Parallelepiped - Volume

Volume

The volume of a parallelepiped is the product of the area of its base A and its height h. The base is any of the six faces of the parallelepiped. The height is the perpendicular distance between the base and the opposite face.

An alternative method defines the vectors a = (a1, a2, a3), b = (b1, b2, b3) and c = (c1, c2, c3) to represent three edges that meet at one vertex. The volume of the parallelepiped then equals the absolute value of the scalar triple product a · (b × c):

This is true because, if we choose b and c to represent the edges of the base, the area of the base is, by definition of the cross product (see geometric meaning of cross product),

A = |b| |c| sin θ = |b × c|,

where θ is the angle between b and c, and the height is

h = |a| cos α,

where α is the internal angle between a and h.

From the figure, we can deduce that the magnitude of α is limited to 0° ≤ α < 90°. On the contrary, the vector b × c may form with a an internal angle β larger than 90° (0° ≤ β ≤ 180°). Namely, since b × c is parallel to h, the value of β is either β = α or β = 180° − α. So

cos α = ±cos β = |cos β|,

and

h = |a| |cos β|.

We conclude that

V = Ah = |a| |b × c| |cos β|,

which is, by definition of the scalar product, equivalent to the absolute value of a · (b × c), Q.E.D..

The latter expression is also equivalent to the absolute value of the determinant of a three dimensional matrix built using a, b and c as rows (or columns):

V = \left| \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \right|.

This is found using Cramer's Rule on three reduced two dimensional matrices found from the original.

If a, b, and c are the parallelepiped edge lengths, and α, β, and γ are the internal angles between the edges, the volume is

V = a b c \sqrt{1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)}.

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