The bond angle for a symmetric tetrahedral molecule such as CH may be calculated using the dot product of two vectors. As shown in the diagram, the molecule can be inscribed in a cube with the tetravalent atom (e.g. carbon) at the cube centre which is the origin of coordinates, O. The four monovalent atoms (e.g. hydrogens) are at four corners of the cube (A, B, C, D) chosen so that no two atoms are at adjacent corners linked by only one cube edge.
If the edge length of the cube is chosen as 2 units, then the two bonds OA and OB correspond to the vectors a = (1, –1, 1) and b = (1, 1, –1), and the bond angle θ is the angle between these two vectors. This angle may be calculated from the dot product of the two vectors, defined as a&thinsp;•&thinsp;b = ||a|| ||b|| cos&thinsp;θ where ||a|| denotes the length of vector a. As shown in the diagram, the dot product here is –1 and the length of each vector is √3, so that cos&thinsp;θ = –1/3 and the tetrahedral bond angle θ = arccos(–1/3) ≃ 109.47°.