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Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

Expand the first component using the definition of the cross product:The added $a_1{b}_1{c}_1$ terms cancel in the last step. Cyclically permuting the component indices gives the other two components, proving the vector triple product identity

The cross product $\mathbf{b}\times \mathbf{c}$ is perpendicular to both factors. ThusCyclic permutations give all the off-diagonal dot products between the original basis and its reciprocal basis.

The vector triple product identity givesEliminate $V$ with iii to obtain the reciprocal basis involution:This formula actually holds without the unit-length assumptions. Under those assumptions it also says $|{\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime }|=1/|V|$; normalizing this cross product alone would lose the sign of $V$ and could return $-\mathbf{a}$.

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.

A normal vector to the plane spanned by $\mathbf{a},\mathbf{b}$ is their cross product:Dividing by this Euclidean norm gives the two possible choices of unit normal,

The two edge vectors based at $A$ are $\mathbf{b}-\mathbf{a}=(-3,-1,3)$ and $\mathbf{c}-\mathbf{a}=(-3,2,2)$. Their cross product is $(-8,-3,-9)$, with squared Euclidean norm $64+9+81=154$. The triangle occupies half the corresponding parallelogram, so

Two directions in the plane are $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})$ is a normal vector. Moreover,so $\mathbf{N}$ cannot vanish. The required unit normal and offset areTaking dot products with $\mathbf{a},\mathbf{b},\mathbf{c}$ gives the same offset for all three points.

Take the dot product of the equation with $a$. Since $a\cdot (a\times r)=0$, we obtain $a\cdot r=(a\cdot c)/\lambda$. Next take its cross product with $a$ and apply the vector triple product identity:Substitute $a\times r=c-\lambda r$ and the known dot product, then rearrange:Here the vectors and $\lambda \ne 0$ are real, so the denominator is positive. The formula also covers $a=0$, giving $r=c/\lambda$. To check existence as well as uniqueness, write its numerator as $N$. The vector triple product identity gives $a\times N=\lambda a\times c-a(a\cdot c)+|a{|}^{2}c$, so $a\times N+\lambda N=({\lambda }^2+|a{|}^2)c$. Substitution verifies the original equation. This is an instance of the inverse of a scalar plus a cross-product operator.

The cross product $\mathbf{b}\times \mathbf{c}$ is orthogonal to both factors. Division by the nonzero scalar $D$ preserves that perpendicularity, giving $\boxed{\mathbf{A}\cdot \mathbf{b}=\mathbf{A}\cdot \mathbf{c}=0}$. Cyclic permutation gives the other off-diagonal pairings needed for the reciprocal basis.

The displacement vectors are $B-A=(-1,2,-2)$ and $C-A=(1,0,1)$. Their cross product is $(2,-1,-2)$, with Euclidean norm three. Thus either sign ofis a unit normal vector to the second plane, whose equation is $2x-y-2z=3$. A direction vector for the intersection is the cross product of the two normal vectors:It is orthogonal to each normal vector. Since both planes contain $A$, the whole intersection line is $r=(2,-1,1)+t(6,2,5)$, $t\in \mathbb{R}$.

Take $P=(2,-1,2)$, $Q=(1,2,3)$ and $R=(4,1,0)$. The two plane directions are $Q-P=(-1,3,1)$ and $R-P=(2,2,-2)$. Their cross product is $(-8,0,-8)$, so a normal vector is $(1,0,1)$. The Cartesian equation of a plane is thereforeEach given point satisfies it, and the nonzero cross product confirms that the points determine a unique plane.

Choose a common origin at vertex $C$, set $\mathbf{a}=\overrightarrow{CB}$ and $\mathbf{b}=\overrightarrow{CA}$, and orient the remaining edge as $\mathbf{c}=\overrightarrow{AB}$. ThenThis convention makes the interior angle between $\mathbf{a}$ and $\mathbf{b}$ equal to $C$, and their lengths equal to the opposite side lengths $a$ and $b$.

Take the dot product of the difference with itself. Since $\mathbf{a}\cdot \mathbf{b}=ab\cos C$, the law of cosines follows:The magnitude of the cross product of two edge vectors is twice the triangle's area $K$. Applying this at each vertex gives $2K=ab\sin C=bc\sin A=ca\sin B$. Divide by the nonzero product $abc$ to obtain the law of sines:

Let $\mathbf{p}=(2,1,1)$, $\mathbf{q}=(1,2,2)$ and $\mathbf{r}=(0,0,3)$. Cross products give plane normals $\mathbf{p}\times \mathbf{r}=(3,-6,0)$ and $\mathbf{q}\times \mathbf{r}=(6,-3,0)$. Their dot product is $36$ and each has norm $3\sqrt{5}$. Thus the acute angle between two planes isThis is also the interior dihedral angle here: perpendicular to the common edge $OR$, the interior wedge lies between the projected directions $(2,1)$ and $(1,2)$. The two outward face normals instead make the supplementary angle.

The face $PQR$ has edge vectors $\mathbf{q}-\mathbf{p}=(-1,1,1)$ and $\mathbf{r}-\mathbf{p}=(-2,-1,2)$. Their cross product is $(3,0,3)$, so choose the outward normal vector $\mathbf{n}=(1,0,1)$. The angle between vectors obeysReversing the chosen normal gives the supplementary angle $150^{\circ }$; the acute angle with the normal line is $30^{\circ }$.

The magnitude of the cross product of two edge vectors is the area of their parallelogram. The triangular face occupies half of it:

Orient the triangle by the path $A\to B\to C\to A$. The cross product $(B-A)\times (C-A)=(0,1,0)$ gives the positive normal vector. In the plane $y=1$, the triangular region is $x,z\ge 0$, $x+z\le 1$. The surface integral in Stokes theorem is thereforeFor the boundary line integral, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\le t\le 1$. Its contribution isThus both oriented integrals equal $1/15$. Reversing the orientation would reverse both signs.

Use the vector triple product identity twice:Subtracting cancels the common term, yieldingIt is generally nonzero: the cross product does not satisfy the associative property.

The vector triple product identity and $\mathbf{n}\cdot \mathbf{n}=1$ giveApply $H$ once more. The term parallel to $\mathbf{n}$ vanishes under its cross product with $\mathbf{n}$, leaving $H^3\mathbf{x}=-{\alpha }^3\mathbf{n}\times \mathbf{x}=-{\alpha }^{2}H\mathbf{x}$. The result includes $\alpha =0$.