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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