Volume of a tetrahedron vectors

Volume of a tetrahedron vectors DEFAULT
Hint: Here, we will use the concept that volume of tetrahedron is given as one – sixth of the modulus of the products of the vectors from which it is formed. So, first we will find the vectors and then calculate the scalar triple product which is equal to the determinant of the coefficients of the vectors.

Complete step-by-step answer:
We know that a tetrahedron is one kind of pyramid, which is a polyhedron with a flat polygon base and triangular faces connecting the base to a common point. In the case of a tetrahedron the base is a triangle and any of the four faces can be considered as the base , so a tetrahedron is also known as a triangular pyramid.
A tetrahedron has four faces, six edges and four vertices.
Its three edges meet at each vertex.

The four vertices that we have been given in the question are A(1,1,0) B(-4,3,6) C(-1,0,3) and D(2,4,-5).
So, the vector AB will be = (-4-1) i + (3-1) j + (6-0) k = -5i + 2j +6k
Similarly vector AC is given as = -2i –j + 3k
And, vector AD = i + 3j – 5k

Now, the formula for the volume of the tetrahedron will be = $\dfrac{1}{6}\times |scalar\,triple\,product\,of\,these\,three\,vectors|$
\[=\dfrac{1}{6}\times |\left\{ \left( \overrightarrow{AB}\times \overrightarrow{AC} \right).\overrightarrow{AD} \right\}|\]
Scalar triple product = $\left( \begin{matrix}
   -5 & 2 & 6 \\
   -2 & -1 & 3 \\
   1 & 3 & -5 \\
\end{matrix} \right)$
$\begin{align}
  & =-5\left( 5-9 \right)-2\left( 10-3 \right)+6\left\{ -6-\left( -1 \right) \right\} \\
 & =-5\left( -4 \right)-2\left( 7 \right)+6\left( -6+1 \right) \\
 & =20-14-30 \\
 & =-6-30=-36 \\
\end{align}$

Therefore, volume of the tetrahedron = $\dfrac{1}{6}\times |-36|=\dfrac{36}{6}=6$
Hence, the volume of the given tetrahedron is 6 cubic units.

Note: Here, it should be noted that we always apply modulus while finding the volume of the tetrahedron. If the scalar triple product is negative, we have to make it positive because volume is always a positive quantity.
Sours: https://www.vedantu.com/question-answer/find-the-volume-of-tetrahedron-whose-vertices-class-10-maths-cbse-5eedf688ccf1522a9847565b

Volume of tetrahedron, build on vectors online calculator

Volume of the tetrahedron equals to (1/6) times scalar triple product of vectors which it is build on:

definition of the volume of tetrahedron, build on vectors

Because of the value of scalar triple vector product can be the negative number and the volume of the tetrahedrom is not, one should find the magnitude of the result of triple vector product when calculating the volume of geometric body.

To calculate the volume of the tetrahedron, build on vectors, one can use our online calculator with step by step solution.

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Volume of Tetrahedron Proof using Box Product-VECTORS - IIT JEE MAIN,ADVANCED

Tetrahedron

Polyhedron with 4 faces

Not to be confused with tetrahedroid or Tetrahedron (journal).

Regular tetrahedron
Tetrahedron.jpg
(Click here for rotating model)
TypePlatonic solid
shortcode3> 2z
ElementsF = 4, E = 6
V = 4 (χ = 2)
Faces by sides4{3}
Conway notationT
Schläfli symbols{3,3}
h{4,3}, s{2,4}, sr{2,2}
Face configurationV3.3.3
Wythoff symbol3 | 2 3
| 2 2 2
Coxeter diagramCDel node 1.pngCDel 3.pngCDel node.pngCDel 3.pngCDel node.png = CDel node h.pngCDel 4.pngCDel node.pngCDel 3.pngCDel node.png
CDel node h.pngCDel 2x.pngCDel node h.pngCDel 4.pngCDel node.png
CDel node h.pngCDel 2x.pngCDel node h.pngCDel 2x.pngCDel node h.png
SymmetryTd, A3, [3,3], (*332)
Rotation groupT, [3,3]+, (332)
ReferencesU01, C15, W1
Propertiesregular, convexdeltahedron
Dihedral angle70.528779° = arccos(1⁄3)
Tetrahedron vertfig.png
3.3.3
(Vertex figure)
Tetrahedron.png
Self-dual
(dual polyhedron)
Tetrahedron flat.svg
Net
3D model of regular tetrahedron.

In geometry, a tetrahedron (plural: tetrahedra or tetrahedrons), also known as a triangular pyramid, is a polyhedron composed of four triangularfaces, six straight edges, and four vertex corners. The tetrahedron is the simplest of all the ordinary convex polyhedra and the only one that has fewer than 5 faces.[1]

The tetrahedron is the three-dimensional case of the more general concept of a Euclideansimplex, and may thus also be called a 3-simplex.

The tetrahedron is one kind of pyramid, which is a polyhedron with a flat polygon base and triangular faces connecting the base to a common point. In the case of a tetrahedron the base is a triangle (any of the four faces can be considered the base), so a tetrahedron is also known as a "triangular pyramid".

Like all convex polyhedra, a tetrahedron can be folded from a single sheet of paper. It has two such nets.[1]

For any tetrahedron there exists a sphere (called the circumsphere) on which all four vertices lie, and another sphere (the insphere) tangent to the tetrahedron's faces.[2]

Regular tetrahedron[edit]

A regular tetrahedron is a tetrahedron in which all four faces are equilateral triangles. It is one of the five regular Platonic solids, which have been known since antiquity.

In a regular tetrahedron, all faces are the same size and shape (congruent) and all edges are the same length.

Five tetrahedra are laid flat on a plane, with the highest 3-dimensional points marked as 1, 2, 3, 4, and 5. These points are then attached to each other and a thin volume of empty spaceis left, where the five edge angles do not quite meet.

Regular tetrahedra alone do not tessellate (fill space), but if alternated with regular octahedra in the ratio of two tetrahedra to one octahedron, they form the alternated cubic honeycomb, which is a tessellation. Some tetrahedra that are not regular, including the Schläfli orthoscheme and the Hill tetrahedron, can tessellate.

The regular tetrahedron is self-dual, which means that its dual is another regular tetrahedron. The compound figure comprising two such dual tetrahedra form a stellated octahedron or stella octangula.

Coordinates for a regular tetrahedron[edit]

The following Cartesian coordinates define the four vertices of a tetrahedron with edge length 2, centered at the origin, and two level edges:

{\displaystyle \left(\pm 1,0,-{\frac {1}{\sqrt {2}}}\right)\quad {\mbox{and}}\quad \left(0,\pm 1,{\frac {1}{\sqrt {2}}}\right)}

Expressed symmetrically as 4 points on the unit sphere, centroid at the origin, with lower face level, the vertices are:

{\displaystyle v_{1}=\left({\sqrt {\frac {8}{9}}},0,-{\frac {1}{3}}\right)}

{\displaystyle v_{2}=\left(-{\sqrt {\frac {2}{9}}},{\sqrt {\frac {2}{3}}},-{\frac {1}{3}}\right)}

{\displaystyle v_{3}=\left(-{\sqrt {\frac {2}{9}}},-{\sqrt {\frac {2}{3}}},-{\frac {1}{3}}\right)}

{\displaystyle v_{4}=(0,0,1)}

with the edge length of {\displaystyle {\sqrt {\frac {8}{3}}}}.

Still another set of coordinates are based on an alternatedcube or demicube with edge length 2. This form has Coxeter diagramCDel node h.pngCDel 4.pngCDel node.pngCDel 3.pngCDel node.png and Schläfli symbol h{4,3}. The tetrahedron in this case has edge length 2√2. Inverting these coordinates generates the dual tetrahedron, and the pair together form the stellated octahedron, whose vertices are those of the original cube.

Tetrahedron: (1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1)
Dual tetrahedron: (−1,−1,−1), (−1,1,1), (1,−1,1), (1,1,−1)
Regular tetrahedron ABCD and its circumscribed sphere

Angles and distances[edit]

For a regular tetrahedron of edge length a:

Face area {\displaystyle A_{0}={\frac {\sqrt {3}}{4}}a^{2}\,}
Surface area[3]A=4\,A_{0}={\sqrt {3}}a^{2}\,
Height of pyramid[4]{\displaystyle h={\frac {\sqrt {6}}{3}}a={\sqrt {\frac {2}{3}}}\,a\,}
Centroid to vertex distance {\displaystyle {\frac {3}{4}}\,h={\frac {\sqrt {6}}{4}}\,a={\sqrt {\frac {3}{8}}}\,a\,}
Edge to opposite edge distance {\displaystyle l={\frac {1}{\sqrt {2}}}\,a\,}
Volume[3]{\displaystyle V={\frac {1}{3}}A_{0}h={\frac {\sqrt {2}}{12}}a^{3}={\frac {a^{3}}{6{\sqrt {2}}}}\,}
Face-vertex-edge angle {\displaystyle \arccos \left({\frac {1}{\sqrt {3}}}\right)=\arctan \left({\sqrt {2}}\right)\,}
(approx. 54.7356°)
Face-edge-face angle, i.e., "dihedral angle"[3]{\displaystyle \arccos \left({\frac {1}{3}}\right)=\arctan \left(2{\sqrt {2}}\right)\,}
(approx. 70.5288°)
Vertex-Center-Vertex angle,[5] the angle between lines from the tetrahedron center to any two vertices. It is also the angle between Plateau borders at a vertex. In chemistry it is called the tetrahedral bond angle. This angle (in radians) is also the arclength of the geodesic segment on the unit sphere resulting from centrally projecting one edge of the tetrahedron to the sphere. {\displaystyle \arccos \left(-{\frac {1}{3}}\right)=2\arctan \left({\sqrt {2}}\right)\,}
(approx. 109.4712°)
Solid angle at a vertex subtended by a face {\displaystyle \arccos \left({\frac {23}{27}}\right)}
(approx. 0.55129 steradians)
(approx. 1809.8 square degrees)
Radius of circumsphere[3]{\displaystyle R={\frac {\sqrt {6}}{4}}a={\sqrt {\frac {3}{8}}}\,a\,}
Radius of insphere that is tangent to faces[3]{\displaystyle r={\frac {1}{3}}R={\frac {a}{\sqrt {24}}}\,}
Radius of midsphere that is tangent to edges[3]{\displaystyle r_{\mathrm {M} }={\sqrt {rR}}={\frac {a}{\sqrt {8}}}\,}
Radius of exspheres{\displaystyle r_{\mathrm {E} }={\frac {a}{\sqrt {6}}}\,}
Distance to exsphere center from the opposite vertex {\displaystyle d_{\mathrm {VE} }={\frac {\sqrt {6}}{2}}a={\sqrt {\frac {3}{2}}}a\,}

With respect to the base plane the slope of a face (2√2) is twice that of an edge (√2), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof).

For a regular tetrahedron with side length a, radius R of its circumscribing sphere, and distances di from an arbitrary point in 3-space to its four vertices, we have[6]

{\displaystyle {\begin{aligned}{\frac {d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}}{4}}+{\frac {16R^{4}}{9}}&=\left({\frac {d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}}{4}}+{\frac {2R^{2}}{3}}\right)^{2};\\4\left(a^{4}+d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}\right)&=\left(a^{2}+d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}\right)^{2}.\end{aligned}}}

Isometries of the regular tetrahedron[edit]

The proper rotations, (order-3 rotation on a vertex and face, and order-2 on two edges) and reflection plane (through two faces and one edge) in the symmetry group of the regular tetrahedron

The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron (see above, and also animation, showing one of the two tetrahedra in the cube). The symmetries of a regular tetrahedron correspond to half of those of a cube: those that map the tetrahedra to themselves, and not to each other.

The tetrahedron is the only Platonic solid that is not mapped to itself by point inversion.

The regular tetrahedron has 24 isometries, forming the symmetry groupTd, [3,3], (*332), isomorphic to the symmetric group, S4. They can be categorized as follows:

  • T, [3,3]+, (332) is isomorphic to alternating group, A4 (the identity and 11 proper rotations) with the following conjugacy classes (in parentheses are given the permutations of the vertices, or correspondingly, the faces, and the unit quaternion representation):
    • identity (identity; 1)
    • rotation about an axis through a vertex, perpendicular to the opposite plane, by an angle of ±120°: 4 axes, 2 per axis, together 8 ((1 2 3), etc.; 1 ± i ± j ± k/2)
    • rotation by an angle of 180° such that an edge maps to the opposite edge: 3 ((1 2)(3 4), etc.; i, j, k)
  • reflections in a plane perpendicular to an edge: 6
  • reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to −x): the rotations correspond to those of the cube about face-to-face axes

Orthogonal projections of the regular tetrahedron[edit]

The regular tetrahedron has two special orthogonal projections, one centered on a vertex or equivalently on a face, and one centered on an edge. The first corresponds to the A2Coxeter plane.

Cross section of regular tetrahedron[edit]

A central cross section of a regular tetrahedronis a square.

The two skew perpendicular opposite edges of a regular tetrahedron define a set of parallel planes. When one of these planes intersects the tetrahedron the resulting cross section is a rectangle.[7] When the intersecting plane is near one of the edges the rectangle is long and skinny. When halfway between the two edges the intersection is a square. The aspect ratio of the rectangle reverses as you pass this halfway point. For the midpoint square intersection the resulting boundary line traverses every face of the tetrahedron similarly. If the tetrahedron is bisected on this plane, both halves become wedges.

A tetragonal disphenoid viewed orthogonally to the two green edges.

This property also applies for tetragonal disphenoids when applied to the two special edge pairs.

Spherical tiling[edit]

The tetrahedron can also be represented as a spherical tiling, and projected onto the plane via a stereographic projection. This projection is conformal, preserving angles but not areas or lengths. Straight lines on the sphere are projected as circular arcs on the plane.

Helical stacking[edit]

Regular tetrahedra can be stacked face-to-face in a chiral aperiodic chain called the Boerdijk–Coxeter helix. In four dimensions, all the convex regular 4-polytopes with tetrahedral cells (the 5-cell, 16-cell and 600-cell) can be constructed as tilings of the 3-sphere by these chains, which become periodic in the three-dimensional space of the 4-polytope's boundary surface.

Other special cases[edit]

Tetrahedral subgroup tree.png
Tetrahedral symmetry subgroup relations
Tetrahedron symmetry tree.png
Tetrahedral symmetries shown in tetrahedral diagrams

An isosceles tetrahedron, also called a disphenoid, is a tetrahedron where all four faces are congruent triangles. A space-filling tetrahedron packs with congruent copies of itself to tile space, like the disphenoid tetrahedral honeycomb.

In a trirectangular tetrahedron the three face angles at one vertex are right angles. If all three pairs of opposite edges of a tetrahedron are perpendicular, then it is called an orthocentric tetrahedron. When only one pair of opposite edges are perpendicular, it is called a semi-orthocentric tetrahedron. An isodynamic tetrahedron is one in which the cevians that join the vertices to the incenters of the opposite faces are concurrent, and an isogonic tetrahedron has concurrent cevians that join the vertices to the points of contact of the opposite faces with the inscribed sphere of the tetrahedron.

Isometries of irregular tetrahedra[edit]

The isometries of an irregular (unmarked) tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3-dimensional point group is formed. Two other isometries (C3, [3]+), and (S4, [2+,4+]) can exist if the face or edge marking are included. Tetrahedral diagrams are included for each type below, with edges colored by isometric equivalence, and are gray colored for unique edges.

Tetrahedron name Edge
equivalence
diagram
Description
Symmetry
Schön.Cox.Orb.Ord.
Regular tetrahedron Regular tetrahedron diagram.png

Four equilateral triangles

It forms the symmetry group Td, isomorphic to the symmetric group, S4. A regular tetrahedron has Coxeter diagramCDel node 1.pngCDel 3.pngCDel node.pngCDel 3.pngCDel node.png and Schläfli symbol {3,3}.
Td
T
[3,3]
[3,3]+
*332
332
24
12
Triangular pyramidIsosceles trigonal pyramid diagram.png

An equilateral triangle base and three equal isosceles triangle sides

It gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C3v, isomorphic to the symmetric group, S3. A triangular pyramid has Schläfli symbol {3}∨( ).
C3v
C3
[3]
[3]+
*33
33
6
3
Mirrored sphenoid Sphenoid diagram.png

Two equal scalene triangles with a common base edge

This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group Cs, also isomorphic to the cyclic group, Z2.
Cs
=C1h
=C1v
[ ]*2
Irregular tetrahedron
(No symmetry)
Scalene tetrahedron diagram.png

Four unequal triangles

Its only isometry is the identity, and the symmetry group is the trivial group. An irregular tetrahedron has Schläfli symbol ( )∨( )∨( )∨( ).

C1[ ]+11
Disphenoids (Four equal triangles)
Tetragonal disphenoidTetragonal disphenoid diagram.png

Four equal isosceles triangles

It has 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D2d. A tetragonal disphenoid has Coxeter diagram CDel node h.pngCDel 2x.pngCDel node h.pngCDel 4.pngCDel node.png and Schläfli symbol s{2,4}.

D2d
S4
[2+,4]
[2+,4+]
2*2
8
4
Rhombic disphenoidRhombic disphenoid diagram.png

Four equal scalene triangles

It has 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-groupV4 or Z22, present as the point group D2. A rhombic disphenoid has Coxeter diagram CDel node h.pngCDel 2x.pngCDel node h.pngCDel 2x.pngCDel node h.png and Schläfli symbol sr{2,2}.

D2[2,2]+2224
Generalized disphenoids (2 pairs of equal triangles)
Digonal disphenoidDigonal disphenoid diagram2.png
Digonal disphenoid diagram.png

Two pairs of equal isosceles triangles

This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C2v, isomorphic to the Klein four-groupV4. A digonal disphenoid has Schläfli symbol { }∨{ }.
C2v
C2
[2]
[2]+
*22
22
4
2
Phyllic disphenoid Half-turn tetrahedron diagram.png
Half-turn tetrahedron diagram2.png

Two pairs of equal scalene or isosceles triangles

This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C2 isomorphic to the cyclic group, Z2.

C2[2]+222

General properties[edit]

Volume[edit]

The volume of a tetrahedron is given by the pyramid volume formula:

{\displaystyle V={\frac {1}{3}}A_{0}\,h\,}

where A0 is the area of the base and h is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apexes to the opposite faces are inversely proportional to the areas of these faces.

For a tetrahedron with vertices a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is 1/6|det(ad, bd, cd)|, or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding

V={\frac {|(\mathbf {a} -\mathbf {d} )\cdot ((\mathbf {b} -\mathbf {d} )\times (\mathbf {c} -\mathbf {d} ))|}{6}}.

If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so

V={\frac {|\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )|}{6}},

where a, b, and c represent three edges that meet at one vertex, and a · (b × c) is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 1/6 of the volume of any parallelepiped that shares three converging edges with it.

The absolute value of the scalar triple product can be represented as the following absolute values of determinants:

6\cdot V={\begin{Vmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{Vmatrix}} or 6\cdot V={\begin{Vmatrix}\mathbf {a} \\\mathbf {b} \\\mathbf {c} \end{Vmatrix}} where {\displaystyle {\begin{cases}\mathbf {a} =(a_{1},a_{2},a_{3}),\\\mathbf {b} =(b_{1},b_{2},b_{3}),\\\mathbf {c} =(c_{1},c_{2},c_{3}),\end{cases}}} are expressed as row or column vectors.

Hence

36\cdot V^{2}={\begin{vmatrix}\mathbf {a^{2}} &\mathbf {a} \cdot \mathbf {b} &\mathbf {a} \cdot \mathbf {c} \\\mathbf {a} \cdot \mathbf {b} &\mathbf {b^{2}} &\mathbf {b} \cdot \mathbf {c} \\\mathbf {a} \cdot \mathbf {c} &\mathbf {b} \cdot \mathbf {c} &\mathbf {c^{2}} \end{vmatrix}} where {\displaystyle {\begin{cases}\mathbf {a} \cdot \mathbf {b} =ab\cos {\gamma },\\\mathbf {b} \cdot \mathbf {c} =bc\cos {\alpha },\\\mathbf {a} \cdot \mathbf {c} =ac\cos {\beta }.\end{cases}}}

which gives

V={\frac {abc}{6}}{\sqrt {1+2\cos {\alpha }\cos {\beta }\cos {\gamma }-\cos ^{2}{\alpha }-\cos ^{2}{\beta }-\cos ^{2}{\gamma }}},\,

where α, β, γ are the plane angles occurring in vertex d. The angle α, is the angle between the two edges connecting the vertex d to the vertices b and c. The angle β, does so for the vertices a and c, while γ, is defined by the position of the vertices a and b.

If we do not require that d = 0 then

{\displaystyle 6\cdot V=\left|\det \left({\begin{matrix}a_{1}&b_{1}&c_{1}&d_{1}\\a_{2}&b_{2}&c_{2}&d_{2}\\a_{3}&b_{3}&c_{3}&d_{3}\\1&1&1&1\end{matrix}}\right)\right|\,.}

Given the distances between the vertices of a tetrahedron the volume can be computed using the Cayley–Menger determinant:

288\cdot V^{2}={\begin{vmatrix}0&1&1&1&1\\1&0&d_{12}^{2}&d_{13}^{2}&d_{14}^{2}\\1&d_{12}^{2}&0&d_{23}^{2}&d_{24}^{2}\\1&d_{13}^{2}&d_{23}^{2}&0&d_{34}^{2}\\1&d_{14}^{2}&d_{24}^{2}&d_{34}^{2}&0\end{vmatrix}}

where the subscripts i, j ∈ {1, 2, 3, 4} represent the vertices {a, b, c, d} and dij is the pairwise distance between them – i.e., the length of the edge connecting the two vertices. A negative value of the determinant means that a tetrahedron cannot be constructed with the given distances. This formula, sometimes called Tartaglia's formula, is essentially due to the painter Piero della Francesca in the 15th century, as a three dimensional analogue of the 1st century Heron's formula for the area of a triangle.[8]

Denote a, b, c be three edges that meet at a point, and x, y, z the opposite edges. Let V be the volume of the tetrahedron; then[9]

{\displaystyle V={\frac {\sqrt {4a^{2}b^{2}c^{2}-a^{2}X^{2}-b^{2}Y^{2}-c^{2}Z^{2}+XYZ}}{12}}}

where

{\displaystyle {\begin{aligned}X&=b^{2}+c^{2}-x^{2},\\Y&=a^{2}+c^{2}-y^{2},\\Z&=a^{2}+b^{2}-z^{2}.\end{aligned}}}

The above formula uses six lengths of edges, and the following formula uses three lengths of edges and three angles.

{\displaystyle V={\frac {abc}{6}}{\sqrt {1+2\cos {\alpha }\cos {\beta }\cos {\gamma }-\cos ^{2}{\alpha }-\cos ^{2}{\beta }-\cos ^{2}{\gamma }}}}

Heron-type formula for the volume of a tetrahedron[edit]

Six edge-lengths of Tetrahedron

If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then[10]

{\displaystyle V={\frac {\sqrt {\,(-p+q+r+s)\,(p-q+r+s)\,(p+q-r+s)\,(p+q+r-s)}}{192\,u\,v\,w}}}

where

{\displaystyle {\begin{aligned}p&={\sqrt {xYZ}},&q&={\sqrt {yZX}},&r&={\sqrt {zXY}},&s&={\sqrt {xyz}},\end{aligned}}}
{\displaystyle {\begin{aligned}X&=(w-U+v)\,(U+v+w),&x&=(U-v+w)\,(v-w+U),\\Y&=(u-V+w)\,(V+w+u),&y&=(V-w+u)\,(w-u+V),\\Z&=(v-W+u)\,(W+u+v),&z&=(W-u+v)\,(u-v+W).\end{aligned}}}

Volume divider[edit]

Any plane containing a bimedian (connector of opposite edges' midpoints) of a tetrahedron bisects the volume of the tetrahedron.[11]

Non-Euclidean volume[edit]

For tetrahedra in hyperbolic space or in three-dimensional elliptic geometry, the dihedral angles of the tetrahedron determine its shape and hence its volume. In these cases, the volume is given by the Murakami–Yano formula.[12] However, in Euclidean space, scaling a tetrahedron changes its volume but not its dihedral angles, so no such formula can exist.

Distance between the edges[edit]

Any two opposite edges of a tetrahedron lie on two skew lines, and the distance between the edges is defined as the distance between the two skew lines. Let d be the distance between the skew lines formed by opposite edges a and bc as calculated here. Then another volume formula is given by

V={\frac {d|(\mathbf {a} \times \mathbf {(b-c)} )|}{6}}.

Properties analogous to those of a triangle[edit]

The tetrahedron has many properties analogous to those of a triangle, including an insphere, circumsphere, medial tetrahedron, and exspheres. It has respective centers such as incenter, circumcenter, excenters, Spieker center and points such as a centroid. However, there is generally no orthocenter in the sense of intersecting altitudes.[13]

Gaspard Monge found a center that exists in every tetrahedron, now known as the Monge point: the point where the six midplanes of a tetrahedron intersect. A midplane is defined as a plane that is orthogonal to an edge joining any two vertices that also contains the centroid of an opposite edge formed by joining the other two vertices. If the tetrahedron's altitudes do intersect, then the Monge point and the orthocenter coincide to give the class of orthocentric tetrahedron.

An orthogonal line dropped from the Monge point to any face meets that face at the midpoint of the line segment between that face's orthocenter and the foot of the altitude dropped from the opposite vertex.

A line segment joining a vertex of a tetrahedron with the centroid of the opposite face is called a median and a line segment joining the midpoints of two opposite edges is called a bimedian of the tetrahedron. Hence there are four medians and three bimedians in a tetrahedron. These seven line segments are all concurrent at a point called the centroid of the tetrahedron.[14] In addition the four medians are divided in a 3:1 ratio by the centroid (see Commandino's theorem). The centroid of a tetrahedron is the midpoint between its Monge point and circumcenter. These points define the Euler line of the tetrahedron that is analogous to the Euler line of a triangle.

The nine-point circle of the general triangle has an analogue in the circumsphere of a tetrahedron's medial tetrahedron. It is the twelve-point sphere and besides the centroids of the four faces of the reference tetrahedron, it passes through four substitute Euler points, one third of the way from the Monge point toward each of the four vertices. Finally it passes through the four base points of orthogonal lines dropped from each Euler point to the face not containing the vertex that generated the Euler point.[15]

The center T of the twelve-point sphere also lies on the Euler line. Unlike its triangular counterpart, this center lies one third of the way from the Monge point M towards the circumcenter. Also, an orthogonal line through T to a chosen face is coplanar with two other orthogonal lines to the same face. The first is an orthogonal line passing through the corresponding Euler point to the chosen face. The second is an orthogonal line passing through the centroid of the chosen face. This orthogonal line through the twelve-point center lies midway between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelve-point center lies at the midpoint of the corresponding Euler point and the orthocenter for that face.

The radius of the twelve-point sphere is one third of the circumradius of the reference tetrahedron.

There is a relation among the angles made by the faces of a general tetrahedron given by[16]

{\begin{vmatrix}-1&\cos {(\alpha _{12})}&\cos {(\alpha _{13})}&\cos {(\alpha _{14})}\\\cos {(\alpha _{12})}&-1&\cos {(\alpha _{23})}&\cos {(\alpha _{24})}\\\cos {(\alpha _{13})}&\cos {(\alpha _{23})}&-1&\cos {(\alpha _{34})}\\\cos {(\alpha _{14})}&\cos {(\alpha _{24})}&\cos {(\alpha _{34})}&-1\\\end{vmatrix}}=0\,

where αij is the angle between the faces i and j.

The geometric median of the vertex position coordinates of a tetrahedron and its isogonic center are associated, under circumstances analogous to those observed for a triangle. Lorenz Lindelöf found that, corresponding to any given tetrahedron is a point now known as an isogonic center, O, at which the solid angles subtended by the faces are equal, having a common value of π sr, and at which the angles subtended by opposite edges are equal.[17] A solid angle of π sr is one quarter of that subtended by all of space. When all the solid angles at the vertices of a tetrahedron are smaller than π sr, O lies inside the tetrahedron, and because the sum of distances from O to the vertices is a minimum, O coincides with the geometric median, M, of the vertices. In the event that the solid angle at one of the vertices, v, measures exactly π sr, then O and M coincide with v. If however, a tetrahedron has a vertex, v, with solid angle greater than π sr, M still corresponds to v, but O lies outside the tetrahedron.

Geometric relations[edit]

A tetrahedron is a 3-simplex. Unlike the case of the other Platonic solids, all the vertices of a regular tetrahedron are equidistant from each other (they are the only possible arrangement of four equidistant points in 3-dimensional space).

A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual.

A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are

(+1, +1, +1);
(−1, −1, +1);
(−1, +1, −1);
(+1, −1, −1).

This yields a tetrahedron with edge-length 2√2, centered at the origin. For the other tetrahedron (which is dual to the first), reverse all the signs. These two tetrahedra's vertices combined are the vertices of a cube, demonstrating that the regular tetrahedron is the 3-demicube.

The volume of this tetrahedron is one-third the volume of the cube. Combining both tetrahedra gives a regular polyhedral compound called the compound of two tetrahedra or stella octangula.

The interior of the stella octangula is an octahedron, and correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e., rectifying the tetrahedron).

The above embedding divides the cube into five tetrahedra, one of which is regular. In fact, five is the minimum number of tetrahedra required to compose a cube. To see this, starting from a base tetrahedron with 4 vertices, each added tetrahedra adds at most 1 new vertex, so at least 4 more must be added to make a cube, which has 8 vertices.

Inscribing tetrahedra inside the regular compound of five cubes gives two more regular compounds, containing five and ten tetrahedra.

Regular tetrahedra cannot tessellate space by themselves, although this result seems likely enough that Aristotle claimed it was possible. However, two regular tetrahedra can be combined with an octahedron, giving a rhombohedron that can tile space.

However, several irregular tetrahedra are known, of which copies can tile space, for instance the disphenoid tetrahedral honeycomb. The complete list remains an open problem.[18]

If one relaxes the requirement that the tetrahedra be all the same shape, one can tile space using only tetrahedra in many different ways. For example, one can divide an octahedron into four identical tetrahedra and combine them again with two regular ones. (As a side-note: these two kinds of tetrahedron have the same volume.)

The tetrahedron is unique among the uniform polyhedra in possessing no parallel faces.

A law of sines for tetrahedra and the space of all shapes of tetrahedra[edit]

Tetra.png

Main article: Trigonometry of a tetrahedron

A corollary of the usual law of sines is that in a tetrahedron with vertices O, A, B, C, we have

Sours: https://en.wikipedia.org/wiki/Tetrahedron

Of vectors volume a tetrahedron

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HOW TO FIND VOLUME OF TETRAHEDRON - VECTOR DIFFERENTIAL CALCULUS SOLVED PROBLEM 6 - LECTURE 6

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