Forces of magnitudes 5 and 3 units acting in the directions ` 6hati + 2hatj + 3hatk and 3 hati - 2hati +6hatk` respectively act on a particle which is displaced from the point ( 2,2,-1) to ( 4,3,1) . The work done by the forces, is

Options

A. 148 unit
B. ` 148/7` unit
C. 296 units
D. none of these

Explanation

To solve the problem step by step, we need to calculate the work done by the forces acting on a particle during its displacement. Here’s how we can do it:

### Step 1: Identify the Forces and Their Directions

We have two forces:
1. Force \( F_1 \) of magnitude 5 units acting in the direction of the vector \( \mathbf{A} = 6\hat{i} + 2\hat{j} + 3\hat{k} \).
2. Force \( F_2 \) of magnitude 3 units acting in the direction of the vector \( \mathbf{B} = 3\hat{i} - 2\hat{j} + 6\hat{k} \).

### Step 2: Calculate the Unit Vectors of the Forces

To find the unit vectors, we first need to calculate the magnitudes of the direction vectors.

**Magnitude of \( \mathbf{A} \)**:
\[ |\mathbf{A}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]
Thus, the unit vector \( \hat{A} \) is:
\[ \hat{A} = \frac{\mathbf{A}}{|\mathbf{A}|} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} \]

**Magnitude of \( \mathbf{B} \)**:
\[ |\mathbf{B}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]
Thus, the unit vector \( \hat{B} \) is:
\[ \hat{B} = \frac{\mathbf{B}}{|\mathbf{B}|} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} \]

### Step 3: Calculate the Force Vectors

Now we can calculate the force vectors by multiplying the magnitudes with the unit vectors.

**Force Vector \( \mathbf{F_1} \)**:
\[ \mathbf{F_1} = 5 \hat{A} = 5 \cdot \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = \frac{30\hat{i} + 10\hat{j} + 15\hat{k}}{7} \]

**Force Vector \( \mathbf{F_2} \)**:
\[ \mathbf{F_2} = 3 \hat{B} = 3 \cdot \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = \frac{9\hat{i} - 6\hat{j} + 18\hat{k}}{7} \]

### Step 4: Calculate the Net Force Vector

Now, we can find the net force vector \( \mathbf{F} \):
\[ \mathbf{F} = \mathbf{F_1} + \mathbf{F_2} = \frac{30\hat{i} + 10\hat{j} + 15\hat{k}}{7} + \frac{9\hat{i} - 6\hat{j} + 18\hat{k}}{7} \]
Combining the components:
\[ \mathbf{F} = \frac{(30 + 9)\hat{i} + (10 - 6)\hat{j} + (15 + 18)\hat{k}}{7} = \frac{39\hat{i} + 4\hat{j} + 33\hat{k}}{7} \]

### Step 5: Calculate the Displacement Vector

The particle is displaced from point \( (2, 2, -1) \) to \( (4, 3, 1) \). The displacement vector \( \mathbf{D} \) is:
\[ \mathbf{D} = (4 - 2)\hat{i} + (3 - 2)\hat{j} + (1 - (-1))\hat{k} = 2\hat{i} + 1\hat{j} + 2\hat{k} \]

### Step 6: Calculate the Work Done

The work done \( W \) by the forces is given by the dot product of the net force vector and the displacement vector:
\[ W = \mathbf{F} \cdot \mathbf{D} = \left(\frac{39\hat{i} + 4\hat{j} + 33\hat{k}}{7}\right) \cdot (2\hat{i} + 1\hat{j} + 2\hat{k}) \]
Calculating the dot product:
\[ W = \frac{1}{7} \left(39 \cdot 2 + 4 \cdot 1 + 33 \cdot 2\right) = \frac{1}{7} (78 + 4 + 66) = \frac{1}{7} \cdot 148 = \frac{148}{7} \]

### Final Answer
The work done by the forces is:
\[ W = \frac{148}{7} \text{ units} \]