7.
A BF$_3$ molecule lies in the xy-plane (so all the z-positions are 0); the xy-positions of the
atoms are (in Å)
$\mathbf{r}_B = \begin{pmatrix} 0.000 \\ 0.000 \end{pmatrix}$, $\mathbf{r}_{F1} = \begin{pmatrix} 1.313 \\ 0.000 \end{pmatrix}$, $\mathbf{r}_{F2} = \begin{pmatrix} -0.657 \\ 1.137 \end{pmatrix}$, $\mathbf{r}_{F3} = \begin{pmatrix} -0.657 \\ -1.137 \end{pmatrix}$
a) Sketch the structure of the molecule in a 3D coordinate system.
b) Calculate the F1-B-F2 bond angle, $\alpha$.
c) Show that rotating the molecule about the z-axis by 2$\alpha$ is the same as rotating
it by -$\alpha$. That is, show that
D($\alpha_z$)^2 = D($\alpha_z$)^-1.
The required rotation matrix is
D($\alpha_z$) = $\begin{pmatrix} cos \alpha & -sin \alpha \\ sin \alpha & cos \alpha \end{pmatrix}$