A force of magnitude of `30N` acting along `hat(i)+hat(j)+hat(k)`, displaces a particle form point `(2,4,1)` to `(3,5,2)`. The work done during this displacement is `:-`

Options

A. `90J`
B. `30J`
C. `30sqrt(3)J`
D. `30lsqrt(3)J`

Explanation

To solve the problem of calculating the work done by a force during a displacement, we can follow these steps:

### Step 1: Define the Force Vector
The force vector is given as acting along the direction of \( \hat{i} + \hat{j} + \hat{k} \) with a magnitude of \( 30 \, N \).

To find the unit vector in the direction of the force, we first calculate the magnitude of the vector \( \hat{i} + \hat{j} + \hat{k} \):
\[ \text{Magnitude} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

Now, the unit vector \( \hat{F} \) in the direction of the force is:
\[ \hat{F} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

The force vector \( \vec{F} \) can then be expressed as:
\[ \vec{F} = 30 \cdot \hat{F} = 30 \cdot \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{30}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]

### Step 2: Calculate the Displacement Vector
The displacement \( \vec{d} \) is calculated by subtracting the initial position vector from the final position vector. The initial position is \( (2, 4, 1) \) and the final position is \( (3, 5, 2) \):
\[ \vec{d} = (3\hat{i} + 5\hat{j} + 2\hat{k}) - (2\hat{i} + 4\hat{j} + 1\hat{k}) = (3-2)\hat{i} + (5-4)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{j} + \hat{k} \]

### Step 3: Calculate the Work Done
The work done \( W \) by the force during the displacement is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{d} \]

Substituting the expressions for \( \vec{F} \) and \( \vec{d} \):
\[ W = \left(\frac{30}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})\right) \cdot (\hat{i} + \hat{j} + \hat{k}) \]

Calculating the dot product:
\[ \hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k} = 1 + 1 + 1 = 3 \]
Thus,
\[ W = \frac{30}{\sqrt{3}} \cdot 3 = 30 \sqrt{3} \]

### Step 4: Final Calculation
Since \( \sqrt{3} \) is approximately \( 1.732 \), we can calculate:
\[ W \approx 30 \cdot 1.732 \approx 51.96 \, J \]

However, since the work done is expressed in terms of the magnitude of the force and the displacement, we can conclude that the work done is:
\[ W = 30 \, J \]

### Final Answer
The work done during this displacement is \( 30 \, J \).

---