Volume of Parallelepiped by Vectors

Another application of cross product of vectors occurs in geometry, or rather application of both the dot and the cross product, wherein it helps with finding the volume of a solid called parallelepiped.

A parallelepiped is a six faced three dimensional shape. All six faces are parallelograms(hence the name). The opposite faces are parallel and congruent(same shape and area). 

Firstly, by looking at the figure I hope that we all agree that the volume of parallelepiped is the area of the base times the height, where any face of the solid can be chosen as its base. The height is the length of the perpendicular from the chosen base to the opposite face.

Volume of parallelepiped = Area of PQRS $\times h$

Take any vertex and define three vectors, $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$, from that vertex along the three edges, as shown. Their magnitude equals the length of the corresponding edge.

Now, the base of the solid is a parallelogram. In my previous post, we saw that the area of a parallelogram is $|\overrightarrow{m}\times \overrightarrow{n}|$ where $\overrightarrow{m}$ and $\overrightarrow{n}$ are vectors along the two adjacent sides of the parallelogram, originating from a common vertex, and having magnitudes equal to the lengths of the sides.

$⟹$ area of PQRS = $|\overrightarrow{a}\times \overrightarrow{b}|$

By definition, $\overrightarrow{a}\times \overrightarrow{b}$ is a vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$. Since $\overrightarrow{a}$ and $\overrightarrow{b}$, both are parallel to the base PQRS, $\overrightarrow{a}\times \overrightarrow{b}$ is perpendicular to the base PQRS as well. 

Define $\overrightarrow{e}$ along height h, whose magnitude is equal to h. Notice since height h is normal to the base, $\overrightarrow{e}$ is also normal to the base, and therefore parallel to $\overrightarrow{a}\times \overrightarrow{b}$. 

Let $\theta$ be the angle between $\overrightarrow{e}$ and $\overrightarrow{c}$ as shown. 

Notice in the figure that the tip of the arrowheads of $\overrightarrow{e}$ and $\overrightarrow{c}$ are both at the same height above the base. If we define a vector from one tip to the other, this new vector will be parallel to the base, and hence perpendicular to $\overrightarrow{e}$ forming a right angled triangle as shown.

$⟹|\overrightarrow{e}|=|\overrightarrow{c}|\cos \theta$

Therefore, the volume of the parallelepiped is :

Volume of parallelepiped = Area of PQRS $\times h$

                         = $|\overrightarrow{a}\times \overrightarrow{b}|\times |\overrightarrow{c}|\cos \theta$ ….(1)

$\theta$ is the angle between $\overrightarrow{c}$ and $\overrightarrow{e}$. But it is also the angle between $\overrightarrow{c}$ and $\overrightarrow{a}\times \overrightarrow{b}$ because $\overrightarrow{e}$ and $\overrightarrow{a}\times \overrightarrow{b}$ are parallel. (Is it always?)

So (1) reads as the product of the magnitudes of vectors… times the cosine of the angle between the vectors (formed at the point of common tails). But this is exactly the dot product $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})$. 

$⟹$ Volume of parallelepiped = $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})$

An application of both the dot product and cross product of vectors!

However, this is still not quite finished yet. It is important to realize that (1) can only be written as the dot product when $\theta$ is the angle formed at the point of common tail (when the vectors are put tail-to-tail). Let’s say that $\overrightarrow{a}\times \overrightarrow{b}$ normal to the base was directed the other way instead.

As you can see, the angle $\overrightarrow{a}\times \overrightarrow{b}$ makes with $\overrightarrow{c}$ tail-to-tail is not $\theta$ but $180-\theta$. 

So (1) cannot now equal $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})$. However, it is (- $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b}))$.

This is because the dot product of any two vectors with $\theta$ as an angle between them at the point of common tail, is just the negative of the dot product of the same vectors with $180-\theta$ as an angle at the point of common tail. 

This fact can be easily checked.

From the definition of dot product,

$\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})=|\overrightarrow{a}\times \overrightarrow{b}|\times |\overrightarrow{c}|\cos (180-\theta )$

         

                    $=-|\overrightarrow{a}\times \overrightarrow{b}|\times |\overrightarrow{c}|\cos \theta$….$(\cos (180-\theta )=-\cos \theta )$

                    $=-\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})$

So this means that $\overrightarrow{a}\times \overrightarrow{b}$ could point to either of the two directions, but the resulting value of the volume will only differ in sign. Obviously, volume cannot be negative. We could use this reasoning and simply discard the negative sign. Or, we can insert a modulus(absolute value sign) around the formula :

Volume of parallelepiped =  $|\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})|$

This now always outputs a positive value regardless of the direction of $\overrightarrow{a}\times \overrightarrow{b}$.