Given that `P=Q=R`. If `vec(P)+vec(Q)=vec(R)` then the angle between `vec(P)` and `vec(R)` is `theta_(1)`. If `vec(P)+vec(Q)+vec(R)=vec(0)` then the angle between `vec(P)` and `vec(R)` is `theta_(2)`. The relation between `theta_(1)` and `theta_(2)` is :-

Options

A. `theta_(1)=theta_(2)`
B. `theta_(1)=theta_(2)/2`
C. `theta_(1)=2theta_(2)`
D. None of these

Explanation

To solve the problem, we need to analyze two cases involving vectors \( \vec{P}, \vec{Q}, \) and \( \vec{R} \) where \( P = Q = R \).

### Case 1: \( \vec{P} + \vec{Q} = \vec{R} \)

1. **Given**: \( \vec{P} + \vec{Q} = \vec{R} \) and \( P = Q = R \).
2. **Substituting**: Since \( P = Q = R \), we can write \( \vec{Q} = \vec{R} - \vec{P} \).
3. **Using the cosine rule**: The magnitude of \( \vec{Q} \) can be expressed as:
\[ |\vec{Q}|^2 = |\vec{R}|^2 + |\vec{P}|^2 - 2 |\vec{R}||\vec{P}| \cos(\theta_1) \]
Since \( |\vec{P}| = |\vec{Q}| = |\vec{R}| = x \):
\[ x^2 = x^2 + x^2 - 2x^2 \cos(\theta_1) \]
4. **Simplifying**:
\[ x^2 = 2x^2 - 2x^2 \cos(\theta_1) \]
\[ 0 = x^2 - 2x^2 \cos(\theta_1) \]
\[ 2x^2 \cos(\theta_1) = x^2 \]
\[ \cos(\theta_1) = \frac{1}{2} \]
5. **Finding \( \theta_1 \)**:
\[ \theta_1 = 60^\circ \]

### Case 2: \( \vec{P} + \vec{Q} + \vec{R} = \vec{0} \)

1. **Given**: \( \vec{P} + \vec{Q} + \vec{R} = \vec{0} \).
2. **Rearranging**: This implies \( \vec{R} = -(\vec{P} + \vec{Q}) \).
3. **Using the cosine rule**: The magnitude of \( \vec{Q} \) can be expressed as:
\[ |\vec{Q}|^2 = |\vec{P}|^2 + |\vec{R}|^2 + 2 |\vec{P}||\vec{R}| \cos(\theta_2) \]
Since \( |\vec{P}| = |\vec{Q}| = |\vec{R}| = x \):
\[ x^2 = x^2 + x^2 + 2x^2 \cos(\theta_2) \]
4. **Simplifying**:
\[ x^2 = 2x^2 + 2x^2 \cos(\theta_2) \]
\[ 0 = x^2 + 2x^2 \cos(\theta_2) \]
\[ 2x^2 \cos(\theta_2) = -x^2 \]
\[ \cos(\theta_2) = -\frac{1}{2} \]
5. **Finding \( \theta_2 \)**:
\[ \theta_2 = 120^\circ \]

### Conclusion: Relation between \( \theta_1 \) and \( \theta_2 \)

From our calculations:
- \( \theta_1 = 60^\circ \)
- \( \theta_2 = 120^\circ \)

Thus, the relation between \( \theta_1 \) and \( \theta_2 \) is:
\[ \theta_1 = \frac{\theta_2}{2} \]