Consider a triangle OAB. We denote by M the point dividing the side OA internally in the ratio 3:1 and denote by N the point dividing the side OB internally in the ratio 1:2. Also we denote by P the point of intersection of the line segment AN and the line segment BM.

(1) When the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ are denoted by $\vec{a}$ and $\vec{b}$ respectively, we are to express the vector $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}$ and $\vec{b}$.

When we set

$$\begin{array}{ll}
\mathrm{AP}:\mathrm{PN} = s:(1-s) & (0<s<1) \\
\mathrm{BP}:\mathrm{PM} = t:(1-t) & (0<t<1)
\end{array}$$

we have

$$\overrightarrow{\mathrm{OP}} = (\mathbf{A} - t)\vec{b} = \frac{\mathbf{D}}{\mathbf{E}}t\vec{a}+\frac{\mathbf{B}}{\mathbf{E}}s\vec{b},$$

from which we obtain

$$s=\frac{\mathbf{G}}{\mathbf{H}}, \quad t=\frac{\square}{\mathbf{I}}.$$

Hence, $\overrightarrow{\mathrm{OP}}$ can be expressed in terms of $\vec{a}$ and $\vec{b}$ as

$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{K}}{\mathbf{L}}\vec{a}+\frac{\mathbf{M}}{\mathbf{N}}\vec{b}.$$

(2) We are to look at the relation between the length of the line segment OP and the size of $\angle\mathrm{AOB}$, when $\mathrm{OA}=6$ and $\mathrm{OB}=9$.

Let us denote the length of OP by $\ell$. When we express $\ell^2$ in terms of $\vec{a}\cdot\vec{b}$, we have

$$\ell^2 = \frac{\mathbf{Q}}{\mathbf{PQ}}\vec{a}\cdot\vec{b}+\mathbf{RS}$$

where $\vec{a}\cdot\vec{b}$ represents the inner product of $\vec{a}$ and $\vec{b}$.

Hence, for instance, if $\ell=4$, then we have

$$\cos\angle\mathrm{AOB}=\frac{\mathbf{TU}}{\mathbf{V}}$$

When the size of $\angle\mathrm{AOB}$ varies, the range of the values which $\ell$ can take is

$$\mathbf{W}<\ell<\mathbf{X}.$$