The volume of a parallelepiped

Vector calculus in plane and space: The Cross Product

The volume of a parallelepiped

The volume of a parallelepiped spanned by the vectors #\vec{a}#, #\vec{b}#, and #\vec{c}# (that is to say: with vertices #\vec{0}#, #\vec{a}#, #\vec{b}#, #\vec{c}#, #\vec{a}+\vec{b}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}#, #\vec{a}+\vec{b}+\vec{c}#), can be expressed in terms of a cross and a dot product:

The volume of the parallelepiped spanned by the vectors #\vec{a}#, #\vec{b}#, and #\vec{c}# in the space is equal to \[\left|\,(\vec{a}\times\vec{b})\cdot \vec{c}\,\right|\]

The volume of the parallelepiped is equal to the area of a base parallellogram, let us say spanned by #\vec{a}# and #\vec{b}#, multiplied by the height. The area of the parallelogram is equal to #\parallel\vec{a}\times \vec{b}\parallel# as we have already seen in the theory The cross product, in terms of two lengths. Because #\vec{a}\times \vec{b}# is perpendicular to the parallelogram, the height is equal to the (length of the) projection of #\vec{c}# on #\vec{a}\times \vec{b}#, i.e., to the absolute value of #\parallel\vec{c}\parallel\cdot \cos (\varphi)#, wherein #\varphi# is the angle between #\vec{c}# and #\vec{a}\times \vec{b}#. Therefore, the volume is \[\parallel\vec{a}\times \vec{b}\parallel \cdot \parallel\vec{c}\parallel \cdot |\cos(\varphi)|=\left |\, (\vec{a}\times\vec{b})\cdot \vec{c}\,\right|\]

Because the volume of the parallelepiped can be calculated in several different ways, we may draw some useful consequences. Here is an example.

Parallelepiped rule

If #\vec{a}#, #\vec{b}#, and #\vec{c}# are vectors in the space, then \[(\vec{a}\times\vec{b})\cdot \vec{c}=-(\vec{a}\times\vec{c})\cdot \vec{b}\]

If #\vec{a}#, #\vec{b}#, #\vec{c}# do not form a basis, then both sides of the equality are equal to #\vec{0}#. Regarding the rest of the proof, we can assume that #\vec{a}#, #\vec{b}#, #\vec{c}# is a basis.

According to the statement above, the content of the parallelepiped spanned by the vectors #\vec{a}#, #\vec{b}#, and #\vec{c}# is equal to #\varepsilon\cdot(\vec{a}\times\vec{b})\cdot \vec{c}#, wherein #\varepsilon=\pm1#, in such a way that the expression is non-negative (and because of the assumption that we are dealing with a basis, is even positive). But, if we switch #\vec{b}# and #\vec{c}# in this formula, then the absolute value of the result is still the volume of the same parallelepiped, while the orientation of #\vec{a}#, #\vec{c}#, #\vec{b}# is opposite to that of #\vec{a}#, #\vec{b}#, #\vec{c}#. This means that #\varepsilon\cdot(\vec{a}\times\vec{b})\cdot \vec{c}=-\varepsilon\cdot(\vec{a}\times\vec{c})\cdot \vec{b}#. Dividing by #\varepsilon# gives #(\vec{a}\times\vec{b})\cdot \vec{c}=-(\vec{a}\times\vec{c})\cdot \vec{b}# , which could be proven.

Let #\vec{a}#, #\vec{b}#, #\vec{c}# be a basis of the space.

What is the volume of the triangular prism with vertices #\vec{0}#, #\vec{a}#, #\vec{b}#, #\vec{c}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}#?

#\frac{1}{2}\cdot\left|\,(\vec{a}\times\vec{b})\cdot \vec{c}\,\right|#

This follows from volume rules 2 and 3:

The volume of the triangular prism with vertices #\vec{0}#, #\vec{a}#, #\vec{b}#, #\vec{c}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}# is equal to the volume of the triangular prism with vertices #\vec{a}+\vec{b}#, #\vec{a}#, #\vec{b}#, #\vec{a}+\vec{b}+\vec{c}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}#. Reflecting the plane through #\vec{a}#, #\vec{b}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}# actually carries one into the other, making rule 2 applicable.
The two prisms intersect in the plane through #\vec{a}#, #\vec{b}#, #\vec{a}+\vec{c}#, #\vec{b}+\vec{c}#. Together, they form the parallelepiped defined by #\vec{a}#, #\vec{b}#, #\vec{c}#. The volume of two triangular prisms as given is, because of rule 3, equal to the volume of said parallelepiped, so, thanks to the theory, equal to #\left|\,(\vec{a}\times\vec{b})\cdot \vec{c}\,\right|#.
Accordingly, the requested volume is equal to #\frac{1}{2}\cdot\left|\,(\vec{a}\times\vec{b})\cdot \vec{c}\,\right|#.

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