Dear Uncle Colin,

Apparently, the volume of a tetrahedron with three edges given by the vectors $\vec{AB}$, $\vec{AC}$ and $\vec{AD}$, is $\frac{1}{6} \left| \vec{AB} \cdot \br{\vec{AC}\times\vec{AD}} \right|$. 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.

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 $\frac{1}{3}x^2h$ – 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 $\phi$ with the base, the height is $c \sin(\phi)$.

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

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 $\vec{AC}$ has magnitude $a$ and $\vec{AD}$ has magnitude $b$, then $\br{\vec{AC}\times\vec{AD}}$ gives $ab \sin(\theta) \bb{\hat{n}}$, where $\bb{\hat{n}}$ is a unit vector perpendicular to the base.

Meanwhile, $|\vec{AB}\cdot \bb{\hat{n}}|$ gives $c \cos(\alpha)$, where $\alpha$ is the angle between $\vec{AB}$ and the perpendicular to the plane^{2}. However, $\alpha = \frac{\pi}{2} – \phi$, so $\cos(\alpha)=\sin(\phi)$.

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

Hope that helps!

– Uncle Colin