Let `DeltaOAB` be an equilateral triangle with side length unity (O being the origin). Also, M and N being closer to A and N being clower to B. position vectors of A, B, M and N are `veca, vecb, vecm and vecn` respectively, then the value of `vecm.vecn` is equal to

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(13)/(18)`
D. `(4)/(9)`

Answer

To solve the problem, we need to find the dot product of the position vectors of points M and N in the equilateral triangle OAB, where O is the origin, and A and B are the vertices of the triangle.

### Step 1: Define the position vectors of points A and B
Since triangle OAB is equilateral with side length 1, we can place the points in a coordinate system:
- Let O (the origin) be at (0, 0).
- Let A be at (1, 0).
- Let B be at \((\frac{1}{2}, \frac{\sqrt{3}}{2})\).

Thus, the position vectors are:
- \(\vec{a} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)
- \(\vec{b} = \begin{pmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}\)

### Step 2: Find the position vector of M
Point M is closer to A than to B. We can express the position vector of M using the section formula. Let M divide the line segment AB in the ratio \(k:1\) (where \(k > 1\) since M is closer to A).

Using the section formula:
\[ \vec{m} = \frac{k \vec{b} + 1 \vec{a}}{k + 1} \]
Substituting the values of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{m} = \frac{k \begin{pmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix} + 1 \begin{pmatrix} 1 \\ 0 \end{pmatrix}}{k + 1} = \frac{\begin{pmatrix} \frac{k}{2} + 1 \\ \frac{k\sqrt{3}}{2} \end{pmatrix}}{k + 1} \]

### Step 3: Find the position vector of N
Point N is closer to B than to A. We can express the position vector of N using the section formula as well. Let N divide the line segment AB in the ratio \(1:k\) (where \(k > 1\) since N is closer to B).

Using the section formula:
\[ \vec{n} = \frac{1 \vec{b} + k \vec{a}}{1 + k} \]
Substituting the values of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{n} = \frac{1 \begin{pmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix} + k \begin{pmatrix} 1 \\ 0 \end{pmatrix}}{1 + k} = \frac{\begin{pmatrix} \frac{1}{2} + k \\ \frac{\sqrt{3}}{2} \end{pmatrix}}{1 + k} \]

### Step 4: Calculate the dot product \(\vec{m} \cdot \vec{n}\)
Now we can compute the dot product \(\vec{m} \cdot \vec{n}\):
\[ \vec{m} \cdot \vec{n} = \left(\frac{\frac{k}{2} + 1}{k + 1}\right) \left(\frac{\frac{1}{2} + k}{1 + k}\right) + \left(\frac{\frac{k\sqrt{3}}{2}}{k + 1}\right) \left(\frac{\frac{\sqrt{3}}{2}}{1 + k}\right) \]

### Step 5: Simplify the expression
This results in:
\[ \vec{m} \cdot \vec{n} = \frac{(\frac{k}{2} + 1)(\frac{1}{2} + k) + \frac{k\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}}{(k + 1)^2} \]
Calculating the terms:
1. \((\frac{k}{2} + 1)(\frac{1}{2} + k) = \frac{k}{4} + \frac{k^2}{2} + \frac{1}{2} + k\)
2. \(\frac{k\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3k}{4}\)

Combining these gives:
\[ \vec{m} \cdot \vec{n} = \frac{\frac{k^2}{2} + \frac{5k}{4} + \frac{1}{2}}{(k + 1)^2} \]

### Step 6: Evaluate for specific values of k
If we take \(k = 2\) (for simplicity), we can find:
\[ \vec{m} \cdot \vec{n} = \frac{\frac{2^2}{2} + \frac{5 \cdot 2}{4} + \frac{1}{2}}{(2 + 1)^2} = \frac{2 + \frac{10}{4} + \frac{1}{2}}{9} = \frac{2 + 2.5 + 0.5}{9} = \frac{5}{9} \]

### Final Result
Thus, the value of \(\vec{m} \cdot \vec{n}\) is \(\frac{5}{9}\).