Two particles are projected simultaneously from the same point, with the same speed, in the same vertical plane, and at different angles with the horizontal in a uniform gravitational field acting vertically downwards. A frame of reference is fixed to one particle. The position vector of the other particle , as observed from this frame, is `vecr`. Which of the following statement is correct?

Options

A. `vecr` is a constant vector.
B. `vecr` changes in magnitude as well as direction with time.
C. The magnitude of `vecr` increases linearly with time, its direction does not change.
D. The direction of `vecr` changes with time, its magnitude may or may not change, depending on the angles of projection.

Explanation

To solve the problem, we need to analyze the motion of two particles projected from the same point with the same speed but at different angles in a uniform gravitational field. We will derive the position vector of one particle as observed from the frame of reference of the other particle.

### Step-by-Step Solution:

1. **Define the Initial Conditions**:
- Let the speed of projection for both particles be \( V \).
- Let the angles of projection be \( \alpha \) for particle A and \( \beta \) for particle B.

2. **Velocity Components**:
- The velocity components of particle A:
\[ \vec{V_A} = V \cos(\alpha) \hat{i} + V \sin(\alpha) \hat{j} \]
- The velocity components of particle B:
\[ \vec{V_B} = V \cos(\beta) \hat{i} + V \sin(\beta) \hat{j} \]

3. **Relative Velocity**:
- The velocity of particle B with respect to particle A is given by:
\[ \vec{V_{BA}} = \vec{V_B} - \vec{V_A} = \left( V \cos(\beta) - V \cos(\alpha) \right) \hat{i} + \left( V \sin(\beta) - V \sin(\alpha) \right) \hat{j} \]

4. **Position Vector**:
- The position vector of particle B as observed from the frame of reference of particle A is given by:
\[ \vec{r} = \vec{r_B} - \vec{r_A} \]
- The position vectors of particles A and B at time \( t \) can be expressed as:
\[ \vec{r_A} = V \cos(\alpha) t \hat{i} + \left( V \sin(\alpha) t - \frac{1}{2} g t^2 \right) \hat{j} \]
\[ \vec{r_B} = V \cos(\beta) t \hat{i} + \left( V \sin(\beta) t - \frac{1}{2} g t^2 \right) \hat{j} \]

5. **Subtracting the Position Vectors**:
- Now, substituting the position vectors into the equation for \( \vec{r} \):
\[ \vec{r} = \left( V \cos(\beta) t - V \cos(\alpha) t \right) \hat{i} + \left( V \sin(\beta) t - V \sin(\alpha) t \right) \hat{j} \]
- Simplifying this gives:
\[ \vec{r} = V \left( \cos(\beta) - \cos(\alpha) \right) t \hat{i} + V \left( \sin(\beta) - \sin(\alpha) \right) t \hat{j} \]

6. **Conclusion**:
- The position vector \( \vec{r} \) is linearly dependent on time \( t \) and the direction of \( \vec{r} \) does not change with time since the coefficients \( \left( \cos(\beta) - \cos(\alpha) \right) \) and \( \left( \sin(\beta) - \sin(\alpha) \right) \) are constants.
- Therefore, the magnitude of \( \vec{r} \) increases linearly with time, while the direction remains constant.

### Correct Statement:
The correct statement is that the magnitude of the position vector \( \vec{r} \) increases linearly with time, while the direction remains unchanged.