Two particles are moving along two long straight lines, in the same plane with same speed equal to `20 cm//s.` The angle between the two linse is `60^@` and their intersection point isO. At a certain moment, the two particles are located at distances 3m and 4m from O and are moving twowards O. Subsequently, the shortest distance between them will be

Options

A. `50 cm`
B. `40sqrt2 cm`
C. `50sqrt2 cm`
D. `50sqrt3 cm`

Explanation

To find the shortest distance between the two particles moving towards each other, we can follow the steps below:

### Step-by-Step Solution

1. **Understand the Initial Setup**:
- Let the two particles be A and B.
- Particle A is initially at a distance of 3 m from point O, and particle B is at a distance of 4 m from point O.
- The angle between the paths of the two particles is \(60^\circ\).

2. **Define the Position Vectors**:
- We can place point O at the origin (0, 0).
- The position of particle A can be represented as:
\[ \vec{R_A} = 3 \hat{i} \quad \text{(moving towards O)} \]
- The position of particle B can be represented as:
\[ \vec{R_B} = 4 \left(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}\right) = 4 \left(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right) = 2 \hat{i} + 2\sqrt{3} \hat{j} \]

3. **Calculate the Relative Position Vector**:
- The relative position vector from A to B is given by:
\[ \vec{R} = \vec{R_A} - \vec{R_B} = (3 \hat{i}) - (2 \hat{i} + 2\sqrt{3} \hat{j}) = (3 - 2) \hat{i} - 2\sqrt{3} \hat{j} = 1 \hat{i} - 2\sqrt{3} \hat{j} \]

4. **Determine the Velocities**:
- Both particles are moving towards point O with a speed of 20 cm/s (0.2 m/s).
- The velocity of particle A is:
\[ \vec{V_A} = -20 \hat{i} \quad \text{(moving towards O)} \]
- The velocity of particle B can be resolved into components:
\[ \vec{V_B} = -20 \left(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}\right) = -20 \left(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right) = -10 \hat{i} - 10\sqrt{3} \hat{j} \]

5. **Calculate the Relative Velocity**:
- The relative velocity of B with respect to A is:
\[ \vec{V_{BA}} = \vec{V_B} - \vec{V_A} = (-10 \hat{i} - 10\sqrt{3} \hat{j}) - (-20 \hat{i}) = (10 \hat{i} - 10\sqrt{3} \hat{j}) \]

6. **Find the Shortest Distance**:
- The shortest distance occurs when the relative position vector is perpendicular to the relative velocity vector.
- The magnitude of the relative position vector is:
\[ |\vec{R}| = \sqrt{(1)^2 + (-2\sqrt{3})^2} = \sqrt{1 + 12} = \sqrt{13} \]
- The magnitude of the relative velocity is:
\[ |\vec{V_{BA}}| = \sqrt{(10)^2 + (-10\sqrt{3})^2} = \sqrt{100 + 300} = \sqrt{400} = 20 \]

7. **Calculate the Perpendicular Component**:
- The perpendicular component of the relative position vector can be calculated using the formula:
\[ R_{\perpendicular} = \sqrt{R^2 - R_{\parallel}^2} \]
- To find \(R_{\parallel}\), we can use the dot product:
\[ R_{\parallel} = \frac{\vec{R} \cdot \vec{V_{BA}}}{|\vec{V_{BA}}|} = \frac{(1)(10) + (-2\sqrt{3})(-10\sqrt{3})}{20} = \frac{10 + 60}{20} = \frac{70}{20} = 3.5 \]
- Now, substituting back:
\[ R_{\perpendicular} = \sqrt{13 - (3.5)^2} = \sqrt{13 - 12.25} = \sqrt{0.75} \approx 0.866 \text{ m} \]

8. **Convert to Centimeters**:
- The shortest distance between the two particles is:
\[ 0.866 \text{ m} \times 100 = 86.6 \text{ cm} \]

### Final Answer
The shortest distance between the two particles will be approximately **86.6 cm**.