Point $M$ divides the side $B C$ of triangle $A B C$ in the ratio $B M: M C=2: 5$, It is known that $\overrightarrow{A B}=\vec{a}, \overrightarrow{A C}=$ $\vec{b}$. Find the vector $\overrightarrow{A M}$.
Note that

$$
\overrightarrow{A M}=\overrightarrow{A B}+\overrightarrow{B M}, \overrightarrow{A M}=\overrightarrow{A C}+\overrightarrow{C M}
$$

Multiply both sides of the first equation by 5, and the second equation by 2, and add the resulting equations term by term. Then

$$
\begin{gathered}
7 \overrightarrow{A M}=5 \overrightarrow{A B}+2 \overrightarrow{A C}+5 \overrightarrow{B M}+2 \overrightarrow{C M}= \\
=5 \overrightarrow{A B}+2 \overrightarrow{A C}+\overrightarrow{0}=5 \overrightarrow{A B}+2 \overrightarrow{A C}
\end{gathered}
$$

Therefore,

$$
\overrightarrow{A M}=\frac{1}{7}(5 \overrightarrow{A B}+2 \overrightarrow{A C})=\frac{5}{7} \overrightarrow{A B}+\frac{2}{7} \overrightarrow{A C}=\frac{5}{7} \vec{a}+\frac{2}{7} \vec{b}
$$

## Answer

$\frac{5}{7} \vec{a}+\frac{2}{7} \vec{b}$