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Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.

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 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.

Set $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. Since these vectors form a basis, write $\mathbf{r}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. The reciprocal basis givesA point is in the closed tetrahedron precisely when it is a convex combination of the four vertices, with weights $1-\alpha -\beta -\gamma ,\alpha ,\beta ,\gamma$. Thus the necessary and sufficient conditions areThese four half-spaces are the four inward sides of its faces. This is tetrahedron membership via scalar triple products; making all inequalities strict describes the interior.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

A tetrahedron whose six edges have equal length; its four faces are equilateral triangles.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

By Melvin A. Neuman, Trinh-Toan, Lawrence F. Dahl.

Journal of the American Chemical Society 94, 3383–3388. Original diffraction study: four bonded Fe vertices form a tetrahedron; four triply bridging carbonyl carbons form an interpenetrating tetrahedron.

The 24-operation molecular point group of an ideal regular tetrahedron, including improper operations. Its irreducible representations are $A_1,A_2,E,T_1,T_2$, with dimensions $1,1,2,3,3$. The polar-vector representation is $T_2$, the axial-vector representation is $T_1$, and symmetric quadratic functions transform as $A_1+E+T_2$.

Four nearest neighbours at tetrahedron vertices around a central site.

A position surrounded by four nearest sites at tetrahedron vertices. A conventional face-centred cubic cell has eight such sites, at all combinations of quarter and three-quarter fractional coordinates. Whether occupancy constitutes a defect depends on the reference crystal.

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.

In tetrahedral molecular geometry, four neighbouring chemical atoms surround a central chemical atom in the directions of the vertices of a tetrahedron. In the regular idealization, the central atomic nucleus lies at the centre of a regular tetrahedron, and every pair of chemical bonds makes the same bond angle. Methane, ${\mathrm{CH}}_4$, provides the standard example: its carbon has four bonding electron domains and no lone pair. See Chemistry 2e, §7.6: tetrahedral methane.

The exact ideal bond angle follows from four directions proportional to $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Each has squared Euclidean norm $3$, and each distinct pair has dot product $-1$. HenceThis is a statement about a regular tetrahedron, rather than a universal exact value for every molecule labelled tetrahedral. It also explains why a cross-shaped flat drawing with $90^{\circ }$ gaps is misleading: four equal directions can spread out more evenly in three dimensions.

The same vectors sum to zero. Consequently, four equal bond-dipole contributions along those directions cancel. The symmetric methane molecule is nonpolar, but a tetrahedral shape alone does not imply a nonpolar molecule: replacing one neighbour can make the electronic distribution and bond contributions unequal. Chemical polarity depends on both molecular geometry and electron density. See McMurry, Organic Chemistry, §2.2: molecular dipoles and methane.

Four equivalent $s$ atomic orbitals at the vertices of a regular tetrahedron carry the tetrahedral symmetry representation $A_1\oplus T_2$. The normalized equal sum is $A_1$; three coordinate-sign combinations span $T_2$. Every normalized coefficient vector has entries $\pm 1/2$ in a suitable orthonormal real basis. These functions simplify a Hamiltonian invariant under the tetrahedral operations, but equivalent symmetry copies on other sites may still mix.

Place the centroid of a tetrahedron at the origin, so its four vertex vectors satisfy $r_1+r_2+r_3+r_4=0$. Midpoints of opposite edges have opposite vectors, hence equal distances from the centroid. For the three opposite-edge pairs, choose midpoint vectors $(r_1+r_j)/2$, $j=2,3,4$. Expanding their squared lengths and using ${\sum }_{j=2}^4{r}_j=-r_1$ givesThe argument uses no regularity or equal-edge assumption on the tetrahedron.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The reciprocal basis gives the coordinates of any position vector asThe point lies in the closed tetrahedron with vertices $0,\mathbf{a},\mathbf{b},\mathbf{c}$ exactly when $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma \le 1$. Indeed, these conditions express it as a convex combination of the four vertices, with fourth weight $1-\alpha -\beta -\gamma$. They also describe the intersection of the four inward half-spaces bounded by its faces. Strict inequalities characterize its interior.

Place one pair of opposite tetrahedron vertices on the line $b+\lambda a$ and the other pair on $d+\mu c$, with fixed nonzero direction vectors $a,c$. Let $\Delta =d-b$ and let $\delta \lambda ,\delta \mu$ be the parameter differences within the two pairs. The scalar triple product of the three edge vectors from one vertex gives the displayed volume. Terms parallel to $a$ or $c$ vanish in the triple product, leaving only $\Delta \cdot (a\times c)$. Sliding either pair along its own line while preserving its separation leaves this volume unchanged. When the two lines are skew lines, it can also be written $V=d_{\perp }{\ell }_1{\ell }_2|\sin \alpha |/6$, where $d_{\perp }$ is their shortest distance, ${\ell }_1,{\ell }_2$ the two edge lengths and $\alpha$ their direction angle. Intersecting or parallel lines give zero volume because all four vertices are coplanar.

The three-uniform tetrahedron is the hypergraph on four vertices with all four possible triple hyperedges. It is the complete uniform hypergraph $K_4^3$, and differs from its geometric realization as a tetrahedron. Its edge pattern encodes four-term arithmetic progressions.