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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.

Put $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$. The vector triple product identity givesTaking its dot product with $\mathbf{a}\times \mathbf{b}$ and using cyclic invariance of the scalar triple product yieldsThere was no division by $V$, so this also proves the identity for coplanar vectors.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

Parametrize the function graph by $\mathbf{R}(x,y)=(x,y,f(x,y))$. Its two tangent vectors are ${\mathbf{R}}_x=(1,0,f_x)$ and ${\mathbf{R}}_y=(0,1,f_y)$. A small parameter rectangle maps, to first order, to a tangent parallelogram with area $|{\mathbf{R}}_x\times {\mathbf{R}}_y|dxdy$. Sincethe surface area of a graph isFor example, $f\in C^1$ ensures a regular surface chart; more general domains can be handled by the corresponding convergent improper integral.

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.

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.

Subtract the two position vectors to obtain the direction $\mathbf{b}-\mathbf{a}=(-3,-1,3)$. The affine line in a vector space through the points is thereforeThe values $t=0,1$ recover the two specified points.

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

Use the normal vector $\mathbf{N}=(-8,-3,-9)$ found from the edge vectors, rather than the normal vector to the origin-based linear span in part (b). The required plane satisfies $\mathbf{N}\cdot (\mathbf{r}-\mathbf{a})=0$. Since $\mathbf{N}\cdot \mathbf{a}=5$,After dividing by $|\mathbf{N}|=\sqrt{154}$, its equation has a unit normal. Its perpendicular distance from the origin is thus $\boxed{5/\sqrt{154}}$.

Adding and subtracting the two coordinates shows that $A(1,1{)}^T=(c+s)(1,1{)}^T$ and $A(1,-1{)}^T=(c-s)(1,-1{)}^T$. The normalized eigenvectors can therefore be chosen asTheir inner product is $(1-1)/2=0$, and their squared norms are one, proving orthonormality. If $s=0$, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

Three vectors in three-dimensional Euclidean space are non-coplanar precisely when their scalar triple product is nonzero. Put $\Delta =a\cdot (b\times c)\ne 0$. In particular, $a\ne 0$. The identity for cross products with a common vector givesFor completeness, this follows from the vector triple product identity: writing $p=a\times b$ gives $p\times (a\times c)=a(p\cdot c)-c(p\cdot a)$, with $p\cdot c=\Delta$ and $p\cdot a=0$. Therefore the new scalar triple product isThus $a,a\times b,a\times c$ are non-coplanar, or equivalently linearly independent.

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.

For the numerical vectors, $\mathbf{b}\times \mathbf{c}=(-2,1,-2)$ and $D=(-1,1,0)\cdot (-2,1,-2)=3\ne 0$, proving noncoplanarity. The reciprocal basis isReciprocal dot products extract the coefficients of any vector in the original basis. Here $\mathbf{A}\cdot \mathbf{d}=-1/3$, $\mathbf{B}\cdot \mathbf{d}=2/3$, and $\mathbf{C}\cdot \mathbf{d}=5/3$, soSubstitution in Cartesian components recovers $(2,1,-1)$.

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 $r=(2,-1,1)+t(6,2,5)$, $t\in \mathbb{R}$.

The matrix product is $Pa=a-a(a^{T}a)-b(b^{T}a)=a-a=0$. Thus $\boxed{Pa=0}$, and the nonzero vector $a$ is in its kernel. Hence $P$ is singular and $\boxed{\det P=0}$. This includes $n=2$, when $a,b$ span the whole space and $P=0$.

The Euclidean norms are $\Vert v_{\pm }\Vert =2$ and $\Vert v_0\Vert =\sqrt{2}$. With normalized eigenvectors as rows,The rows form an orthonormal basis, so $R{R}^T=I$ and $R$ is an orthogonal matrix. The zero eigenvalue means $M{v}_0=0$ for a nonzero vector; thus $M$ is singular and has no matrix inverse.

The characteristic polynomial is $\det (\lambda I-M)=\lambda ({\lambda }^2-2)$. Solving the three eigenvector equations givesEach eigenspace consists of the scalar multiples of its displayed nonzero vector. Their pairwise dot products vanish: $v_{+}\cdot v_{-}=1-2+1=0$ and $v_{\pm }\cdot v_0=1-1=0$. This also agrees with the orthogonal diagonalization of a real symmetric matrix.

The coordinate intercepts suggest the normal vectorIndeed $\mathbf{n}\cdot \mathbf{a}=\mathbf{n}\cdot \mathbf{b}=\mathbf{n}\cdot \mathbf{c}=1$, so the two independent edge vectors in the plane are both orthogonal to $\mathbf{n}$.

The vector $\mathbf{a}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$. Also $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}=1/S$, hence $\mathbf{d}\cdot (\mathbf{b}-\mathbf{c})=0$. Subtraction gives $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Both vectors are nonzero, soThis is the altitude property of the orthogonal projection onto an intercept plane: the projection point is the triangle's orthocenter.

Write $P=\mathbf{a}{\mathbf{a}}^T$. Since $\mathbf{a}$ is a unit vector, $P^2=\mathbf{a}({\mathbf{a}}^T\mathbf{a}){\mathbf{a}}^T=P$ and $P^T=P$. Thus this Householder reflection satisfiesGeometrically, the mirror preserves the component tangent to its plane and reverses the normal component; doing it twice restores the original vector.