To solve the problem, we need to find the value of the dot product of the position vectors of points M and N in the equilateral triangle OAB, where O is the origin, A and B are the vertices of the triangle, and M and N are points on the line segment AB.
### Step-by-Step Solution:
1. **Position Vectors of Points A and B**:
- Let the position vector of point A be \( \vec{a} \) and the position vector of point B be \( \vec{b} \).
- Since OAB is an equilateral triangle with side length 1, we can place the points as follows:
- \( \vec{a} = (1, 0) \) (point A)
- \( \vec{b} = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \) (point B)
- \( \vec{o} = (0, 0) \) (point O)
2. **Finding the Position Vector of M**:
- M is closer to A than to B. We can express the position vector of M as a weighted average of \( \vec{a} \) and \( \vec{b} \).
- Let the ratio of division be \( 2:1 \) (since M is closer to A).
- Thus, the position vector of M is given by:
\[
\vec{m} = \frac{2\vec{a} + 1\vec{b}}{2 + 1} = \frac{2\vec{a} + \vec{b}}{3}
\]
3. **Calculating \( \vec{m} \)**:
- Substituting the values of \( \vec{a} \) and \( \vec{b} \):
\[
\vec{m} = \frac{2(1, 0) + \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}{3} = \frac{\left(2 + \frac{1}{2}, 0 + \frac{\sqrt{3}}{2}\right)}{3} = \frac{\left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right)}{3} = \left(\frac{5}{6}, \frac{\sqrt{3}}{6}\right)
\]
4. **Finding the Position Vector of N**:
- N is closer to B than to A. We can express the position vector of N similarly.
- Let the ratio of division be \( 1:2 \) (since N is closer to B).
- Thus, the position vector of N is given by:
\[
\vec{n} = \frac{1\vec{a} + 2\vec{b}}{1 + 2} = \frac{\vec{a} + 2\vec{b}}{3}
\]
5. **Calculating \( \vec{n} \)**:
- Substituting the values of \( \vec{a} \) and \( \vec{b} \):
\[
\vec{n} = \frac{(1, 0) + 2\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}{3} = \frac{\left(1 + 1, 0 + \sqrt{3}\right)}{3} = \frac{(2, \sqrt{3})}{3} = \left(\frac{2}{3}, \frac{\sqrt{3}}{3}\right)
\]
6. **Calculating the Dot Product \( \vec{m} \cdot \vec{n} \)**:
- Now we can calculate the dot product:
\[
\vec{m} \cdot \vec{n} = \left(\frac{5}{6}, \frac{\sqrt{3}}{6}\right) \cdot \left(\frac{2}{3}, \frac{\sqrt{3}}{3}\right)
\]
- This expands to:
\[
\vec{m} \cdot \vec{n} = \frac{5}{6} \cdot \frac{2}{3} + \frac{\sqrt{3}}{6} \cdot \frac{\sqrt{3}}{3} = \frac{10}{18} + \frac{3}{18} = \frac{13}{18}
\]
### Final Answer:
The value of \( \vec{m} \cdot \vec{n} \) is \( \frac{13}{18} \).
