Calculating bond angles of a symmetrical tetrahedral molecule such as methane using a dot product

Consider the vectors **U** and **V ** (with respective magnitudes |**U**| and |**V**|). If those vectors enclose an angle θ then the dot product of those vectors can be written as:

- $\mathbf {U} \cdot \mathbf {V} =|\mathbf {U} ||\mathbf {V} |\cos(\theta )$

If the vectors can be written as:

- $\mathbf {U} =(U_{x},U_{y},U_{z})$
- $\mathbf {V} =(V_{x},V_{y},V_{z})$

then the *dot product* is given by:

- $\mathbf {U} \cdot \mathbf {V} =U_{x}V_{x}+U_{y}V_{y}+U_{z}V_{z}$

For example,

- $(1,2,3)\cdot (2,2,2)=1(2)+2(2)+3(2)=12.$

and

- $(0,5,0)\cdot (4,0,0)=0.$

We can interpret the last case by noting that the product is zero because the angle between the two vectors is 90 degrees.

Since

- $|\mathbf {U} |={\sqrt {U_{x}^{2}+U_{y}^{2}+U_{z}^{2}}}$

and

- $|\mathbf {V} |={\sqrt {V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}}$

this means that

- $\cos(\theta )={\frac {U_{x}V_{x}+U_{y}V_{y}+U_{z}V_{z}}{{\sqrt {U_{x}^{2}+U_{y}^{2}+U_{z}^{2}}}{\sqrt {V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}}}}$