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This plane contains the direction $\mathbf{b}-\mathbf{a}$ and is parallel to $\mathbf{c}$. Put $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times \mathbf{c}$. Its scalar triple product with $\mathbf{a}$ is again $D\ne 0$, proving $\mathbf{N}\ne 0$. HenceThe dot product with $\mathbf{c}$ vanishes, while the dot products with $\mathbf{a}$ and $\mathbf{b}$ equal $\mu$ after normalization. These verify both incidence and parallelism.

The normal vector must be parallel to $\mathbf{b}$, and the plane must contain $\mathbf{a}$. Therefore one choice isThe nonzero scalar triple product ensures $\mathbf{b}\ne 0$. Replacing both $\mathbf{n}$ and $\mu$ by their negatives describes the same affine hyperplane.

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.

Put $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is unchanged by cyclic permutation, soThese are the diagonal pairings of a reciprocal basis.

The volume of a parallelepiped is the absolute scalar triple product of its edge vectors. HereThe absolute value removes the orientation of the chosen edge ordering.

In components, the scalar triple product isIt is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$. A cyclic permutation of the rows has positive sign, so $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$. Equivalently the Levi-Civita symbol expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by that cyclic relabeling.

The scalar triple product is a determinant:A cyclic permutation of its three columns is an even permutation, so it leaves the determinant unchanged. Thus $\det (\mathbf{a},\mathbf{b},\mathbf{c})=\det (\mathbf{b},\mathbf{c},\mathbf{a})=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, proving the identity.

With $s=\sin {\theta }_0$ and $c=\cos {\theta }_0$, conversion from spherical coordinates gives $\mathbf{a}=r_0(s,0,c)$, $\mathbf{b}=r_0(-s/2,\sqrt{3}s/2,c)$ and $\mathbf{c}=r_0(-s/2,-\sqrt{3}s/2,c)$. The volume of the parallelepiped is the absolute value of their scalar triple product:The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

Write $\mathbf{\Delta }=\mathbf{d}-\mathbf{b}$. At an intersection, $\mathbf{\Delta }=\lambda \mathbf{a}-\mu \mathbf{c}$, so its scalar triple product with $\mathbf{a},\mathbf{c}$ vanishes. The expression in the question is exactlyUnder a shift of origin by $\mathbf{h}$, the direction vectors $\mathbf{a},\mathbf{c}$ stay fixed, while $\mathbf{b},\mathbf{d}$ both change by $-\mathbf{h}$. Thus $\mathbf{\Delta }$ and the condition are invariant. For nonparallel lines this is also sufficient for intersection; for parallel lines it only asserts coplanarity and need not imply intersection.

Put $\delta \lambda ={\lambda }_2-{\lambda }_1$ and $\delta \mu ={\mu }_2-{\mu }_1$. Then ${\mathbf{x}}_2-{\mathbf{x}}_1=\delta \lambda \mathbf{a}$ and ${\mathbf{x}}_4-{\mathbf{x}}_1=({\mathbf{x}}_3-{\mathbf{x}}_1)+\delta \mu \mathbf{c}$. Bilinearity of the cross product givesThe tetrahedron volume is one sixth of the absolute value of this triple product. Sliding either vertex pair along its line preserves $|\delta \lambda |$ or $|\delta \mu |$, because its separation is fixed and the direction vectors are fixed. Thus the volume is unchanged, including the zero-volume coplanar case. This is tetrahedron volume under sliding opposite edges.

The identity $\nabla \cdot (\mathbf{a}\times \mathbf{v})=\mathbf{v}\cdot (\nabla \times \mathbf{a})-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ reduces to $-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ when $\mathbf{a}$ is constant. In components, this follows by differentiating ${\epsilon }_{ijk}{a}_j{v}_k$ and using the antisymmetry of the Levi-Civita symbol. The divergence theorem and the cyclic scalar triple product identity now giveSince $\mathbf{a}$ is arbitrary,The order of the cross product determines the minus sign, as in the vector curl form of the divergence theorem.

Three vectors in real three-dimensional space have zero determinant exactly when they are linearly dependent. Their linear span then has dimension at most two, so they lie in a common plane through the origin. Conversely vectors in such a plane span at most two dimensions, so their scalar triple product vanishes. This includes parallel nonzero vectors: they need not determine a unique plane, but they lie in at least one. Even zero vectors satisfy the determinant characterization, though the question excludes them.

The scalar triple product is the determinant of the matrix with columns $a,b,c$. A cyclic permutation of three columns is a product of two swaps, so its determinant sign is positive. Therefore $[a,b,c]=[b,c,a]$. This uses the alternating multilinearity of the determinant, and remains true for linearly dependent vectors.

At an intersection, $a+tu=b+t^{\prime }v$, so $b-a=tu-t^{\prime }v$. Alternating multilinearity of the scalar triple product then gives $[v,b-a,u]=t[v,u,u]-t^{\prime }[v,v,u]=0$, establishing $[v,b,u]=[v,a,u]$.

For the location, write $b=a+tu-t^{\prime }v$ and take its scalar triple product with $a$ and $v$:Under the additional stated assumption $[a,u,v]\ne 0$, division givesThese are intersection parameters from scalar triple products. The formula presupposes an intersection; the first coplanarity condition by itself also holds for distinct parallel lines. Intersections can exist with the displayed denominator zero, in which case a different auxiliary vector or a component calculation is needed.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Put $D=[a,m,n]\ne 0$ and rewrite the intersection equation as $b=a+sm-tn$. Take scalar triple products with $a$ and $n$:Taking the triple product with $a$ and $m$ instead givesThus the intersection parameters from scalar triple products areThe assumed nonzero denominator makes these expressions valid and guarantees nonparallel directions. The resulting point is $r=a+sm=b+tn$; these formulas are conditional on the intersection assumed in the question.

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.

The cross product is $\mathbf{a}\times \mathbf{b}=(1,-7,-3)$. Taking its dot product with $\mathbf{c}$ gives the scalar triple product

The nonparallel direction unit vectors have nonzero cross product. LetFor any joining vector $\mathbf{s}=\mathbf{c}-\mathbf{a}+\mu \hat{\mathbf{d}}-\lambda \hat{\mathbf{b}}$, its normal component is fixed:Thus its length is at least the absolute value of that component. Since the two line directions span ${\mathbf{n}}^{\perp }$, suitable $\lambda ,\mu$ cancel the entire component in that plane, attaining the bound. Therefore the distance between skew lines, including the zero-distance intersecting case, isThe numerator is a scalar triple product. The nonparallel hypothesis is essential to the denominator and to the spanning argument.

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.

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.