant under change of basis,This proves that $A$ is the required axial rotation. The trace determines the unsigned angle; the action in the oriented perpendicular plane determines its sign. The identity case has angle zero and any axis.

For the stated angle, $q=\cos (2\pi /3)=-1/2$, so $\Vert u+v\Vert =\sqrt{2+2q}=1$. The minimizing unit vector is therefore $w=-(u+v)$, andIndeed all three pairwise dot products equal $-1/2$, so the vectors lie in one plane, separated by $120$ degrees. This is also the global minimum pairwise dot-product sum of three unit vectors: for any three unit vectors,so $S\ge -3/2$, with equality exactly when $u+v+w=0$.

Two nonzero vectors are linearly independent when $ax+by=0$ implies $a=b=0$. Their span then has dimension two. If they are linearly dependent, one is a scalar multiple of the other, so their span is a nonzero line and has dimension one. Thus the requested dimensions are two and one, respectively.

The dot product and corresponding Euclidean norm areThe Cauchy-Schwarz inequality states $\boxed{|x\cdot y|\le \Vert x\Vert \Vert y\Vert }$. If $y=0$ this is immediate. Otherwise the nonnegativity of a squared norm, with $t=(x\cdot y)/\Vert y{\Vert }^2$, givesMultiplying by $\Vert y{\Vert }^2$ and taking square roots proves the inequality. Equality holds precisely when the vectors are linearly dependent. Expanding another squared norm and using the inequality givesand hence the triangle inequality $\boxed{\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert }$.

In ${\mathbb{R}}^3$, the two vectors lie in a common plane. If their directions make angle $\alpha$, the law of cosines gives $\Vert x-y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2-2\Vert x\Vert \Vert y\Vert \cos \alpha$. Comparing with the dot-product expansion yieldsThe sketch shows the same relation as a projection: multiply the length of $x$ by the signed component of $y$ in its direction. That component is negative for obtuse $\alpha$.

A unit vector is a vector of Euclidean norm one, equivalently $u\cdot u=1$. In particular the dot product of two unit vectors is the cosine of their mutual angle.

Use suffix notation, summing repeated indices. The contraction of two Levi-Civita symbols givesAdding the directional derivative $(F\cdot \nabla )G_i=F_j{\partial }_j{G}_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j{\partial }_i{F}_j$. The total right-hand side is thereforethe $i$th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

Move the origin to the centroid and write the four vertex vectors as $r_1,r_2,r_3,r_4$. Their sum is zero. The midpoints of the first opposite-edge pair have vectors $(r_1+r_2)/2$ and $(r_3+r_4)/2=-(r_1+r_2)/2$, so their distances from the centroid are equal. The same argument applies to the other two opposite-edge pairs.

Choose the three midpoint distances as $u=|r_1+r_2|/2$, $v=|r_1+r_3|/2$, $w=|r_1+r_4|/2$. Expanding the dot products,Identifying these four norms with the vertex distances proves the tetrahedron centroid-to-midpoint sum of squares:The proof uses no orthogonality or regularity assumption on the tetrahedron.

Write the row components as $E_{ai}=(e_a{)}_i$. Expanding the cross product in components,equals the alternating six-term expansion of the determinant of $E$. Therefore $(e_1\times e_2)\cdot e_3=\det E$.

Geometrically, $|e_1\times e_2|$ is the area of the parallelogram spanned by $e_1,e_2$, and its direction is perpendicular to that plane. A nonzero scalar triple product means both that this area is nonzero and that $e_3$ has a nonzero component normal to the plane. Thus the vectors are linearly independent and form a basis of ${\mathbb{R}}^3$, without needing to be orthogonal. Equivalently, their parallelepiped has nonzero volume.

The entries of $E{\hat{E}}^T$ are $e_a\cdot {\hat{e}}_b$, so the required duality is exactly $E{\hat{E}}^T=I$. Since $E$ is invertible, this has the unique solutionIts determinant is $1/\det E\ne 0$, so its rows are again a basis. Put $\Delta =(e_1\times e_2)\cdot e_3$. The cross product $e_2\times e_3$ is perpendicular to $e_2,e_3$, and its dot product with $e_1$ is $\Delta$. Consequently the reciprocal basis isThese vectors have precisely the required dot products, so uniqueness identifies them with the rows of $E^{-T}$.

For the final request, an orthonormal basis of rows gives $E{E}^T=I$. If instead the reciprocal rows are orthonormal, then $\hat{E}{\hat{E}}^T=I$ and substitution of $\hat{E}=E^{-T}$ gives $E^{-T}{E}^{-1}=I$, again implying $E{E}^T=I$. Thus in either case $E$ is an orthogonal matrix and $\hat{E}=E$.

The further printed claim that $E$ must be a rotation matrix is false without an orientation assumption. The matrix $E=\mathrm{diag}(1,1,-1)$ has orthonormal rows and is its own reciprocal basis matrix, but $\det E=-1$: it is a reflection. By the orientation of an orthonormal reciprocal basis, the corrected conclusion isA right-handed original ordered basis supplies exactly this missing condition.

Orient the triangular boundary from $P=(1,0,0)$ to $Q=(0,1,0)$ to $R=(0,0,1)$ and back to $P$. This orientation corresponds to the positive normal vector $(Q-P)\times (R-P)=(1,1,1)$. The cyclic coordinate rotation $(x,y,z)\mapsto (z,x,y)$ sends the three directed edges to one another and transforms the vector field by the same rotation. Since rotations preserve dot products, the three edge contributions to the line integral are equal.

On the first edge use $x=1-t$, $y=t$, $z=0$, $0\le t\le 1$. ThenIts integral is $2/3$, so the required circulation is

For a sufficiently smooth vector field on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the Stokes theorem statesTherefore the requested surface integral, with all normal components positive, is also $2$.

To verify it directly, parametrize the triangle by $\mathbf{r}(y,z)=(1-y-z,y,z)$ on $y\ge 0$, $z\ge 0$, $y+z\le 1$. Its two tangent vectors are ${\mathbf{r}}_y=(-1,1,0)$ and ${\mathbf{r}}_z=(-1,0,1)$, giving the vector area elementDifferentiation of the vector field gives $\nabla \times A=2(y+z,z+x,x+y)$. On the triangle $x+y+z=1$, so its dot product with $(1,1,1)$ is $4$. Consequentlyconfirming both the magnitude and the orientation sign.

About a fixed origin, the total momentum and angular momentum areA time-independent translation ${\mathbf{x}}_i^{\prime }={\mathbf{x}}_i-\mathbf{b}$ leaves every velocity unchanged. ThereforeThe cross product $\mathbf{b}\times \mathbf{P}$ is perpendicular to $\mathbf{P}$, giving the invariantThe other requested transformation follows from the vector triple-product identity:The actual PDF asks about $\mathbf{L}\times \mathbf{P}$; the TeX's occurrence of $\mathbf{F}$ is a transcription error.

If $\mathbf{P}=0$, the translation term vanishes and angular momentum is independent of origin. If $\mathbf{P}\ne 0$, chooseThen $\mathbf{b}\times \mathbf{P}=\mathbf{L}-(\mathbf{L}\cdot \mathbf{P})\mathbf{P}/|\mathbf{P}{|}^2$, and soThus the new angular momentum is parallel to the total momentum. The freedom along $\mathbf{P}$ gives the central axis of a momentum system. If the invariant dot product is zero, the new angular momentum is the zero vector. All expressions assume a translated fixed origin, not a moving reference frame.

For a linear code $X\le {\mathbb{F}}^n$, its dual code isThe standard bilinear form is nondegenerate. The map ${\mathbb{F}}^n\to X^{\ast }$ obtained by restricting the dot-product functional to $X$ is surjective: extend a functional from a basis of $X$ to a basis of ${\mathbb{F}}^n$, then use the nondegenerate dot product to represent it. Its kernel is $X^{\perp }$. The rank-nullity theorem therefore gives

The first proposed description of the dual code is false without self-orthogonality. At length three take $X=⟨(1,0,0)⟩$. It has the specified dimension, butwhereas $⟨X,(1,1,1)⟩$ contains $(1,0,0)$ and $(1,1,1)$, neither of which is in $X^{\perp }$. The natural corrected statement is the odd-length binary self-orthogonal dual extension: if also $X\subseteq X^{\perp }$, every word of $X$ has even Hamming weight, the all-ones word lies in $X^{\perp }$ but not in $X$, and dimensions then give $X^{\perp }=X\oplus ⟨1⟩$.

For a binary self-dual code, $x\cdot x=0$ for every $x\in X$. In ${\mathbb{F}}_2$, this equals ${\sum }_i{x}_i$, so every codeword has even Hamming weight. Thus $1\cdot x=0$ for all $x$, giving $1\in X^{\perp }=X$. The dimension identity yields $2\dim X=n$, so $n$ is even. Alternatively $1\in X$ and $1\cdot 1=0$ itself gives evenness. Hence the second assertion is true.

For every $n=2m$, useThis linear code has dimension $m$, and $(u,u)\cdot (v,v)=2(u\cdot v)=0$. Therefore $X\subseteq X^{\perp }$; both have dimension $m$, so equality holds. This constructs a binary self-dual code at every even length. Conversely the dimension identity already proves every self-dual length-$n$ linear code has dimension $n/2$.

For a nonbinary example, over ${\mathbb{F}}_5$ take $X=⟨(1,2)⟩$. Its generator has self-product $1+2^2=0$ in ${\mathbb{F}}_5$, so $X\subseteq X^{\perp }$. Both are one-dimensional, proving

The smallest admissible axial stretch is ${\lambda }_{\min }=\sin 2\theta$, where the two branches merge and ${\alpha }^2=\tan \theta =s/c$. The current unit fibre vectors are $m_{\pm }=U{l}_{\pm }$. Their dot product isAt the minimum, both terms in parentheses equal $cs$, soHence the two families are orthogonal at the maximum possible contraction. There is also a geometric proof of the bound: their reference parallelogram area is $\sin 2\theta$, and its current area is $\sin 2\theta /\lambda$, because the in-plane area stretch is $1/\lambda$. Two unit vectors span area at most one, with equality precisely when perpendicular. This gives $\lambda \ge \sin 2\theta$ and the same equality condition without solving the quadratic. At $\theta =\pi /4$ the minimum is one, so no axial contraction is possible and the fibres are already orthogonal.

Write $A={\sum }_i{a}_i^2$, $B={\sum }_i{b}_i^2$, $C={\sum }_i{a}_i{b}_i$. If $B=0$, every $b_i=0$ and the claimed inequality is immediate. Otherwise the sum of squaresholds for every real $t$. Choosing $t=C/B$ gives $0\le A-C^2/B$, and hence the Cauchy-Schwarz inequality in its stronger absolute-value form:In particular $C\le \sqrt{A}\sqrt{B}$, as requested. For unit vectors, $A=B=1$, so this same argument givesThe absolute value is essential for opposite-pointing vectors; it follows from the squared inequality without a sign assumption on their dot product.

Yes. Suppose ${\alpha }_1{a}_1+{\alpha }_2{a}_2+{\alpha }_3{a}_3=0$. Taking its dot product with $a_2\times a_3$ eliminates the last two terms and givesThe nonzero scalar triple product forces ${\alpha }_1=0$. Cyclically using $a_3\times a_1$ and $a_1\times a_2$ forces ${\alpha }_2={\alpha }_3=0$ as well. This is the definition of linear independence. Equivalently the three-column matrix has nonzero determinant.

Represent a subset by the coordinate vector of its indicator function in ${\mathbb{F}}_2^n$. Symmetric difference then becomes coordinatewise addition, and the proposed pairing is the dot productIt is a symmetric bilinear form, and it is nondegenerate because its Gram matrix in the singleton basis is the identity. Coordinate permutations preserve the sum, so the natural $S_n$ action preserves the bilinear form.

Let $j=(1,\dots ,1)$, the vector representing the full set. Thenis the even-weight subspace. For even $n$, $B(j,j)=n=0$ in ${\mathbb{F}}_2$, so $j\in W$. Moreover $B(x,x)={\sum }_i{x}_i^2={\sum }_i{x}_i=0$ for $x\in W$, making the restricted bilinear form alternating. Nondegeneracy on the ambient space implies $W^{\perp }=⟨j⟩$, so the radical of $B{|}_W$ is exactlyThus it descends to an alternating bilinear form on $W/⟨j⟩$, independent of the representatives, and that quotient form has zero radical. The induced form is nondegenerate, and its dimension is $n-2$. This is the symplectic quotient of the binary subset module; at $n=2$ it is the zero-dimensional nondegenerate space.

Use vectors in three-dimensional Euclidean space, with the usual dot product and cross product. The following explicit assignments compute the circumcenter of a tetrahedron:General position means $d\ne 0$. The reciprocal cross-product identities give $x\cdot u=|u{|}^2/2$, $x\cdot v=|v{|}^2/2$ and $x\cdot w=|w{|}^2/2$. Consequentlyand likewise for $C,D$. Thus $O$ is the unique point equidistant from the four vertices. It can lie outside the tetrahedron; no inside constraint should be imposed.

For $\mathbf{b}\ne 0$, define the orthogonal projection of $\mathbf{a}$ onto its span byThe first is parallel or antiparallel to $\mathbf{b}$, or zero, and ${\mathbf{a}}_{\perp }\cdot \mathbf{b}=0$. This proves the decomposition. If $\mathbf{a}\ne 0$ as well, comparison with ${\mathbf{a}}_{\parallel }=|\mathbf{a}|\cos \theta \hat{\mathbf{b}}$ yieldsIndeed $|\mathbf{a}{|}^2=|{\mathbf{a}}_{\parallel }{|}^2+|{\mathbf{a}}_{\perp }{|}^2$ shows that the ratio lies in $[-1,1]$. There is a unique $\theta \in [0,\pi ]$ with this cosine. The definition is symmetric in the two vectors, unchanged by orthogonal coordinate changes and positive rescaling, and agrees with ordinary plane geometry in their span: projecting a unit vector onto the other gives its cosine. Parallel, perpendicular and antiparallel directions give $0,\pi /2,\pi$ respectively. The zero vector has no defined angle; if $\mathbf{b}=0$, the nonzero-axis projection formula itself must not be used.

The vertices of the centered hypercube are sign vectors $\mathbf{u}\in \{-1,1{\}}^n$. A body diagonal identifies $\mathbf{u}$ and $-\mathbf{u}$, hence there are $2^{n-1}$ such diagonals. If two sign vectors disagree in $h$ positions, their dot product is $n-h-h=n-2h$. For odd $n$ this integer is odd and cannot be zero. Thus no two body diagonals are perpendicular in odd dimension, by the orthogonality of hypercube body diagonals.

In four dimensions, mutually orthogonal nonzero vectors are linearly independent, so at most four mutually perpendicular diagonals are possible. Four are achieved by the directionsEach has squared norm four and every distinct pair has zero dot product; these are the rows of a Hadamard matrix. Therefore the maximum is four.

For distinct diagonals, the Hamming distance is $h=1,2,3$; $h=0,4$ would represent the same line. Oriented vertex directions therefore give cosines $1/2,0,-1/2$. The angles areFor unoriented diagonals, the usual smaller angle between the lines is $\pi /3$ or $\pi /2$; the supplementary intersection angle $2\pi /3$ also occurs when the two rays are chosen oppositely. Stating this convention avoids counting a line and its reverse as different diagonals.

Let $e=\frac{1}{2}(|A{|}^2+|B{|}^2)$. The time equations and the dot product differentiation rule give the local identityThe divergence of a cross product isCombining them yields ${\partial }_{t}e+\nabla \cdot (A\times B)=0$. On a fixed bounded region with a boundary regular enough for the divergence theorem, differentiating under the integral and applying that theorem givesHere $n$ is the outward unit normal, so the sign expresses loss of the interior quadratic quantity by outward flux. No vanishing-boundary assumption is required.