Editing Parallelepiped (section)

==Volume==
[[File:Parallelepiped-bf.svg|thumb|upright=1.2|Parallelepiped, generated by three vectors]]
A parallelepiped is a [[Prism (geometry)|prism]] with a [[parallelogram]] as base.
Hence the volume <math>V</math> of a parallelepiped is the product of the base area <math>B</math> and the height <math>h</math> (see diagram). With 
*<math>B = \left|\mathbf a\right| \cdot \left|\mathbf b\right| \cdot \sin \gamma = \left|\mathbf a \times \mathbf b\right|</math> (where <math>\gamma</math> is the angle between vectors <math>\mathbf a</math> and <math>\mathbf b</math>), and
*<math>h = \left|\mathbf c\right| \cdot \left|\cos \theta\right|</math> (where <math>\theta</math> is the angle between vector <math>\mathbf c</math> and the [[Normal (geometry)|normal]] to the base), one gets:
<math display="block">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|.</math>
The mixed product of three vectors is called [[triple product]]. It can be described by a [[determinant]]. Hence for <math>\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},</math> the volume is:
{{NumBlk||<math display="block">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| . </math>|{{EquationRef|V1}}}}

Another way to prove '''({{EquationNote|V1}})''' is to use the scalar component in the direction of <math>\mathbf a\times\mathbf b</math> of vector <math>\mathbf c</math>:
<math display="block">\begin{align}
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{align}</math>
The result follows.

An alternative representation of the volume uses geometric properties (angles and edge lengths) only:
{{NumBlk||<math display="block">V = abc\sqrt{1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)},</math>|{{EquationRef|V2}}}}
where <math>\alpha = \angle(\mathbf b, \mathbf c)</math>, <math>\beta = \angle(\mathbf a,\mathbf c)</math>, <math>\gamma = \angle(\mathbf a,\mathbf b) </math>, and <math>a,b,c </math> are the edge lengths.  

{{math proof | title = Proof of ({{EquationNote|V2}})
| proof =
The proof of '''({{EquationNote|V2}})''' uses [[Determinant#Properties of the determinant|properties of a determinant]] and the [[dot product#geometric definition|geometric interpretation of the dot product]]:

Let <math>M</math> be the 3×3-matrix, whose columns are the vectors <math>\mathbf a, \mathbf b, \mathbf c</math> (see above). Then the following is true: 
<math display="block">\begin{align}
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^2c^2 - b^2c^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^2b^2c^2-a^2b^2c^2\cos^2(\alpha) \\
&\quad-a^2b^2c^2\cos^2(\gamma)+a^2b^2c^2\cos(\alpha)\cos(\beta)\cos(\gamma) \\
&\quad+a^2b^2c^2\cos(\alpha)\cos(\beta)\cos(\gamma)-a^2b^2c^2\cos^2(\beta) \\

&=\ a^2b^2c^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^2b^2c^2\;\left(1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)\right).
\end{align}</math>

(The last steps use <math>\mathbf a \cdot \mathbf a=a^2</math>, ..., <math>\mathbf a\cdot \mathbf b=ab\cos\gamma</math>, <math>\mathbf a \cdot \mathbf c = ac\cos\beta</math>, <math>\mathbf b\cdot \mathbf c=bc\cos\alpha</math>, ...)}}

;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 [[Tetrahedron#Volume|proof]]).