Parallelepiped (original) (raw)

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Hexahedron with parallelogram faces

Parallelepiped
Parallelepiped
Type PrismPlesiohedron
Faces 6 parallelograms
Edges 12
Vertices 8
Symmetry group C i, [2+,2+], (×), order 2
Properties convex, zonohedron

In geometry, a parallelepiped is a three-dimensional figure formed by six parallelograms (the term rhomboid is also sometimes used with this meaning). By analogy, it relates to a parallelogram just as a cube relates to a square.[a]

Three equivalent definitions of parallelepiped are

The rectangular cuboid (six rectangular faces), cube (six square faces), and the rhombohedron (six rhombus faces) are all special cases of parallelepiped.

"Parallelepiped" is now usually pronounced or ;[1] traditionally it was PARR-ə-lel-EP-ih-ped[2] because of its etymology in Greek παραλληλεπίπεδον parallelepipedon (with short -i-), a body "having parallel planes".

Parallelepipeds are a subclass of the prismatoids.

Any of the three pairs of parallel faces can be viewed as the base planes of the prism. A parallelepiped has three sets of four parallel edges; the edges within each set are of equal length.

Parallelepipeds result from linear transformations of a cube (for the non-degenerate cases: the bijective linear transformations).

Since each face has point symmetry, a parallelepiped is a zonohedron. Also the whole parallelepiped has point symmetry Ci (see also triclinic). Each face is, seen from the outside, the mirror image of the opposite face. The faces are in general chiral, but the parallelepiped is not.

A space-filling tessellation is possible with congruent copies of any parallelepiped.

Parallelepiped, generated by three vectors

A parallelepiped is a prism with a parallelogram as base. Hence the volume V {\displaystyle V} {\displaystyle V} of a parallelepiped is the product of the base area B {\displaystyle B} {\displaystyle B} and the height h {\displaystyle h} {\displaystyle h} (see diagram). With

V = B ⋅ h = ( | a | | b | sin ⁡ γ ) ⋅ | c | | cos ⁡ θ | = | a × b | | c | | cos ⁡ θ | = | ( a × b ) ⋅ c | . {\displaystyle V=B\cdot h=\left(\left|\mathbf {a} \right|\left|\mathbf {b} \right|\sin \gamma \right)\cdot \left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\mathbf {a} \times \mathbf {b} \right|\left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.} {\displaystyle V=B\cdot h=\left(\left|\mathbf {a} \right|\left|\mathbf {b} \right|\sin \gamma \right)\cdot \left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\mathbf {a} \times \mathbf {b} \right|\left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.}The mixed product of three vectors is called triple product. It can be described by a determinant. Hence for a = ( a 1 , a 2 , a 3 ) T , b = ( b 1 , b 2 , b 3 ) T , c = ( c 1 , c 2 , c 3 ) T , {\displaystyle \mathbf {a} =(a_{1},a_{2},a_{3})^{\mathsf {T}},~\mathbf {b} =(b_{1},b_{2},b_{3})^{\mathsf {T}},~\mathbf {c} =(c_{1},c_{2},c_{3})^{\mathsf {T}},} {\displaystyle \mathbf {a} =(a_{1},a_{2},a_{3})^{\mathsf {T}},~\mathbf {b} =(b_{1},b_{2},b_{3})^{\mathsf {T}},~\mathbf {c} =(c_{1},c_{2},c_{3})^{\mathsf {T}},} the volume is:

| V = | det [ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ] | . {\displaystyle V=\left|\det {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}\right|.} {\displaystyle V=\left|\det {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}\right|.} | (V1) | | ----------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------- |

Another way to prove (V1) is to use the scalar component in the direction of a × b {\displaystyle \mathbf {a} \times \mathbf {b} } {\displaystyle \mathbf {a} \times \mathbf {b} } of vector c {\displaystyle \mathbf {c} } {\displaystyle \mathbf {c} }: V = | a × b | | scal a × b ⁡ c | = | a × b | | ( a × b ) ⋅ c | | a × b | = | ( a × b ) ⋅ c | . {\displaystyle {\begin{aligned}V=\left|\mathbf {a} \times \mathbf {b} \right|\left|\operatorname {scal} _{\mathbf {a} \times \mathbf {b} }\mathbf {c} \right|=\left|\mathbf {a} \times \mathbf {b} \right|{\frac {\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|}{\left|\mathbf {a} \times \mathbf {b} \right|}}=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.\end{aligned}}} {\displaystyle {\begin{aligned}V=\left|\mathbf {a} \times \mathbf {b} \right|\left|\operatorname {scal} _{\mathbf {a} \times \mathbf {b} }\mathbf {c} \right|=\left|\mathbf {a} \times \mathbf {b} \right|{\frac {\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|}{\left|\mathbf {a} \times \mathbf {b} \right|}}=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.\end{aligned}}}The result follows.

An alternative representation of the volume uses geometric properties (angles and edge lengths) only:

V = a b c 1 + 2 cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) − cos 2 ⁡ ( α ) − cos 2 ⁡ ( β ) − cos 2 ⁡ ( γ ) , {\displaystyle V=abc{\sqrt {1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}},} {\displaystyle V=abc{\sqrt {1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}},} (V2)

where α = ∠ ( b , c ) {\displaystyle \alpha =\angle (\mathbf {b} ,\mathbf {c} )} {\displaystyle \alpha =\angle (\mathbf {b} ,\mathbf {c} )}, β = ∠ ( a , c ) {\displaystyle \beta =\angle (\mathbf {a} ,\mathbf {c} )} {\displaystyle \beta =\angle (\mathbf {a} ,\mathbf {c} )}, γ = ∠ ( a , b ) {\displaystyle \gamma =\angle (\mathbf {a} ,\mathbf {b} )} {\displaystyle \gamma =\angle (\mathbf {a} ,\mathbf {b} )}, and a , b , c {\displaystyle a,b,c} {\displaystyle a,b,c} are the edge lengths.

Proof of (V2)

The proof of (V2) uses properties of a determinant and the geometric interpretation of the dot product:

Let M {\displaystyle M} {\displaystyle M} be the 3×3-matrix, whose columns are the vectors a , b , c {\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} } {\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} } (see above). Then the following is true: V 2 = ( det M ) 2 = det M det M = det M T det M = det ( M T M ) = det [ a ⋅ a a ⋅ b a ⋅ c b ⋅ a b ⋅ b b ⋅ c c ⋅ a c ⋅ b c ⋅ c ] = a 2 ( b 2 c 2 − b 2 c 2 cos 2 ⁡ ( α ) ) − a b cos ⁡ ( γ ) ( a b cos ⁡ ( γ ) c 2 − a c cos ⁡ ( β ) b c cos ⁡ ( α ) ) + a c cos ⁡ ( β ) ( a b cos ⁡ ( γ ) b c cos ⁡ ( α ) − a c cos ⁡ ( β ) b 2 ) = a 2 b 2 c 2 − a 2 b 2 c 2 cos 2 ⁡ ( α ) − a 2 b 2 c 2 cos 2 ⁡ ( γ ) + a 2 b 2 c 2 cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) + a 2 b 2 c 2 cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) − a 2 b 2 c 2 cos 2 ⁡ ( β ) = a 2 b 2 c 2 ( 1 − cos 2 ⁡ ( α ) − cos 2 ⁡ ( γ ) + cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) + cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) − cos 2 ⁡ ( β ) ) = a 2 b 2 c 2 ( 1 + 2 cos ⁡ ( α ) cos ⁡ ( β ) cos ⁡ ( γ ) − cos 2 ⁡ ( α ) − cos 2 ⁡ ( β ) − cos 2 ⁡ ( γ ) ) . {\displaystyle {\begin{aligned}V^{2}&=\left(\det M\right)^{2}=\det M\det M=\det M^{\mathsf {T}}\det M=\det(M^{\mathsf {T}}M)\\&=\det {\begin{bmatrix}\mathbf {a} \cdot \mathbf {a} &\mathbf {a} \cdot \mathbf {b} &\mathbf {a} \cdot \mathbf {c} \\\mathbf {b} \cdot \mathbf {a} &\mathbf {b} \cdot \mathbf {b} &\mathbf {b} \cdot \mathbf {c} \\\mathbf {c} \cdot \mathbf {a} &\mathbf {c} \cdot \mathbf {b} &\mathbf {c} \cdot \mathbf {c} \end{bmatrix}}\\&=\ a^{2}\left(b^{2}c^{2}-b^{2}c^{2}\cos ^{2}(\alpha )\right)\\&\quad -ab\cos(\gamma )\left(ab\cos(\gamma )c^{2}-ac\cos(\beta )\;bc\cos(\alpha )\right)\\&\quad +ac\cos(\beta )\left(ab\cos(\gamma )bc\cos(\alpha )-ac\cos(\beta )b^{2}\right)\\&=\ a^{2}b^{2}c^{2}-a^{2}b^{2}c^{2}\cos ^{2}(\alpha )\\&\quad -a^{2}b^{2}c^{2}\cos ^{2}(\gamma )+a^{2}b^{2}c^{2}\cos(\alpha )\cos(\beta )\cos(\gamma )\\&\quad +a^{2}b^{2}c^{2}\cos(\alpha )\cos(\beta )\cos(\gamma )-a^{2}b^{2}c^{2}\cos ^{2}(\beta )\\&=\ a^{2}b^{2}c^{2}\left(1-\cos ^{2}(\alpha )-\cos ^{2}(\gamma )+\cos(\alpha )\cos(\beta )\cos(\gamma )+\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\beta )\right)\\&=\ a^{2}b^{2}c^{2}\;\left(1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )\right).\end{aligned}}} {\displaystyle {\begin{aligned}V^{2}&=\left(\det M\right)^{2}=\det M\det M=\det M^{\mathsf {T}}\det M=\det(M^{\mathsf {T}}M)\\&=\det {\begin{bmatrix}\mathbf {a} \cdot \mathbf {a} &\mathbf {a} \cdot \mathbf {b} &\mathbf {a} \cdot \mathbf {c} \\\mathbf {b} \cdot \mathbf {a} &\mathbf {b} \cdot \mathbf {b} &\mathbf {b} \cdot \mathbf {c} \\\mathbf {c} \cdot \mathbf {a} &\mathbf {c} \cdot \mathbf {b} &\mathbf {c} \cdot \mathbf {c} \end{bmatrix}}\\&=\ a^{2}\left(b^{2}c^{2}-b^{2}c^{2}\cos ^{2}(\alpha )\right)\\&\quad -ab\cos(\gamma )\left(ab\cos(\gamma )c^{2}-ac\cos(\beta )\;bc\cos(\alpha )\right)\\&\quad +ac\cos(\beta )\left(ab\cos(\gamma )bc\cos(\alpha )-ac\cos(\beta )b^{2}\right)\\&=\ a^{2}b^{2}c^{2}-a^{2}b^{2}c^{2}\cos ^{2}(\alpha )\\&\quad -a^{2}b^{2}c^{2}\cos ^{2}(\gamma )+a^{2}b^{2}c^{2}\cos(\alpha )\cos(\beta )\cos(\gamma )\\&\quad +a^{2}b^{2}c^{2}\cos(\alpha )\cos(\beta )\cos(\gamma )-a^{2}b^{2}c^{2}\cos ^{2}(\beta )\\&=\ a^{2}b^{2}c^{2}\left(1-\cos ^{2}(\alpha )-\cos ^{2}(\gamma )+\cos(\alpha )\cos(\beta )\cos(\gamma )+\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\beta )\right)\\&=\ a^{2}b^{2}c^{2}\;\left(1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )\right).\end{aligned}}}

(The last steps use a ⋅ a = a 2 {\displaystyle \mathbf {a} \cdot \mathbf {a} =a^{2}} {\displaystyle \mathbf {a} \cdot \mathbf {a} =a^{2}}, ..., a ⋅ b = a b cos ⁡ γ {\displaystyle \mathbf {a} \cdot \mathbf {b} =ab\cos \gamma } {\displaystyle \mathbf {a} \cdot \mathbf {b} =ab\cos \gamma }, a ⋅ c = a c cos ⁡ β {\displaystyle \mathbf {a} \cdot \mathbf {c} =ac\cos \beta } {\displaystyle \mathbf {a} \cdot \mathbf {c} =ac\cos \beta }, b ⋅ c = b c cos ⁡ α {\displaystyle \mathbf {b} \cdot \mathbf {c} =bc\cos \alpha } {\displaystyle \mathbf {b} \cdot \mathbf {c} =bc\cos \alpha }, ...)

Corresponding tetrahedron

The volume of any tetrahedron that shares three converging edges of a parallelepiped is equal to one sixth of the volume of that parallelepiped (see proof).

The surface area of a parallelepiped is the sum of the areas of the bounding parallelograms: A = 2 ⋅ ( | a × b | + | a × c | + | b × c | ) = 2 ( a b sin ⁡ γ + b c sin ⁡ α + c a sin ⁡ β ) . {\displaystyle {\begin{aligned}A&=2\cdot \left(|\mathbf {a} \times \mathbf {b} |+|\mathbf {a} \times \mathbf {c} |+|\mathbf {b} \times \mathbf {c} |\right)\\&=2\left(ab\sin \gamma +bc\sin \alpha +ca\sin \beta \right).\end{aligned}}} {\displaystyle {\begin{aligned}A&=2\cdot \left(|\mathbf {a} \times \mathbf {b} |+|\mathbf {a} \times \mathbf {c} |+|\mathbf {b} \times \mathbf {c} |\right)\\&=2\left(ab\sin \gamma +bc\sin \alpha +ca\sin \beta \right).\end{aligned}}}(For labeling: see previous section.)

Special cases by symmetry

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Form Cube Square cuboid Trigonal trapezohedron Rectangular cuboid Right rhombic prism Right parallelogrammic prism Oblique rhombic prism
Constraints a = b = c {\displaystyle a=b=c} {\displaystyle a=b=c} α = β = γ = 90 ∘ {\displaystyle \alpha =\beta =\gamma =90^{\circ }} {\displaystyle \alpha =\beta =\gamma =90^{\circ }} a = b {\displaystyle a=b} {\displaystyle a=b} α = β = γ = 90 ∘ {\displaystyle \alpha =\beta =\gamma =90^{\circ }} {\displaystyle \alpha =\beta =\gamma =90^{\circ }} a = b = c {\displaystyle a=b=c} {\displaystyle a=b=c} α = β = γ {\displaystyle \alpha =\beta =\gamma } {\displaystyle \alpha =\beta =\gamma } α = β = γ = 90 ∘ {\displaystyle \alpha =\beta =\gamma =90^{\circ }} {\displaystyle \alpha =\beta =\gamma =90^{\circ }} a = b {\displaystyle a=b} {\displaystyle a=b} α = β = 90 ∘ {\displaystyle \alpha =\beta =90^{\circ }} {\displaystyle \alpha =\beta =90^{\circ }} α = β = 90 ∘ {\displaystyle \alpha =\beta =90^{\circ }} {\displaystyle \alpha =\beta =90^{\circ }} a = b {\displaystyle a=b} {\displaystyle a=b} α = β {\displaystyle \alpha =\beta } {\displaystyle \alpha =\beta }
Symmetry Ohorder 48 D4horder 16 D3dorder 12 D2horder 8 C2horder 4
Image
Faces 6 squares 2 squares,4 rectangles 6 rhombi 6 rectangles 4 rectangles,2 rhombi 4 rectangles,2 parallelograms 2 rhombi,4 parallelograms

Perfect parallelepiped

[edit]

A perfect parallelepiped is a parallelepiped with integer-length edges, face diagonals, and space diagonals. In 2009, dozens of perfect parallelepipeds were shown to exist,[3] answering an open question of Richard Guy. One example has edges 271, 106, and 103, minor face diagonals 101, 266, and 255, major face diagonals 183, 312, and 323, and space diagonals 374, 300, 278, and 272.

Some perfect parallelepipeds having two rectangular faces are known. But it is not known whether there exist any with all faces rectangular; such a case would be called a perfect cuboid.

Coxeter called the generalization of a parallelepiped in higher dimensions a parallelotope. In modern literature, the term parallelepiped is often used in higher (or arbitrary finite) dimensions as well.[4]

Specifically in _n_-dimensional space it is called _n_-dimensional parallelotope, or simply n-parallelotope (or n-parallelepiped). Thus a parallelogram is a 2-parallelotope and a parallelepiped is a 3-parallelotope.

The diagonals of an _n_-parallelotope intersect at one point and are bisected by this point. Inversion in this point leaves the _n_-parallelotope unchanged. See also Fixed points of isometry groups in Euclidean space.

The edges radiating from one vertex of a _k_-parallelotope form a _k_-frame ( v 1 , … , v n ) {\displaystyle (v_{1},\ldots ,v_{n})} {\displaystyle (v_{1},\ldots ,v_{n})} of the vector space, and the parallelotope can be recovered from these vectors, by taking linear combinations of the vectors, with weights between 0 and 1.

The _n_-volume of an _n_-parallelotope embedded in R m {\displaystyle \mathbb {R} ^{m}} {\displaystyle \mathbb {R} ^{m}} where m ≥ n {\displaystyle m\geq n} {\displaystyle m\geq n} can be computed by means of the Gram determinant. Alternatively, the volume is the norm of the exterior product of the vectors: V = ‖ v 1 ∧ ⋯ ∧ v n ‖ . {\displaystyle V=\left\|v_{1}\wedge \cdots \wedge v_{n}\right\|.} {\displaystyle V=\left\|v_{1}\wedge \cdots \wedge v_{n}\right\|.}

If m = n, this amounts to the absolute value of the determinant of matrix formed by the components of the n vectors.

A formula to compute the volume of an n-parallelotope P in R n {\displaystyle \mathbb {R} ^{n}} {\displaystyle \mathbb {R} ^{n}}, whose n + 1 vertices are V 0 , V 1 , … , V n {\displaystyle V_{0},V_{1},\ldots ,V_{n}} {\displaystyle V_{0},V_{1},\ldots ,V_{n}}, is V o l ( P ) = | det ( [ V 0 1 ] T , [ V 1 1 ] T , … , [ V n 1 ] T ) | , {\displaystyle \mathrm {Vol} (P)=\left|\det \left(\left[V_{0}\ 1\right]^{\mathsf {T}},\left[V_{1}\ 1\right]^{\mathsf {T}},\ldots ,\left[V_{n}\ 1\right]^{\mathsf {T}}\right)\right|,} {\displaystyle \mathrm {Vol} (P)=\left|\det \left(\left[V_{0}\ 1\right]^{\mathsf {T}},\left[V_{1}\ 1\right]^{\mathsf {T}},\ldots ,\left[V_{n}\ 1\right]^{\mathsf {T}}\right)\right|,}where [ V i 1 ] {\displaystyle [V_{i}\ 1]} {\displaystyle [V_{i}\ 1]} is the row vector formed by the concatenation of the components of V i {\displaystyle V_{i}} {\displaystyle V_{i}} and 1.

Similarly, the volume of any _n_-simplex that shares n converging edges of a parallelotope has a volume equal to one 1/n! of the volume of that parallelotope.

The term parallelepiped stems from Ancient Greek παραλληλεπίπεδον (parallēlepípedon, "body with parallel plane surfaces"), from parallēl ("parallel") + epípedon ("plane surface"), from epí- ("on") + pedon ("ground"). Thus the faces of a parallelepiped are planar, with opposite faces being parallel.[5][6]

In English, the term parallelipipedon is attested in a 1570 translation of Euclid's Elements by Henry Billingsley. The spelling parallelepipedum is used in the 1644 edition of Pierre Hérigone's Cursus mathematicus. In 1663, the present-day parallelepiped is attested in Walter Charleton's Chorea gigantum.[5]

Charles Hutton's Dictionary (1795) shows parallelopiped and parallelopipedon, showing the influence of the combining form parallelo-, as if the second element were pipedon rather than epipedon. Noah Webster (1806) includes the spelling parallelopiped. The 1989 edition of the Oxford English Dictionary describes parallelopiped (and parallelipiped) explicitly as incorrect forms, but these are listed without comment in the 2004 edition, and only pronunciations with the emphasis on the fifth syllable pi (/paɪ/) are given.

  1. ^ In Euclidean geometry, the four concepts—parallelepiped and cube in three dimensions, parallelogram and square in two dimensions—are defined, but in the context of a more general affine geometry, in which angles are not differentiated, only parallelograms and parallelepipeds exist.

  2. ^ "parallelepiped". Dictionary.com Unabridged (Online). n.d.

  3. ^ Oxford English Dictionary 1904; Webster's Second International 1947

  4. ^ Sawyer, Jorge F.; Reiter, Clifford A. (2011). "Perfect Parallelepipeds Exist". Mathematics of Computation. 80 (274): 1037–1040. arXiv:0907.0220. doi:10.1090/s0025-5718-2010-02400-7. S2CID 206288198..

  5. ^ Morgan, C. L. (1974). Embedding metric spaces in Euclidean space. Journal of Geometry, 5(1), 101–107. https://doi.org/10.1007/bf01954540

  6. ^ a b "parallelepiped". Oxford English Dictionary. 1933.

  7. ^ parallhlepi/pedon. Liddell, Henry George; Scott, Robert; A Greek–English Lexicon at the Perseus Project.