Let `veca, vecb,vecc` be unit vectors, equally inclined to each other at an angle `theta, (pi/3 lt theta lt pi/2)`. If these are the poisitions vector of the vertices of a trinalge and `vecg` is the position vector of the centroid of the triangle, then:

Options

A. `|vecg| gt 1`
B. `|vecg| le sqrt(3/2)`
C. `|vecg| ge sqrt(3/2)`
D. `|vecg| le sqrt(2/3)`

Answer

To solve the problem, we need to find the position vector of the centroid \( \vec{g} \) of a triangle formed by the unit vectors \( \vec{a}, \vec{b}, \vec{c} \) which are equally inclined to each other at an angle \( \theta \) (where \( \frac{\pi}{3} < \theta < \frac{\pi}{2} \)).

### Step-by-Step Solution:

1. **Understanding the Centroid**:
The position vector of the centroid \( \vec{g} \) of a triangle with vertices at \( \vec{a}, \vec{b}, \vec{c} \) is given by:
\[ \vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \]

2. **Finding the Magnitude of \( \vec{a} + \vec{b} + \vec{c} \)**:
We need to find the magnitude of \( \vec{a} + \vec{b} + \vec{c} \). To do this, we calculate:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \]

3. **Expanding the Dot Product**:
Using the properties of dot products:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors:
\[ \vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 1 \]
Let \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos \theta \). Thus:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 3 + 2(3 \cos \theta) = 3 + 6 \cos \theta \]

4. **Finding the Magnitude**:
Therefore, we have:
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3 + 6 \cos \theta} \]

5. **Finding the Magnitude of \( \vec{g} \)**:
Now, substituting back into the formula for \( \vec{g} \):
\[ |\vec{g}| = \left|\frac{\vec{a} + \vec{b} + \vec{c}}{3}\right| = \frac{1}{3} |\vec{a} + \vec{b} + \vec{c}| = \frac{1}{3} \sqrt{3 + 6 \cos \theta} \]

6. **Evaluating the Range of \( \cos \theta \)**:
Given \( \frac{\pi}{3} < \theta < \frac{\pi}{2} \), we know:
\[ \cos \frac{\pi}{3} = \frac{1}{2} \quad \text{and} \quad \cos \frac{\pi}{2} = 0 \]
Thus, \( \frac{1}{2} > \cos \theta > 0 \).

7. **Finding the Range of \( |\vec{g}| \)**:
Therefore, we find the range of \( |\vec{g}| \):
- When \( \cos \theta = \frac{1}{2} \):
\[ |\vec{g}| = \frac{1}{3} \sqrt{3 + 6 \cdot \frac{1}{2}} = \frac{1}{3} \sqrt{6} \]
- When \( \cos \theta = 0 \):
\[ |\vec{g}| = \frac{1}{3} \sqrt{3 + 0} = \frac{1}{3} \sqrt{3} \]

Thus, \( |\vec{g}| \) varies from \( \frac{1}{3} \sqrt{3} \) to \( \frac{1}{3} \sqrt{6} \).

### Final Result:
The final expression for the magnitude of the centroid \( \vec{g} \) is:
\[ |\vec{g}| = \frac{1}{3} \sqrt{3 + 6 \cos \theta} \]