4. Let's fix a point $O$ on the plane. Any other point $A$ is uniquely determined by its radius vector $\overrightarrow{O A}$, which we will simply denote as $\vec{A}$. We will use this concept and notation repeatedly in the following. The fixed point $O$ is conventionally called the initial point. We will also introduce the symbol $(A B C)$ - the simple ratio of three points $A, B, C$ lying on the same line; points $A$ and $B$ are called the base points, and point $C$ is the dividing point. The symbol $(A B C)$ is a number determined by the formula:

$$
\overrightarrow{C A}=(A B C) \cdot \overrightarrow{C B}
$$

Furthermore, for two collinear vectors $\vec{a}$ and $\vec{b}$, there is a unique number $\lambda$ such that $\vec{a}=\lambda \vec{b}$. We introduce the notation $\lambda=\frac{\vec{a}}{\vec{b}}$ (the ratio of collinear vectors). In this case, we can write:

$$
(A B C)=\frac{\overrightarrow{C A}}{\overrightarrow{C B}}
$$

Remark. The notation for the ratio of vectors is not related to the scalar product of vectors! (For example, the relation $\frac{\vec{a}}{\vec{b}}=\frac{\vec{c}}{\vec{d}}$ implies $\vec{a}=\lambda \vec{b}$ and $\vec{c}=\lambda \vec{d}$, from which $\vec{a} \cdot \vec{d}=(\lambda \vec{b}) \cdot \left(\frac{1}{\lambda} \vec{c}\right)=\vec{b} \cdot \vec{c}$, but the converse is not true - from the relation $\vec{a} \cdot \vec{d}=\vec{b} \cdot \vec{c}$ it does not follow that $\frac{\vec{a}}{\vec{b}}=\frac{\vec{c}}{\vec{d}}$ : vectors $\vec{a}$ and $\vec{b}$ do not have to be collinear. Provide a corresponding example.)

Derive a formula expressing the vector $\vec{C}$ in terms of the vectors $\vec{A}$ and $\vec{B}$, given that $(A B C)=k$.
4. Since $(A B C)=k$, then $\frac{\overrightarrow{C A}}{\overrightarrow{C B}}=k$, or $\left(\frac{\overrightarrow{O A}-\overrightarrow{O C}}{\overrightarrow{O B}-\overrightarrow{O C}}=k\right) \Rightarrow(\overrightarrow{O A}-$ $-\overrightarrow{O C}=k \overrightarrow{O B}-k \overrightarrow{O C}) \Rightarrow((k-1) \overrightarrow{O C}=k \overrightarrow{O B}-\overrightarrow{O A}) \Rightarrow\left(\vec{C}=\frac{k \vec{B}-\vec{A}}{k-1}\right)$.