The resultant velocity of the swimmer is the vector sum of the river's velocity and the swimmer's velocity relative to the water.

Given:
- River velocity = 6 km/h
- Swimmer's velocity relative to water = 9 km/h

Using the Pythagorean theorem:

\[
v_{\text{resultant}} = \sqrt{(9^2 + 6^2)} = \sqrt{81 + 36} = \sqrt{117}
\]

Thus, the resultant velocity of the swimmer is:

\[
{\sqrt{117} \text{km/h}}
\]

To find the velocity of the monkey as observed by a man standing on the ground, we need to add the velocity of the monkey relative to the train to the velocity of the train.

 Given:
- Velocity of the train (north direction) = 54 km/h
- Velocity of the monkey relative to the train (opposite to train's motion) = 18 km/h

Monkey's velocity relative to the ground:
Since the monkey is running against the train’s motion, the monkey's velocity relative to the ground will be:

\[
\vec{v}_{\text{monkey}} = \vec{v}_{\text{train}} - \vec{v}_{\text{monkey relative to train}}
\]

\[
v_{\text{monkey}} = 54 \, \text{km/h} - 18 \, \text{km/h} = 36 \, \text{km/h}
\]

Thus, the velocity of the monkey as observed by a man on the ground is:

\[
{36 \, \text{km/h} \text{ north} or 10 m/s \text{ north}}
\]

The relative velocity of the boat with respect to the water is given by subtracting the velocity of the water from the velocity of the boat.

\[
\vec{v}_{\text{bw}} = \vec{v}_{\text{boat}} - \vec{v}_{\text{water}}
\]

Given:
\[
\vec{v}_{\text{boat}} = 3\hat{i} + 4\hat{j}, \quad \vec{v}_{\text{water}} = -3\hat{i} - 4\hat{j}
\]

Now, subtract:

\[
\vec{v}_{\text{bw}} = (3\hat{i} + 4\hat{j}) - (-3\hat{i} - 4\hat{j})
\]
\[
\vec{v}_{\text{bw}} = 3\hat{i} + 4\hat{j} + 3\hat{i} + 4\hat{j}
\]
\[
\vec{v}_{\text{bw}} = 6\hat{i} + 8\hat{j}
\]

Thus, the relative velocity of the boat with respect to the water is:

\[ {6\hat{i} + 8\hat{j}} \]

Boat speed in still water \(v_b = 6\text{ km/h}\), river speed \(v_r = 2\text{ km/h}\). Upstream speed \(v_u = v_b - v_r = 4\text{ km/h}\). Downstream speed \(v_d = v_b + v_r = 8\text{ km/h}\). Time upstream \(t_u = 2\text{ km} / 4\text{ km/h} = 0.5\) hours. Time downstream \(t_d = 2\text{ km} / 8\text{ km/h} = 0.25\) hours. Total time = \(0.5 + 0.25 = 0.75\) hours = \(45\) minutes.

Relative velocity is given by \(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (3\hat{i}+4\hat{j}) - (7\hat{i}-3\hat{j}) = -4\hat{i}+7\hat{j}\) m/s. Therefore, option D is correct.

For minimum crossing time, the swimmer must head perpendicular to the river bank. The net ground velocity is the vector sum of swimmer's velocity and river velocity: \(v_g = \sqrt{v_{\text{sw}}^2 + v_r^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\text{ m/s} \approx 9\text{ m/s}\).

The time to cross a river is \(t = \frac{d}{v \sin \theta}\), where \(theta\) is the angle with the river flow. For \(t\) to be minimum, \(sin \theta\) must be maximum, which occurs at \(\theta = 90^\circ = \frac{\pi}{2}\).

Assertion (A): For particles to meet, their relative perpendicular velocity component must be zero, meaning their perpendicular velocities must be equal. Otherwise, they would move apart perpendicular to the line joining them. So (A) is true.

Reason (R): Relative velocity is a vector difference and can be calculated in any direction, not exclusively along the line joining two bodies. So (R) is false.

The time to cross the river depends on the component of velocity perpendicular to the river flow (\(v_{\text{normal}}\)). For simultaneous crossing, \(v_{\text{normal}}\) must be equal for both boats. If total speed relative to river is same, and \(v_{\text{normal}}\) is same, then the magnitude of the parallel component is also same. Different paths result from different directions of the parallel component. Thus, (A) is true and (R) is true and explains (A).