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The cross product of the two in-plane directions is $\mathbf{u}\times \mathbf{v}=(-5,1,2)$, with Euclidean norm $\sqrt{30}$. Either sign gives a unit normal vector:The plane equation is $\hat{\mathbf{n}}\cdot (\mathbf{r}-\mathbf{A})=0$, so the distance from a point to a plane gives

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. Thus

Put $L=5\pi /2$. The curve tangent is ${\mathbf{r}}^{\prime }=(-b\sin \theta ,b\cos \theta ,c)$. For the first line integral, the scalar triple product identity gives $(\mathbf{a}\times \mathbf{r})\cdot {\mathbf{r}}^{\prime }=\mathbf{a}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })$. Its cross product isThe needed definite integrals areand ${\int }_0^L(-\theta \sin \theta -\cos \theta )d\theta =[\theta \cos \theta -2\sin \theta {]}_0^L=-2$. HenceFor the second line integral, substitute the curve into ${\mathbf{v}}_2\cdot {\mathbf{r}}^{\prime }$. The cubic terms combine to $b^3\cos \theta$, leavingThe fundamental theorem of calculus therefore givesBoth calculations retain the full path from $\theta =0$ to $5\pi /2$, rather than inadvertently closing the helix after one turn.

The independent vectors $\mathbf{a},\mathbf{b}$ span the plane through the origin, and their cross product is a nonzero normal vector. Thus the required nonparametric equation is

Take two edge vectors in the face, $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product is the face normal vectorThe face plane is $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Every term in $\mathbf{a}\cdot \mathbf{n}$ except $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ vanishes, givingThe scalar triple product is nonzero, so this normal cannot vanish.

Use the embedded surface parametrizationIts tangent vectors have cross productwhose magnitude is $r\sqrt{1+4a^2{r}^2/b^4}$. Thus the surface area isWith $u=1+4a^2{r}^2/b^4$, the radial integral becomes $b^4(8a^2{)}^{-1}\int u^{1/2}du$. Evaluating its endpoints givesThis is the scalar-area part of the paraboloid sector surface measures.

Evaluate the cross product componentwise and use $\sin u\cos v-\sin v\cos u=\sin (u-v)$:

Each component of the cross product in part (a) is independent of the coordinate with respect to which the divergence differentiates it. Thus

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.

The direction vectors are $\mathbf{u}=(1,-2,1)$ and $\mathbf{v}=(0,-2,1)$. Their cross product is $(0,-1,-2)$, with norm $\sqrt{5}$. A unit normal to the plane is thereforeThe two intersecting directions span the plane, so a vector orthogonal to both is a normal vector to that plane.

Two directions within the plane areTheir cross product is $(1,1,1)/4$, so a unit vector normal to the plane isIts negative is equally valid. An equation of a plane through three points in vector form isor, parametrically, $\mathbf{r}=\mathbf{e}+s(\mathbf{f}-\mathbf{e})+t(\mathbf{g}-\mathbf{e})$. Since the coordinate sum of $E$ is one, the Cartesian equation of a plane isThe distance from a point to a plane is the absolute normal displacement, giving

Let $\mathbf{d}=\mathbf{f}-\mathbf{e}$. A point on the line has position vector $\mathbf{r}=\mathbf{e}+t\mathbf{d}$. By bilinearity and antisymmetry of the cross product,For a point with position vector $\mathbf{c}$, the area of the parallelogram spanned by $\mathbf{c}-\mathbf{e}$ and $\mathbf{d}$ equals base length times perpendicular Euclidean distance. Hence the shortest Euclidean distance to the line isFor $\mathbf{c}=(0,0,1)$, these quantities areThus $\boxed{d=\sqrt{6}/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

For a point $\mathbf{p}={\mathbf{r}}_2(t)$, the magnitude of $(\mathbf{p}-{\mathbf{a}}_1)\times {\mathbf{n}}_1$ is the area of a parallelogram. Dividing by its base length $\Vert {\mathbf{n}}_1\Vert$ gives the perpendicular Euclidean distance to the first line. Explicitly,HenceThis use of the cross product removes any component parallel to the first line automatically.

The boundary order fixes the orientation. Two edges issuing from $A$ are $\overrightarrow{AB}=(0,0,1)$ and $\overrightarrow{AD}=(1,1,0)$. Their cross product gives the oriented vector area:Its norm is $\sqrt{2}$, the ordinary rectangle area. The sign agrees with traversing $A\to B\to C\to D\to A$ by the right-hand orientation rule.

Expanding the cross product givesTaking its divergence component by component,

Take the cross product with $\mathbf{c}$. Since $\mathbf{c}\times \mathbf{c}=0$, the equation impliesAs $\mathbf{c}\ne 0$, the displacement is parallel to $\mathbf{c}$. ThereforeThis is a necessary condition; substitution must still determine which points on this straight line satisfy the full equation.