Figure 2

A given triangle on the convex hull for the PDB:1ABT structure. The three vertices are labeled as 1, 2, and 3. The point p is determined by the extreme values of x, y, and z of these three vertices. The distance of atom i to the triangle is obtained as follows: first obtain the normal vector to the triangle, N, N = (x ₂  − x ₁ ) × (x ₃  − x ₁ ), where x₁, x₂, and x₃ are the vectors from the origin of the systems of Cartesian coordinates to the three vertices. Then, calculate the angle between the normal vector and the line passing through atom i and one of the vertices of this triangle using the following relation: $\mathbf{cos}\mathit{\theta }=\frac{\left({\mathbf{x}}_{\mathbf{1}}-{\mathit{x}}_{\mathbf{i}}\right)·\mathbf{N}}{\left|{\mathbf{x}}_{\mathbf{1}}-{\mathbf{x}}_{\mathit{i}}\right|\left|\mathbf{N}\right|}$, Finally, we compute this distance by d _i  = |x ₁  − x _i | cos θ, where x _i is a vector joining the origin and a given point in this volume.