# Parallelepiped

Parallelepiped

Under a parallelepiped (from the Greek επίπεδον epipedon "surface"; synonyms: Spat , parallelepiped , parallelotope ) refers to a geometrical body , the six pairs of congruent (coincident) in parallel planes lying parallelograms is limited. The term spat comes from Kalkspat ( calcite , chemically: CaCO 3 ), the crystals of which have the shape of a parallelepiped.

A parallelepiped has twelve edges, four of which are parallel and of equal length, and eight corners in which these edges converge at a maximum of three different angles. If these three edges meeting at a corner point are represented as vectors , then the volume of the parallelepiped results from the amount of the spatial product (mixed scalar and cross product ). ${\ displaystyle {\ vec {a}}, {\ vec {b}}, {\ vec {c}}}$

## volume

Parallelepiped generated from 3 vectors

The volume is the product of the base area ( parallelogram ) and the parallelepiped height . With (where the angle between and is) and the height ( is the angle between and the normal to the base) results ${\ displaystyle V}$${\ displaystyle G}$${\ displaystyle h}$${\ displaystyle G = | {\ vec {a}} | \ cdot | {\ vec {b}} | \ cdot \ sin \ gamma = | {\ vec {a}} \ times {\ vec {b}} | }$${\ displaystyle \ gamma}$${\ displaystyle {\ vec {a}}}$${\ displaystyle {\ vec {b}}}$${\ displaystyle h = | {\ vec {c}} | \ cdot | \ cos \ theta |}$${\ displaystyle \ theta}$${\ displaystyle {\ vec {c}}}$

${\ displaystyle V = G \ cdot h = (| {\ vec {a}} || {\ vec {b}} | \ sin \ gamma) \ cdot | {\ vec {c}} || \ cos \ theta | = | {\ vec {a}} \ times {\ vec {b}} | \; | {\ vec {c}} | \; | \ cos \ theta |}$
${\ displaystyle = | ({\ vec {a}} \ times {\ vec {b}}) \ cdot {\ vec {c}} | \.}$

The mixed product is called a late product . It can be written as a determinant. For then the volume is: ${\ displaystyle {\ vec {a}} = (a_ {1}, a_ {2}, a_ {3}) ^ {T}, {\ vec {b}} = (b_ {1}, b_ {2} , b_ {3}) ^ {T}, {\ vec {c}} = (c_ {1}, c_ {2}, c_ {3}) ^ {T}}$

(V1) ${\ displaystyle \ quad V = \ left | \ det {\ begin {bmatrix} a_ {1} & b_ {1} & c_ {1} \\ a_ {2} & b_ {2} & c_ {2} \\ a_ {3} & b_ {3} & c_ {3} \ end {bmatrix}} \; \ right |}$

A formula for the volume that only depends on the geometric properties (edge ​​lengths, angles between adjacent edges) is:

(V2)${\ displaystyle \ quad V = abc {\ sqrt {1 + 2 \ cos (\ alpha) \ cos (\ beta) \ cos (\ gamma) - \ cos ^ {2} (\ alpha) - \ cos ^ {2 } (\ beta) - \ cos ^ {2} (\ gamma)}}.}$

Where and are the edge lengths. ${\ displaystyle \ \ alpha = \ angle ({\ vec {b}}, {\ vec {c}}), \; \ beta = \ angle ({\ vec {a}}, {\ vec {c}} ), \; \ gamma = \ angle ({\ vec {a}}, {\ vec {b}}) \}$${\ displaystyle a, b, c}$

The proof of (V2) can be done with the properties of a determinant and the geometric interpretation of the scalar product . Let it be the 3x3 matrix whose column vectors are the vectors . Then applies ${\ displaystyle M}$${\ displaystyle {\ vec {a}}, {\ vec {b}}, {\ vec {c}}}$

${\ displaystyle V ^ {2} = (\ det M) ^ {2} = \ det M \ det M = \ det M ^ {T} \ det M = \ det (M ^ {T} M)}$
${\ displaystyle = \ det {\ begin {bmatrix} {\ vec {a}} \ cdot {\ vec {a}} & {\ vec {a}} \ cdot {\ vec {b}} & {\ vec { a}} \ cdot {\ vec {c}} \\ {\ vec {b}} \ cdot {\ vec {a}} & {\ vec {b}} \ cdot {\ vec {b}} & {\ vec {b}} \ cdot {\ vec {c}} \\ {\ vec {c}} \ cdot {\ vec {a}} & {\ vec {c}} \ cdot {\ vec {b}} & {\ vec {c}} \ cdot {\ vec {c}} \ end {bmatrix}} = \ a ^ {2} b ^ {2} c ^ {2} \; \ left (1 + 2 \ cos ( \ alpha) \ cos (\ beta) \ cos (\ gamma) - \ cos ^ {2} (\ alpha) - \ cos ^ {2} (\ beta) - \ cos ^ {2} (\ gamma) \ right ).}$

(In the last step was used.) ${\ display style \ {\ vec {a}} \ cdot {\ vec {a}} = a ^ {2}, ..., \; {\ vec {a}} \ cdot {\ vec {b}} = from \ cos \ gamma, \; {\ vec {a}} \ cdot {\ vec {c}} = ac \ cos \ beta, \; {\ vec {b}} \ cdot {\ vec {c}} = bc \ cos \ alpha \}$

## surface

The area of the surface results from the sum of the individual parallelogram areas

${\ displaystyle A_ {O} = 2 \ cdot \ left (| {\ vec {a}} \ times {\ vec {b}} | + | {\ vec {a}} \ times {\ vec {c}} | + | {\ vec {b}} \ times {\ vec {c}} | \ right)}$
${\ displaystyle = 2 (from \ sin \ gamma + bc \ sin \ alpha + ca \ sin \ beta) \}$.

(For the designations: see previous section.)

## Remarks

• Cuboid (all angles 90 °) and rhombohedra (all edges the same length) are special forms of the parallelepiped. The cube combines both special forms in one figure.
• The parallelepiped is a special (oblique) prism with a parallelogram as its base.
• Every parallelepiped is a space filler , that is, the space can be covered with parallel shifted copies of P in such a way that two of them have at most edge points in common.

## Generalization to the n -dimensional space ( n > 1)

The generalization of the parallelepiped into the n -dimensional space is called for parallelotope or n -parallelotope. The two-dimensional analog of the parallelepiped is the parallelogram . ${\ displaystyle n> 2}$

### definition

An n -parallelotope P is the image of the unit cube E under an affine mapping . The unit cube is a set of points whose coordinates take a value between 0 and 1, that is ${\ displaystyle I ^ {n}}$

${\ displaystyle I ^ {n}: = \ left \ {(x_ {1}, \ dots, x_ {n}) \ mid 0 \ leq x_ {i} \ leq 1 \ right \} \ ,.}$

The parallelotope P is a convex polytope with corners. For its m -dimensional sides are themselves m-dimensional parallelotopes. ${\ displaystyle 2 ^ {n}}$${\ displaystyle m

### volume

An affine mapping can be written as , where is the mapping matrix and the displacement. The volume of the unit cube is one. In order to determine the volume of the parallelotope , it must be investigated how much the affine mapping changes the volume. Since a volume is independent of a displacement, this value is contained in the mapping matrix alone. By calculating the determinant of this matrix, one also obtains the factor by which the volume changes. The dashes denote the amount . If one multiplies this factor by the volume of the unit cube, then trivially applies , therefore ${\ displaystyle f \ colon \ mathbb {R} ^ {n} \ to \ mathbb {R} ^ {n}}$${\ displaystyle f (x) = A \ cdot x + t}$${\ displaystyle A}$${\ displaystyle t}$${\ displaystyle P}$${\ displaystyle | \ det (A) |}$${\ displaystyle | \ cdot |}$${\ displaystyle | \ det (A) | \ cdot 1 = | \ det (A) |}$

${\ displaystyle \ operatorname {vol} (P) = | \ det (A) |}$,

where the mapping matrix is ​​the affine mapping that defines the parallelotope . ${\ displaystyle A}$${\ displaystyle P}$

## Parallelotopes located in higher-dimensional spaces

The parallelotope can , with also be embedded in a higher dimensional space. You can bet again without restricting the generality . The matrix is no longer square for, which makes the calculation via the determinant seem impossible. However, a general formula can be found which contains the formula for square matrices as a special case. ${\ displaystyle f \ colon \ mathbb {R} ^ {n} \ to \ mathbb {R} ^ {m}}$${\ displaystyle f (x): = Ax + t}$${\ displaystyle n \ leq m}$${\ displaystyle t = 0}$${\ displaystyle A \ in \ mathbb {R} ^ {m \ times n}}$${\ displaystyle n

The outer product is a vector space, which can be equipped with a canonical scalar product by ${\ displaystyle \ Lambda ^ {n} \ mathbb {R} ^ {m}}$

${\ displaystyle \ langle {\ vec {v}} _ {1} \ wedge \ ldots \ wedge {\ vec {v}} _ {n}, {\ vec {w}} _ {1} \ wedge \ ldots \ wedge {\ vec {w}} _ {n} \ rangle: = \ det (\ langle {\ vec {v}} _ {i}, {\ vec {w}} _ {j} \ rangle)}$

is defined for blades. The scalar product of multivectors is reduced to that of blades via bilinearity.

As with any scalar product is through

${\ displaystyle \ | X \ |: = {\ sqrt {\ langle X, X \ rangle}}}$

given a norm.

The volume of the parallelotope spanned by the vectors is precisely the norm of the blade; H. ${\ displaystyle {\ vec {v}} _ {1}, \ ldots, {\ vec {v}} _ {n}}$

${\ displaystyle \ operatorname {vol} (P) = \ | {\ vec {v}} _ {1} \ wedge \ ldots \ wedge {\ vec {v}} _ {n} \ | = {\ sqrt {\ det (\ langle {\ vec {v}} _ {i}, {\ vec {v}} _ {j} \ rangle)}}.}$

If now holds , where is the canonical base, then results ${\ displaystyle {\ vec {v}} _ {k} = A {\ vec {e}} _ {k}}$${\ displaystyle ({\ vec {e}} _ {1}, \ ldots, {\ vec {e}} _ {n})}$

${\ displaystyle \ operatorname {vol} (P) = {\ sqrt {\ det (A ^ {T} A)}}.}$

It is called the Gram determinant of the matrix . ${\ displaystyle \ det (A ^ {T} A)}$${\ displaystyle A}$

This is also a geometrical consideration can the linear dependence of make. The vectors are linearly dependent if and only if the parallelotope coincides flat, i.e. if the following applies. The easiest way to do this is to first imagine the case and in which a parallelogram collapses into a segment. ${\ displaystyle ({\ vec {v}} _ {1}, \ ldots, {\ vec {v}} _ {n})}$${\ displaystyle \ operatorname {vol} (P) = 0}$${\ displaystyle n = 2}$${\ displaystyle m = 3}$

The linear mapping is injective if and only if its Gramian determinant does not vanish, i.e. H. if applies. According to the equivalence of and , the mapping is also injective if and only if the outer product of the column vectors of does not vanish, i.e. H. ${\ displaystyle f (x) = Ax}$${\ displaystyle \ det (A ^ {T} A) \ neq 0}$${\ displaystyle X = 0}$${\ displaystyle \ | X \ | = 0}$${\ displaystyle A}$

${\ displaystyle {\ vec {v}} _ {1} \ wedge \ ldots \ wedge {\ vec {v}} _ {n} \ neq 0.}$