Work done by a force `F=(hati+2hatj+3hatk)` acting on a particle in displacing it from the point `r_(1)=hati+hatj+hatk` to the point `r_(2)=hati-hatj+hat2k` is

Options

A. `-3 J`
B. `-1 J`
C. zero
D. 2 J

Explanation

To find the work done by the force \( \mathbf{F} = \hat{i} + 2\hat{j} + 3\hat{k} \) as it displaces a particle from point \( \mathbf{r_1} = \hat{i} + \hat{j} + \hat{k} \) to point \( \mathbf{r_2} = \hat{i} - \hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Determine the Displacement Vector
The displacement vector \( \mathbf{S} \) is given by the difference between the final position vector \( \mathbf{r_2} \) and the initial position vector \( \mathbf{r_1} \).

\[ \mathbf{S} = \mathbf{r_2} - \mathbf{r_1} \]

Substituting the values:

\[ \mathbf{S} = (\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) \]

### Step 2: Simplify the Displacement Vector
Now, we simplify the expression:

\[ \mathbf{S} = \hat{i} - \hat{j} + 2\hat{k} - \hat{i} - \hat{j} - \hat{k} \]

Combining like terms:

\[ \mathbf{S} = (1 - 1)\hat{i} + (-1 - 1)\hat{j} + (2 - 1)\hat{k} = 0\hat{i} - 2\hat{j} + 1\hat{k} \]

Thus, the displacement vector is:

\[ \mathbf{S} = -2\hat{j} + \hat{k} \]

### Step 3: Calculate the Work Done
The work done \( W \) by the force \( \mathbf{F} \) is given by the dot product of the force vector and the displacement vector:

\[ W = \mathbf{F} \cdot \mathbf{S} \]

Substituting the values of \( \mathbf{F} \) and \( \mathbf{S} \):

\[ W = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (0\hat{i} - 2\hat{j} + 1\hat{k}) \]

### Step 4: Perform the Dot Product
Calculating the dot product:

\[ W = (1 \cdot 0) + (2 \cdot -2) + (3 \cdot 1) \]

This simplifies to:

\[ W = 0 - 4 + 3 = -1 \]

### Final Result
Thus, the work done by the force is:

\[ W = -1 \text{ joules} \]