Ask Uncle Colin: About A Tetrahedron

Dear Uncle Colin,

Apparently, the volume of a tetrahedron with three edges given by the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$, is $\frac{1}{6}|\overrightarrow{AB}\cdot (\overrightarrow{AC}\times \overrightarrow{AD})|$. Where does that come from?

- Very Obviously Lacklustre Understanding of My Exam

Hi, VOLUME, and thanks for your message!

I think there are two questions wrapped up in that: a) how (conceptually) do you find the volume of a tetrahedron?, and b) how does that vector nonsense fit in?

Let’s start with the first.

The volume of a tetrahedron

I don’t know if this idea has a name, and I’ve no idea what to Google to look it up, but I think of it as the pointy-shape rule:

If a shape tapers nicely to a point, keeping a similar cross-section to the base all the way up, its volume is a third of the volume of the prism ¹ enclosing it. (That can be stated more precisely, but it’s good enough for now.)

For example: the volume of a pyramid with a square base of side length $x$, and a height of $h$, is $\frac{1}{3}{x}^{2}h$ - a third of the volume of the cuboid with the same base and height. A cone? That’s $\frac{1}{3}\pi r^{2}h$, a third of the volume of the surrounding cylinder.

We’re interested in a tetrahedron, the volume of which is a third of the volume of the triangular prism that surrounds it. If the base of the tetrahedron has two side lengths of $a$ and $b$, with the angle between them being $\theta$, the area of the base is $\frac{1}{2}ab\sin (\theta )$.

The height, meanwhile, can be found from the third side originating at the same point as the $a$ and $b$ sides - if it has length $c$ and forms an angle $\varphi$ with the base, the height is $c\sin (\varphi )$.

So the volume of the triangular prism is $\frac{1}{2}abc\sin (\theta )\sin (\varphi )$, and the volume of the tetrahedron is a third of that: $V=\frac{1}{6}abc\sin (\theta )\sin (\varphi )$.

The vectory nonsense

There’s already quite a lot in common between the formula we just found and the vector formula - all three side lengths are in there, and the $\frac{1}{6}$. Let’s break the vector formula down carefully:

If we say $\overrightarrow{AC}$ has magnitude $a$ and $\overrightarrow{AD}$ has magnitude $b$, then $(\overrightarrow{AC}\times \overrightarrow{AD})$ gives $ab\sin (\theta )\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is a unit vector perpendicular to the base.

Meanwhile, $|\overrightarrow{AB}\cdot \hat{\mathbf{n}}|$ gives $c\cos (\alpha )$, where $\alpha$ is the angle between $\overrightarrow{AB}$ and the perpendicular to the plane ². However, $\alpha =\frac{\pi }{2}-\varphi$, so $\cos (\alpha )=\sin (\varphi )$.

We can put that all together to say that $\frac{1}{6}abc\sin (\theta )\sin (\varphi )=\frac{1}{6}|\overrightarrow{AB}\cdot (\overrightarrow{AC}\times \overrightarrow{AD})|$, the volume of a tetrahedron.

Hope that helps!

- Uncle Colin

Footnotes:

1. Or prism-like shape; strictly, prisms have polygonal sides.

2. The absolute value is needed because $\hat{\mathbf{n}}$ could be pointed downwards