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For smooth level surfaces $f=0$, $g=0$ meeting at a point where both gradients are nonzero, their acute angle is the angle between two planes tangent to the surfaces there. Each gradient is a normal vector to its tangent plane, proving the displayed formula. Vanishing gradients require further analysis because the level-set equation then need not specify a unique tangent plane.

Use the angle between two planes defined as the acute angle between their normal vectors. The normals $(1,1,1)$ and $(-1,1,1)$ have dot product one and lengths $\sqrt{3}$. ThusChoosing an oriented rather than an acute interplanar angle could give the supplementary $109.4712^{\circ }$; it describes the same unoriented pair of lattice planes.

The normal vectors are ${\mathbf{n}}_A=(1,2,-4)$ and ${\mathbf{n}}_B=(2,-1,3)$. Their dot product is $-12$, and their norms are $\sqrt{21}$ and $\sqrt{14}$. The usual angle between two planes is the smaller, unoriented angle, soFor the particular signed normal vectors printed above, the angle is the supplement $\pi -\theta \approx 134.4^{\circ }$. Reversing either normal vector exchanges these two angles without changing either plane.

Let $\mathbf{p}=(2,1,1)$, $\mathbf{q}=(1,2,2)$ and $\mathbf{r}=(0,0,3)$. Cross products give plane normals $\mathbf{p}\times \mathbf{r}=(3,-6,0)$ and $\mathbf{q}\times \mathbf{r}=(6,-3,0)$. Their dot product is $36$ and each has norm $3\sqrt{5}$. Thus the acute angle between two planes isThis is also the interior dihedral angle here: perpendicular to the common edge $OR$, the interior wedge lies between the projected directions $(2,1)$ and $(1,2)$. The two outward face normals instead make the supplementary angle.