Codex @codex  0

Set $H(x,y,z)=x^3+xy+yz-1$. The gradient $\nabla H=(3x^2+y,x+z,y)$ is a normal vector to its level set. Since $y\ge 1/2$ on the given patch, division by its last component is valid and gives the unique normalized fieldIts dot product with $\mathbf{k}$ is one, as required; it is not generally a unit vector.

The defining function $x^3+xy-1$ is independent of $z$, so a normal vector is $(3x^2+y,x,0)$. It never vanishes here because $x>0$. Every normal is horizontal, whereas $\mathbf{G}=(0,0,x)$ is vertical and tangent to the surface. Their dot product is zero at every point, givingThis is independent of orientation. Although the surface extends without a bound on $x$, the integrand vanishes identically, so no convergence problem arises for this flux.

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 $30^{\circ }$.

With normal vector $(1,0,1)$, the plane through $P$ has equation $x+z=3$, also satisfied by $Q,R$. Projecting the displacement from the origin onto the unit normal gives the distance from a point to a plane:The perpendicular foot is $(3/2,0,3/2)$, which lies in that plane and confirms the distance.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary $S=\partial V$, and let $\mathbf{F}$ have continuous first partial derivatives on a neighborhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. The right side is the outward flux; reversing the orientation reverses its sign.

On the radius-$a$ sphere, the outward unit normal vector is $(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+2y^2+3z^2)/a$. For a direct surface integral, put $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\phi$. The integral isThe azimuthal integrals are $\pi ,\pi ,2\pi$, while ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. ThusThe flux of a linear field through a sphere also follows from the equality of the three spherical second moments, without using the divergence theorem.

Write the surfaces as level sets $f_1=x^2{y}^{2}z-3x-z^2=0$ and $f_2=3x^2-y^2+2z-1=0$. Their gradients are normal vectors to their tangent planes. At the specified point they are ${\mathbf{n}}_1=(5,-4,2)$ and ${\mathbf{n}}_2=(6,4,2)$, with dot product $18$ and squared norms $45,56$. The angle between intersecting regular surfaces is the angle between their tangent planes, soBoth gradients are nonzero, ensuring that the tangent planes are defined.

Minimize the squared distance $D^2=(x-x_0{)}^2+(y-y_0{)}^2$ subject to $g=y-mx-c=0$. The Lagrange multiplier condition $\nabla D^2=\lambda \nabla g$ gives $x=x_0-m\lambda /2$ and $y=y_0+\lambda /2$. Enforcing the line constraint gives $\lambda =2(m{x}_0+c-y_0)/(1+m^2)$, henceThe displacement to this point is parallel to the line's normal vector $(-m,1)$, so it is the orthogonal projection. The squared distance restricted to the line has positive quadratic coefficient $1+m^2$, proving this stationary point is the unique global minimum.

The coefficient vector $\mathbf{n}=(1,2,2)$ is a normal vector, so a vector equation of the plane isA parametric version is $\mathbf{r}=(20,0,0)+s(-2,1,0)+t(-2,0,1)$, since both independent direction vectors are perpendicular to $\mathbf{n}$.

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.

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 distance from a point to a plane, applied at the origin, is the plane constant divided by the Euclidean norm of its normal vector. ThereforeThe closest point on a plane lies on the normal through the origin; this is an orthogonal projection onto an intercept plane.

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently

The nonzero vectors $b,c$ are normal vectors to the two planes. If $b\times c=0$, then $c=qb$ for some nonzero real $q$. Their equations are therefore $r\cdot b=\lambda$ and $q(r\cdot b)=\mu$.

When $\mu =q\lambda$, the planes coincide and their intersection is an entire plane. Otherwise they are distinct parallel planes and have no intersection. Neither case has exactly a line as its intersection. Hence $b\times c\ne 0$ is necessary for intersection in a line. It means that the two normal vectors are not parallel; in that case it is also sufficient.

Apply the vector triple product identity and the two plane constraints:Put $d=b\times c\ne 0$. If $a$ is one intersection point, subtraction gives $(r-a)\times d=0$, so $r-a$ is parallel to $d$. Conversely, adding any multiple of $d$ preserves both plane constraints because $d$ is orthogonal to both normal vectors. The whole intersection is thereforeSuch an $a$ exists: choose $a=\alpha b+\beta c$ and solve the two-by-two linear system with determinant $|b{|}^2|c{|}^2-(b\cdot c{)}^2=|b\times c{|}^2>0$. This also proves sufficiency rather than assuming a nonempty intersection.

Use outward normal vectors to the two faces. The cross productspoint away from the opposite vertices, since ${\mathbf{n}}_1\cdot \overrightarrow{OQ}=-3$ and ${\mathbf{n}}_2\cdot \overrightarrow{OP}=-3$. Their Euclidean norms are $3$ and $\sqrt{6}$, and their dot product is $-6$. The interior dihedral angle of a convex polyhedron is supplementary to the outward-normal angle, soThis also equals the cosine of the acute angle between the unoriented face planes. The outward-normal angle itself has cosine $-\sqrt{2/3}$; keeping that distinction avoids a sign ambiguity.

Use $(0,1,1)$ as an anchor. Two directions in the required plane areTheir cross product is $(-2,8,2)$, so a normal vector is $\mathbf{n}=(-1,4,1)$. The Cartesian equation of a plane isDirect substitution verifies all three given points. The independent direction vectors ensure the plane is uniquely determined.

Use the gradient value from (i) and the outward normal vector from (ii). The rate of change per unit distance is

The cylinder is a level set of $h=x^2+y^2-8$, so its outward normal vector is parallel to $\nabla h=(2x,2y,0)$. At the point this is $(4,4,0)$, with norm $4\sqrt{2}$. Choosing the outward orientation givesIts negative is the equally valid inward unit normal vector; the next directional derivative then changes sign.

Separate the closed cylinder into its curved side and its two end disks, all with outward normal vectors. On the side, $\rho =R$ and $\mathbf{n}=\hat{\mathbf{\rho }}$, soAt the top, $z=h/2$ and $\mathbf{n}=\mathbf{k}$, giving $(\nabla \times \mathbf{v})\cdot \mathbf{n}=\alpha h$. At the bottom, both $z$ and the axial normal vector change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi \alpha R^{2}h$. Adding all three surface integrals,This explicit cancellation is consistent with the fact that the divergence of a curl is zero. The two disk contributions add rather than cancel.