# Geometrical interpretation of the Triple Product Rule

Most of us who have done multivariable calculus are familiar with the rule $$\left( \frac{\partial x}{\partial y} \right)_{z} \left( \frac{\partial y}{\partial z} \right)_{x} \left( \frac{\partial z}{\partial x} \right)_{y} = -1,$$ particularly useful in thermodynamics. The normal way to prove this (as shown in either of these Wikipedia articles is by mucking about with the algebra of differentials and substituting, and there's nothing as geometrically unintuitive as pushing algebraic symbols around. The $-1$ is also famously counterintuitive to people seeing it for the first time.

Therefore, is there a nice intuitive geometrical derivation or interpretation of this result? (And, indeed, its more-variabled generalisations?)

## 1 Answer

This is probably not direct enough, but it's something.

Every nice surface has tangent planes. When any tangent plane is intersected with the three coordinate planes, suitably oriented, we get three lines whose slopes are the partial derivatives in the triple product rule. So imagine, for instance, a plane in 3D which intersects the positive x, y and z axes at points A, B and C. Notice that scaling any of these three points away from the origin by a positive scalar λ>0 scales one slope by λ, another slope by 1/λ, and leaves the last slope unchanged, so the triple product is invariant. Flipping any of the points across the origin (i.e. changing the sign of one of A, B or C) changes the sign of two of the slopes but that's it, so the triple product remains unchanged. Finally, if A=B=C=1, the triple product is (-1)(-1)(-1)=-1.

The argument can be generalized to n dimensions, where the product should be (-1)^n.

(This triple product invariant with respect to reciprocal scalings of any two factors reminds me of the cohomology of impossible figures discussion, but I don't know that it's actually relevant.)

• This is a great geometric explanation. Indeed, for any plane $ax+by+cz+d=0$ with $a$, $b$ and $c$ all nonzero, a direct computation shows that the product of the slopes of its intersections with the coordinate planes, suitably oriented, is $-1$. – amd Apr 7 '18 at 1:41