A plane body has perpendicular axes OX and OY marked on it and is acted on by following forces 5P in the direction OY, 4P in the direction OX, 10P in the direction OA where A is the point `(3a, 4a)` 15 P in the direction AB where B is the point `(-a, a)`.
Express each force in the vector form & calculate the magnitude & direction of the vector sum of these forces.

Explanation

To solve the problem step by step, we will express each force in vector form and then calculate the vector sum of these forces. Finally, we will determine the magnitude and direction of the resultant vector.

### Step 1: Express each force in vector form

1. **Force F1 (5P in the direction OY)**:
- This force acts along the Y-axis.
- Vector form: \( \vec{F_1} = 5P \hat{j} \)

2. **Force F2 (4P in the direction OX)**:
- This force acts along the X-axis.
- Vector form: \( \vec{F_2} = 4P \hat{i} \)

3. **Force F3 (10P in the direction OA)**:
- Point A is given as (3a, 4a). The direction vector OA can be expressed as \( \vec{OA} = 3\hat{i} + 4\hat{j} \).
- The magnitude of OA is \( \sqrt{(3^2 + 4^2)} = 5 \).
- Therefore, the unit vector in the direction of OA is \( \frac{3\hat{i} + 4\hat{j}}{5} \).
- Thus, the vector form of F3 is:
\[ \vec{F_3} = 10P \cdot \frac{3\hat{i} + 4\hat{j}}{5} = 6P \hat{i} + 8P \hat{j} \]

4. **Force F4 (15P in the direction AB)**:
- Point B is given as (-a, a). The direction vector AB can be expressed as \( \vec{AB} = (-a - 3a)\hat{i} + (a - 4a)\hat{j} = -4a\hat{i} - 3a\hat{j} \).
- The magnitude of AB is \( \sqrt{(-4a)^2 + (-3a)^2} = 5a \).
- Therefore, the unit vector in the direction of AB is \( \frac{-4\hat{i} - 3\hat{j}}{5} \).
- Thus, the vector form of F4 is:
\[ \vec{F_4} = 15P \cdot \frac{-4\hat{i} - 3\hat{j}}{5} = -12P \hat{i} - 9P \hat{j} \]

### Step 2: Calculate the vector sum of the forces

Now, we sum all the force vectors:
\[ \vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} \]
Substituting the values:
\[ \vec{F} = (5P \hat{j}) + (4P \hat{i}) + (6P \hat{i} + 8P \hat{j}) + (-12P \hat{i} - 9P \hat{j}) \]
Combining like terms:
\[ \vec{F} = (4P + 6P - 12P) \hat{i} + (5P + 8P - 9P) \hat{j} \]
\[ \vec{F} = (-2P) \hat{i} + (4P) \hat{j} \]

### Step 3: Calculate the magnitude of the resultant vector

The magnitude of the vector \( \vec{F} \) is given by:
\[ |\vec{F}| = \sqrt{(-2P)^2 + (4P)^2} = \sqrt{4P^2 + 16P^2} = \sqrt{20P^2} = \sqrt{20}P = 2\sqrt{5}P \]

### Step 4: Calculate the direction of the resultant vector

The direction (angle \( \alpha \)) can be found using the tangent function:
\[ \tan \alpha = \frac{F_y}{F_x} = \frac{4P}{-2P} = -2 \]
Thus,
\[ \alpha = \tan^{-1}(-2) \]

### Final Answer

The vector sum of the forces is:
\[ \vec{F} = -2P \hat{i} + 4P \hat{j} \]
The magnitude of the resultant vector is:
\[ |\vec{F}| = 2\sqrt{5}P \]
The direction of the resultant vector is:
\[ \alpha = \tan^{-1}(-2) \]