Given an equilateral triangle ABC with side length equal to 'a'. Let M and N be two points respectivelyАВIn the side AB and AC such that `vec(AN) = Kvec(AC) and vec(AM) = vec(AB)/3` If `vec(BN) and vec(CM)` are orthogonalthen the value of K is equal to

Options

A. `(1)/(5)`
B. `(1)/(4)`
C. `(1)/(3)`
D. `(1)/(2)`

Answer

To solve the problem, we need to determine the value of \( K \) given the conditions of the equilateral triangle \( ABC \) and the points \( M \) and \( N \) on sides \( AB \) and \( AC \) respectively.

### Step-by-Step Solution:

1. **Define the Position Vectors**:
Let the position vectors of points \( A \), \( B \), and \( C \) be represented as:
\[ \vec{A} = \vec{0}, \quad \vec{B} = \vec{a}, \quad \vec{C} = \vec{a} \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \]
Here, \( \vec{B} \) is at \( (a, 0) \) and \( \vec{C} \) is at \( \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right) \).

2. **Determine Points \( M \) and \( N \)**:
- Point \( M \) on \( AB \):
\[ \vec{M} = \frac{1}{3} \vec{B} = \frac{1}{3} \vec{a} \]
- Point \( N \) on \( AC \):
\[ \vec{N} = K \vec{C} = K \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right) \]

3. **Vectors \( \vec{BN} \) and \( \vec{CM} \)**:
- The vector \( \vec{BN} \):
\[ \vec{BN} = \vec{N} - \vec{B} = K \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right) - \vec{a} = \left( \frac{Ka}{2} - a, \frac{Ka\sqrt{3}}{2} \right) \]
- The vector \( \vec{CM} \):
\[ \vec{CM} = \vec{M} - \vec{C} = \frac{1}{3} \vec{a} - \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right) = \left( \frac{1}{3}a - \frac{a}{2}, \frac{1}{3} \cdot 0 - \frac{a\sqrt{3}}{2} \right) \]

4. **Set Up the Orthogonality Condition**:
The vectors \( \vec{BN} \) and \( \vec{CM} \) are orthogonal, so their dot product is zero:
\[ \vec{BN} \cdot \vec{CM} = 0 \]

5. **Calculate the Dot Product**:
\[ \left( \frac{Ka}{2} - a \right) \left( \frac{1}{3}a - \frac{a}{2} \right) + \left( \frac{Ka\sqrt{3}}{2} \right) \left( -\frac{a\sqrt{3}}{2} \right) = 0 \]

6. **Simplify the Equation**:
- The first term simplifies to:
\[ \left( \frac{Ka}{2} - a \right) \left( \frac{1}{3}a - \frac{a}{2} \right) = \left( \frac{Ka - 2a}{2} \right) \left( \frac{2 - 3}{6}a \right) = \left( \frac{Ka - 2a}{2} \right) \left( -\frac{a}{6} \right) \]
- The second term simplifies to:
\[ -\frac{Ka^2 \cdot 3}{4} = -\frac{3Ka^2}{4} \]

7. **Combine and Solve for \( K \)**:
Set the combined equation to zero and solve for \( K \):
\[ \left( \frac{Ka - 2a}{2} \right) \left( -\frac{a}{6} \right) - \frac{3Ka^2}{4} = 0 \]

8. **Final Calculation**:
After simplifying and solving, we find:
\[ K = \frac{1}{5} \]

### Conclusion:
The value of \( K \) is \( \frac{1}{5} \).