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Use a parameter $t$. The characteristic equations areFor a graph of a function $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The chain rule and the partial differential equation give $df(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting linear ordinary differential equation system shows that its characteristic through the initial point agrees with this curve on the graph of a function. This proves invariance as long as the graph of a function remains defined along the curve. A claim about the entire curve additionally requires a surface continued along that flow; a restricted graph of a function patch alone cannot imply global continuation.

Let $\mathbf{k}=(a,b,c)$, $\mathbf{\ell }=(l,m,n)$ and $\mathbf{r}=(x,y,z)$. Then $\dot{\mathbf{r}}=\mathbf{r}-\mathbf{k}\times \mathbf{r}$. Since $|\mathbf{k}|=1$,Along the characteristic starting from $s\mathbf{\ell }$ these give $\mathbf{k}\cdot \mathbf{r}=e^{t}s(\mathbf{k}\cdot \mathbf{\ell })$ and $|\mathbf{r}|=e^t|s|$. Consequently the characteristic surface generated by the given line satisfiesTo see that the whole displayed cone is generated, not just contained in it, decompose $\mathbf{\ell }=\kappa \mathbf{k}+\mathbf{v}$ with $\mathbf{v}\perp \mathbf{k}$, $\kappa =\mathbf{k}\cdot \mathbf{\ell }$. The full solution isWhen $0<|\kappa |<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular cone, withThe edge cases matter: $\kappa =0$ gives the plane $ax+by+cz=0$ (semi-angle $\pi /2$), and $|\kappa |=1$ generates only the axis (semi-angle zero), not a two-dimensional surface. In the latter case the initial line is itself characteristic, so it does not determine a unique integral surface. The nondegenerate cone is smooth away from its vertex and can be used as a graph of a function only on patches whose tangent plane is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-surface meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth graph of a function without them.

In the solid-modelling convention used here, a rigid-body transformation is an orientation-preserving rigid transformation, represented by a proper rotation and a translation. The proposed upper-left block $S$ is orthogonal but hasIt is a reflection in the plane $x=y$. It preserves distances, but reverses handedness, so it cannot describe continuous motion of a rigid solid through proper rotations and translations. Some mathematical uses of rigid transformation include this orientation-reversing case; the determinant is the distinction relevant to the warning.

In a solid modeller, triangle winding and outward normal vectors must agree. A reflection reverses the winding: for edge vectors $a,b$,Simply transforming stored normal vectors by $S$ while retaining the original vertex order therefore makes the geometric and stored normals disagree. Back-face culling, inside/outside tests and Phong reflection model calculations can then use the wrong side. Code assuming a positive oriented volume can likewise interpret the reflected solid incorrectly. A modeller can support reflections if it deliberately corrects these orientation conventions.

A suitable alternative rotates by $-\pi /2$ about $z$ and then translates upward by height. Writing $h=\text{height}$,Its rotation matrix has determinant $+1$. The source bounds $0\le x\le h$, $0\le y\le w$, $z=0$ become $0\le x^{\prime }\le w$, $0\le y^{\prime }\le h$, $z^{\prime }=0$, where $w=\text{width}$. A horizontal stripe $y=k\text{step}$ becomes the vertical stripe $x^{\prime }=k\text{step}$, so even the stripe positions from the left edge are preserved. The face's $+z$ normal vector is also preserved.

For solid modelling, a polyhedron is a solid whose boundary consists of finitely many planar polygons, called faces, meeting along straight edges and at vertices of a polyhedron. A face is not merely a list of arbitrary spatial points: its boundary must lie in one plane and bound a valid planar region. The boundary is closed and consistently oriented so that inside and outside are defined.

For example, a tetrahedron has four triangular faces, six edges and four vertices of a polyhedron. A more general polyhedron need not be convex and can have handles or cavities. The concise definition is

A boundary representation separates geometry, which supplies locations and supporting planes, from topology, which specifies incidence and orientation. A practical winged-edge data structure uses three related record types:

This explicitly stores adjacency rather than recovering it by searching all faces. A half-edge data structure expresses much the same information with two oppositely directed records per undirected edge: each has an origin vertex of a polyhedron, an incident face, its next boundary record and its opposite record; a previous record is often stored as well. The destination can be obtained from the opposite record's origin. This mesh record is distinct from a dangling half-edge of a graph.

The compact implementation answer isThere is deliberate redundancy: the same adjacency can be reached from both an edge and a face, a stored previous edge is recoverable by walking the loop, and a face normal vector can be computed from its ordered noncollinear vertices of a polyhedron. These copies make neighbourhood queries and local edits cheap, and cached plane equations speed geometric tests. They impose consistency obligations: an edit must update reciprocal references, loops, orientations and affected geometric caches together. A half-edge data structure should share the undirected edge's geometry or keep its paired records synchronized.

Insert the points in their given order, starting from $ABCD$. Use outward face orientation and regard $f\le 0$ as the interior side. The initial tetrahedron is nondegenerate becauseThus its volume is $|2|/6=1/3$. The following table gives consistently oriented faces, conveniently scaled plane equations, and the omitted old vertex of a polyhedron test:

Every opposite vertex of a polyhedron is strictly on the interior side, so these orientations are correct. The intersection of these four half-spaces is exactly the initial tetrahedron.

Now test $E=(1,0,0)$. In the same row order the face expressions giveTherefore $ACB$ and $ABD$ are visible and must be removed; $ADC$ and $BCD$ are retained. The common edge $AB$ lies inside the deleted patch and is removed. The directed horizon isCone this cycle to $E$, giving outward face orders $ACE,CBE,BDE,DAE$. Their equations and tests of the two old vertices of a polyhedron not belonging to each face are:

Each new plane contains its three named vertices of a polyhedron and has all remaining input points strictly on the interior side. Together with the two retained half-spaces, these six supporting inequalities describe the final convex hull:Equivalently assemble the closed oriented boundary from $ADC,BCD,ACE,CBE,BDE,DAE$, with adjacent face loops traversing shared edges in opposite directions. It is a triangular bipyramid with base triangle $CDE$ and apices $A,B$; $CDE$ is internal, so it is not a boundary face. There are five vertices of a polyhedron, nine edges and six faces, giving the check $5-9+6=2$. The initial tetrahedron similarly has $4-6+4=2$.

Eliminating $2y-z$ from the two plane equations gives $x=2$, followed by $2y-z=2$. Taking $y=1$ gives the point $\boxed{p=(2,1,0)}$. The entire intersection line is $\mathbf{r}=p+tu$, where $u=(0,1,2)$.

First assume $a\ne 0$. The first equation is a plane, and the second is a sphere centred at $c$. Define its signed normal separation and projected centre byThe intersection of a plane and a sphere isThe circle lies in the given plane. Its radius follows from the Pythagorean theorem, splitting $x-c=(x-q)+(q-c)$ into orthogonal components. If $a=0$, the first equation is the identity $0=0$, so the locus is the entire sphere, rather than a planar section.

Normalize the normal vector $(2,4,3)$, whose Euclidean norm is $\sqrt{29}$. The plane has the formThe scalar $d$ is the signed perpendicular distance from the origin in the chosen unit normal direction; the unsigned distance is $|d|$. The nearest point is $d\mathbf{n}$.

Noncoplanarity makes the scalar triple product $\Delta$ nonzero, so the three displayed vectors form a reciprocal basis. Their scalar products obeywith all six cross-pairings zero. Therefore, assuming the nonzero $\alpha ,\beta ,\gamma$ needed to define the three given points,Every difference of two of these points is orthogonal to $\mathbf{X}$. The two independent differences span their plane, so $\mathbf{X}$ is a normal to the plane, whose equation is $\mathbf{X}\cdot \mathbf{r}=1$. Also $\mathbf{X}\ne 0$, because its scalar product with each specified point is $1$.

The equations are an eigenvalue problem for the matrixA nonzero solution requires $\det (\lambda I-A)=0$. Expanding givesRow reduction gives the complete eigenspaces:For $\lambda =1$ the solutions form the plane $x-2y+z=0$ through the origin; for $\lambda =-3$ they form a line through the origin. Exclude the zero vector when requiring a nontrivial solution. The repeated eigenvalue $1$ has a two-dimensional eigenspace; there are three independent eigenvectors in total. This matrix is not symmetric, so its different eigenspaces need not be mutually orthogonal.

A plane through $r_0$ with nonzero normal vector $n$ consists of the position vectors $r$ satisfying $(r-r_0)\cdot n=0$. Thus $r\cdot n=d$, with $d=r_0\cdot n$. When $n$ is a unit vector, $d$ is the signed distance from the origin in the direction $n$; for an unnormalized normal vector, that distance is $d/\Vert n\Vert$. The displacement within the plane is perpendicular to $n$.

The cross product of the two normal vectors is $(2,6,-3)\times (5,2,-1)=(0,-13,-26)$, so their intersection of two planes has direction $(0,1,2)$. Eliminating $2y-z$ gives $x=2$, and substitution then gives $z=2y-2$. Thus the Cartesian constraints and parametric equation of a straight line areThere is no valid division by a zero $x$ component in a symmetric ratio form; retaining $x=2$ explicitly handles that component.

A normal vector to the plane is $n=(3,4,12{)}^T$, with Euclidean norm $\Vert n\Vert =\sqrt{9+16+144}=13$. For $p=(1,1,1{)}^T$, $n\cdot p=19$, so the distance from a point to a plane isTo verify that this is the minimum, the displacement to the closest point on a plane is parallel to $n$. Put $q=p+tn$ and require $n\cdot q=26$. This gives $t=7/169$ andFor every other point $r$ on the plane, the displacement $r-q$ is orthogonal to $q-p$. The Pythagorean theorem gives $\Vert r-p{\Vert }^2=\Vert q-p{\Vert }^2+\Vert r-q{\Vert }^2\ge (7/13{)}^2$, proving minimality.

The normal vectors are ${\mathbf{n}}_A=(1,2,-4)$ and ${\mathbf{n}}_B=(2,-1,3)$. Their dot product is $-12$, and their norms are $\sqrt{21}$ and $\sqrt{14}$. The usual angle between two planes is the smaller, unoriented angle, soFor the particular signed normal vectors printed above, the angle is the supplement $\pi -\theta \approx 134.4^{\circ }$. Reversing either normal vector exchanges these two angles without changing either plane.

A parallel plane has the same normal vector, so write $C:x+2y-4z=d$. Substitution of the given point gives $d=4+4-4=4$. ThusA displacement between the two planes has a component of magnitude $|5-4|/\Vert {\mathbf{n}}_A\Vert$ along their common unit vector normal. The distance between parallel planes is consequently

Write $\Delta =⟨\mathbf{u},\mathbf{v},\mathbf{w}⟩=\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$. The displayed definitions require $\Delta \ne 0$, equivalently linear independence of the original vectors. The vector triple product identity givesHere the first scalar triple product is $\Delta$ by cyclic permutation, and the second is zero because a cross product is orthogonal to both its factors. ThusThis is the reciprocal basis involution identity. When $\Delta =0$, the proposed reciprocal basis is undefined; for example, three vectors in the $xy$ plane cannot be inserted into the printed quotients.

Use $\mathbf{w}=-\mathbf{u}-\mathbf{v}$ and bilinearity and antisymmetry of the cross product:Consequently,If $\mathbf{u},\mathbf{v}$ are linearly independent, this linear span is the plane through the origin with normal vector $\mathbf{u}\times \mathbf{v}$. If they are dependent, all three nonzero vectors lie on one line through the origin, and their linear span has dimension one. For example, $\mathbf{u}=\mathbf{v}=(1,0,0)$ and $\mathbf{w}=(-2,0,0)$ obey the hypothesis but span only a line. Nonzero vectors alone do not force a two-dimensional linear span.

The Stokes theorem states that for an oriented smooth surface $S$ with compatible oriented boundary $C$ and a continuously differentiable vector field,Here $\nabla \times F=(-2,-2,-2)$, whereas a normal vector to the specified plane is $(1,0,-1)$. Their dot product is zero. Thus every closed curve in that plane has zero circulation by the Stokes theorem.

The source says “any curve” and displays an ordinary line integral, without specifying that it is closed or giving endpoints. An open curve need not give zero. On the plane, put $z=x-1$ and $dz=dx$; the pulled-back differential one-form becomesTherefore the complete endpoint-dependent answer iswhich specializes to zero for a closed curve. For example, the path from $(1,0,0)$ to $(2,0,1)$ lies in the plane and has integral $1$. This illustrates a conservative restriction of a vector field to a plane despite nonzero ambient curl.

Let $D=y(3y-2)$. Implicit differentiation of the two constraints gives the linear systemAdding them, then substituting back, gives at regular pointsThe relevant Jacobian determinant is $2(1+a)yz(3y-2)$. Thus the special parameter value is $\boxed{a=-1}$, assuming the other factors are nonzero. At that parameter, adding the original constraints gives $(x+1{)}^2=0$: the intersection curve lies wholly in the plane $x=-1$ and cannot be parametrized by $x$ on an interval. Calling its derivatives “infinite” describes a vertical projection, not a finite-valued differentiable pair $y(x),z(x)$ at $a=-1$; the displayed ratios then have simultaneous zero numerators on the intersection and are undefined.

There are also point-dependent failures when $y=0$, $y=2/3$ or $z=0$. These can produce vertical tangents or require a different local parameter, and simultaneous numerator zeros need separate analysis. The implicit function theorem guarantees finite $y^{\prime },z^{\prime }$ only where the stated Jacobian determinant is nonzero. This distinction is the geometry of a singular projection of an implicit curve.

The arc length of the helix follows from its parametrization:Thus ${\ell }_h=15\pi \sqrt{5}\mathrm{m}$. Its endpoint is $(0,20,15\pi )$, at height $h=15\pi \mathrm{m}$.

Under the usual intended interpretation that the descending quarter of a circle drops through one radius, its radius is $R=h$. Its arc length is ${\ell }_c=\pi R/2=15{\pi }^2/2\mathrm{m}$, givingThe printed smoothness condition is inconsistent. The terminal tangent vector of the helix is ${\mathbf{r}}^{\prime }(3\pi )=(-10,0,5)$. Every tangent vector to a curve confined to the $yz$ plane has zero $x$ component, so no such regular curve can join this helix with a common tangent line. Moreover, the helix is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.

Even after allowing a corner, a quarter of a circle reaching the ground does not alone determine its radius: the orientation of its endpoint radii must also be specified. For example, in the $yz$ plane, the circular arctraversed toward decreasing $\theta$, descends monotonically for $0\le {\theta }_0\le \pi /2$ and reaches $z=0$ when $R=h/(\sin {\theta }_0+\cos {\theta }_0)$. Choosing ${\theta }_0=\pi /4$ gives $R=h/\sqrt{2}$, whereas ${\theta }_0=\pi /2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.

Let $S$ be a compact piecewise smooth oriented surface, $\mathbf{n}$ its chosen unit normal vector, and $\partial S$ its entire oriented boundary. Let $\mathbf{A}$ be a continuously differentiable vector field on a neighborhood of $S$. Stokes theorem statesHere $\nabla \times \mathbf{A}$ is the curl, $dS$ is the surface area element, and $d\mathbf{r}$ is the directed line integral element. The boundary follows the right-hand rule relative to $\mathbf{n}$; equivalently the surface lies to the left when walking along its boundary with one's head toward $\mathbf{n}$.

The topological requirement is orientability, together with the regularity and boundary assumptions above. Simple connectivity is unnecessary. An annulus or cylindrical side is allowed, provided every boundary component is included with its induced direction. For a concrete counterexample to a disk-only restriction, take the planar annulus $1\le x^2+y^2\le 4$ and $\mathbf{A}=(-y/2,x/2,0)$. The curl is ${\mathbf{e}}_z$, the surface integral is $3\pi$, and the outer counterclockwise and inner clockwise line integrals sum to $4\pi -\pi =3\pi$.

For $S$ in the $xy$ plane with upward normal vector and $\mathbf{A}=(P,Q,0)$, its curl has vertical component $Q_x-P_y$. Thus Stokes theorem becomes Green's theorem:The outer boundary is counterclockwise and any inner boundaries are clockwise.