# Supplementary mathematics/Parallelepiped

In geometry, a Parallelepiped is a three-dimensional figure formed by six parallelograms (the term rhombus is also sometimes used with this meaning). By analogy, it is related to parallelograms as a cube is related to a square. In Euclidean geometry, four concepts - parallelogram and cube in three dimensions, parallelogram and square in two dimensions - are defined, but in the context of a more general associative geometry, where angles are not distinguished, only parallelograms are. There are parallelograms. There are three equivalent definitions of parallelogram. Parallelogram is also one of the geometric volumes and it is of the type of polyhedra, but it is not of the regular type.

## properties of Parallelepiped

• Each of the three pairs of parallel faces can be seen as the base planes of the prism. A parallelogram has three sets of four parallel edges. The edges of each set are of equal length.
• Parallelograms are obtained from linear transformations of a cube (for non-degenerate cases: two-way linear transformation).
• Since every face has point symmetry, the parallelogram is a zonohedron. Also, all parallelograms have point symmetry "Ci". Every face, from the outside, is a mirror image of the opposite face. The faces are generally chiral, but not parallelograms.
• It is possible to fill the space with consonant transcripts from any base parallel.
• If we diagonalize all the sides of the cube to an equal angle, it becomes a parallelogram, all of whose faces are rhombuses, and the area of the parallelogram faces is equal to the area of the cube faces.
• Parallelogram is one of the types of prisms
• The parallelogram can be said to be one of the polyhedra

## relationship

### Corresponding tetrahedron

The volume of any tetrahedron that has three converging edges of parallelograms has a volume equal to one-sixth of the volume of that parallelogram.

### parallelogram relationship with cube

The volume of a parallelogram with equal sides is equal to that of a cube. Also, the areas of both are equal.

### Volume

A parallelogram is a volume created from three three-dimensional vectors a, b, c and created by external multiplication of the vectors.

## Volume

A parallelogram is a volume created from three three-dimensional vectors a, b, c and created by external multiplication of the vectors.

### volume calculation

First, we draw a parallelogram that is in the vector space and place it in the three-dimensional space R3. Its vectors are as follows:

1. $S=\left|\mathbf {a} \right|\cdot \left|\mathbf {b} \right|\cdot \sin \gamma =\left|\mathbf {a} \times \mathbf {b} \right|$ 2. $h=\left|\mathbf {c} \right|\cdot \left|\cos \theta \right|$ The calculation of the volume is like this: the area of the base is obtained based on the area of the parallelogram and its height is obtained based on the Pythagorean relation. So the volume of the parallelogram is equal to this relation.

$V=B\cdot h=\left(\left|\mathbf {a} \right|\left|\mathbf {b} \right|\sin \gamma \right)\cdot \left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\mathbf {a} \times \mathbf {b} \right|\left|\mathbf {c} \right|\left|\cos \theta \right|=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.$ Cosine theta and sine theta are equal to one in calculating the absolute value, the absolute value of the area of vectors a, b, c is equal to themselves.

Its volume can be obtained in a deeper way, the inner product of the external vectors that obtain the parallel plane with the external product of these three vectors is as follows. {T}, ~\mathbf b=(b_1,b_2,b_3)^\mathsf{T}, ~\mathbf c=(c_1,c_2,c_3)^\mathsf{T},[/itex]The volume is

$V=\left|\det {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}\right|.$ which is equal to this relation.

$V=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|.=\left|\det {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}\right|.}$ Another way to prove ( V1 ) is to use the scalar component in the direction a×b

From the vector: a, b, c

{\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 volume of a parallelogram is equal to the above expression in coordinate form, but in another way, which is not written in coordinate form, but is like the volume of a rectangular cube, which is equal to the product of length vectors, but a necessary expression has it.

The result is this.

By using the absolute value method and calculating the internal and external multiplication of vectors, a value called k is needed. k is a value that is obtained based on the angles of the edge of the parallel plane. Its square root is used in the volume of the parallel plane.

The value of k is obtained based on this relationship.

$K={1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )},$ its root value is like this
${\sqrt {K}}={\sqrt {1+2\ cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}},$ Its volume is written based on this relationship.
$V=abc{\sqrt {1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}},$ which can be written like this
$V=abc{\sqrt {K}}$ which is equal to the coordinate volume of the parallelogram.

## area

The area of a parallelogram is obtained based on the sum of the area of six parallelograms, which is written based on this relationship

{\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}} The area of a parallelogram is obtained like the area of a rectangular cube.

In another way, its area is found, which is obtained as the area of parallelograms

To find the area of a parallelogram based on a,b,c is like this.

$2{(ah+bh'+ch'')}$ h is equal to the height of the parallel plane in terms of theta angle, which h is written based on the Pythagorean relation.

${h^{2}=a^{2}-x^{2}}$ ${h'^{2}=a^{2}-y^{2}}$ ${h''^{2}=a^{2}-z^{2}}$ $x,y,z$ =a value based on theta below is a component of the lengths b,c respectively

If we calculate these two relationships, we get this result

{\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}}=2{(ah+bh'+ch'')}} ## formulasion

### dihedral angle

In the corner where the vectors ${\vec {a}},{\vec {b}},{\vec {c}}$ meet are the interior angles $\alpha =\angle ({\vec {b}},{\vec {c}}),\quad \beta =\angle ({\vec {a}},{\vec {c}}),\quad \gamma =\angle ({\vec {a}},{\vec {b}})$ . This corner together with the 3 neighboring corners forms a tetrahedron. If one considers the surrounding sphere of this tetrahedron, then according to the law of cosines for spherical triangles the equation

$\cos(\alpha )=\cos(\beta )\cdot \cos(\gamma )+\sin(\beta )\cdot \sin(\gamma )\cdot \cos(\beta _{a})[/itex]Where\beta _{a}$ is the face angle between the two side facesn that lie at the vector ${\vec {a}}$ . It follows $\beta _{a}=\arccos \left({\frac {\cos(\alpha )-\cos(\beta )\cdot \cos(\gamma )}{\sin(\beta )\cdot \sin(\ gamma)}}\right)$ ### solid angle The spatial angles at each vertex and each corner of a regular polyhedron can be calculated using L'Huilier's geometric theorem. This method is one of the easiest and best methods to calculate the angles of the corners of a polyhedron. This method only uses trigonometric methods. and equations and geometry that are combined are used. To calculate the spatial angles, we use the interior angles of its apex corners.$\alpha =\angle ({\vec {b}},{\vec {c}}),\quad \beta =\angle ({\vec {a}},{\vec {c}}),\quad \gamma =\angle ({\vec {a}},{\vec {b}})$ lies, applies {\begin{aligned}\Omega _{1}&=4\cdot \arctan \left({\sqrt {\tan \left({\frac {\theta _{s}}{2}}\right)\cdot \tan \left({\frac {\theta _{s}-\theta _{a}}{2}}\right)\cdot \tan \left({\frac {\theta _{s}-\theta _{b}}{2}}\right)\cdot \tan \left({\frac {\theta _{s}-\theta _{c}}{2}}\right)}}\right)\\&=4\cdot \arctan \left({\sqrt {\tan \left({\frac {\alpha +\beta +\gamma }{4}}\right)\cdot \tan \left({\frac {-\alpha +\beta +\gamma }{4}}\right)\cdot \tan \left({\frac {\alpha -\beta +\gamma }{4}}\right)\cdot \tan \left({\frac {\alpha +\beta -\gamma }{4}}\right)}}\right)\end{aligned}} wobei $\theta _{s}={\frac {\alpha +\beta +\gamma }{2}}$ , $\theta _{a}=\alpha$ , $\theta _{b}=\beta$ und $\theta _{c}=\gamma$ ist. Two diagonal spatial angles opposite in direction at the corners of parallel vertices are equal because three adjacent angles are equal. The other three solid angles are: • $\theta _{a}=\alpha ,\quad \theta _{b}=180^{\circ }-\beta ,\quad \theta _{c}=180^{\circ }-\gamma$ • $\theta _{a}=180^{\circ }-\alpha ,\quad \theta _{b}=\beta ,\quad \theta _{c}=180^{\circ }-\gamma$ • $\theta _{a}=180^{\circ }-\alpha ,\quad \theta _{b}=180^{\circ }-\beta ,\quad \theta _{c}=\gamma$ ### Table: Summary The dimensions of a parallelogram with the lengths of the edges a, b, c and the interior angles $\alpha,[itex]\beta$ , $\gamma$ ""parallel cotyledons""
'volume' $V=a\cdot b\cdot c\cdot {\sqrt {1+2\cdot \cos(\alpha )\cdot \cos(\beta )\cdot \cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}}$ 'level content $A=2\cdot a\cdot b\cdot \sin(\gamma )+2\cdot a\cdot c\cdot \sin(\beta )+2\cdot b\cdot c\cdot \sin(\alpha )$ 'height' $h={\frac {a}{\sin(\alpha )}}\cdot {\sqrt {1+2\cdot \cos(\alpha )\cdot \cos(\beta )\cdot \cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}}$ 'diagonal distance

$|{\vec {a}}+{\vec {b}}+{\vec {c}}|$ $d={\sqrt {a^{2}+b^{2}+c^{2}+2\cdot a\cdot b\cdot \sin(\gamma )+2\cdot a\cdot c\cdot \sin(\ beta)+2\cdot b\cdot c\cdot \sin(\alpha )}}$ ""angle between""

$\beta _{a}=\arccos \left({\frac {\cos(\alpha )-\cos(\beta )\cdot \cos(\gamma )}{\sin(\beta )\cdot \sin(\ gamma)}}\right)$ ''solid angle in the corners $\Omega _{1}=4\cdot \arctan \left({\sqrt {\tan \left({\tfrac {\alpha +\beta +\gamma }{4}}\right)\cdot \tan \left({\tfrac {-\alpha +\beta +\gamma }{4}}\right)\cdot \tan \left({\tfrac {\alpha -\beta +\gamma }{4}}\right)\cdot \tan \left(\ tfrac{\alpha +\beta -\gamma }{4}\right)}}\right)$ 