Forces acting on a particle have magnitude of 14N, 7N and 7N act in the direction of vectors
`6hat(i)+2hat(j)+3hat(k),3hat(i)-2hat(j) +6hat(k),2hat(i)-3hat(j)-6hat(k)`
respectively. The forces remain constant while the particle is displaced from point A: (2,-1,-3) to B: (5,-1,1). Find the work done. The coordinates are specified in meters.

Options

A. 75 J
B. 55 J
C. 85 J
D. 65 J

Answer

To find the work done on the particle as it moves from point A to point B under the influence of the given forces, we will follow these steps:

### Step 1: Identify the Forces and Their Directions
We have three forces acting on the particle:
1. **Force F1**: Magnitude = 14 N, Direction = \(6\hat{i} + 2\hat{j} + 3\hat{k}\)
2. **Force F2**: Magnitude = 7 N, Direction = \(3\hat{i} - 2\hat{j} + 6\hat{k}\)
3. **Force F3**: Magnitude = 7 N, Direction = \(2\hat{i} - 3\hat{j} - 6\hat{k}\)

### Step 2: Calculate the Unit Vectors for Each Force
To find the unit vectors, we need to calculate the magnitude of each direction vector and then divide the vector by its magnitude.

**Magnitude of F1's direction vector**:
\[ \text{Magnitude} = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]
**Unit vector for F1**:
\[ \hat{u_1} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = \frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k} \]
**Force F1**:
\[ F_1 = 14 \cdot \hat{u_1} = 14 \left(\frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k}\right) = 12\hat{i} + 4\hat{j} + 6\hat{k} \, \text{N} \]

**Magnitude of F2's direction vector**:
\[ \text{Magnitude} = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]
**Unit vector for F2**:
\[ \hat{u_2} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \]
**Force F2**:
\[ F_2 = 7 \cdot \hat{u_2} = 7 \left(\frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k}\right) = 3\hat{i} - 2\hat{j} + 6\hat{k} \, \text{N} \]

**Magnitude of F3's direction vector**:
\[ \text{Magnitude} = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
**Unit vector for F3**:
\[ \hat{u_3} = \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \]
**Force F3**:
\[ F_3 = 7 \cdot \hat{u_3} = 7 \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 2\hat{i} - 3\hat{j} - 6\hat{k} \, \text{N} \]

### Step 3: Calculate the Net Force
Now we sum the forces:
\[ F_{\text{net}} = F_1 + F_2 + F_3 \]
\[ F_{\text{net}} = (12\hat{i} + 4\hat{j} + 6\hat{k}) + (3\hat{i} - 2\hat{j} + 6\hat{k}) + (2\hat{i} - 3\hat{j} - 6\hat{k}) \]
\[ F_{\text{net}} = (12 + 3 + 2)\hat{i} + (4 - 2 - 3)\hat{j} + (6 + 6 - 6)\hat{k} \]
\[ F_{\text{net}} = 17\hat{i} - 1\hat{j} + 6\hat{k} \, \text{N} \]

### Step 4: Calculate the Displacement Vector
The displacement vector \( \vec{dr} \) from point A (2, -1, -3) to point B (5, -1, 1) is:
\[ \vec{dr} = (5 - 2)\hat{i} + (-1 + 1)\hat{j} + (1 + 3)\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k} \]

### Step 5: Calculate the Work Done
The work done \( W \) is given by the dot product of the net force and the displacement vector:
\[ W = F_{\text{net}} \cdot \vec{dr} = (17\hat{i} - 1\hat{j} + 6\hat{k}) \cdot (3\hat{i} + 0\hat{j} + 4\hat{k}) \]
\[ W = (17 \cdot 3) + (-1 \cdot 0) + (6 \cdot 4) = 51 + 0 + 24 = 75 \, \text{J} \]

### Final Answer
The work done is \( \boxed{75 \, \text{J}} \).

---