Work done when a force `F=(hati + 2hat(j) + 3hatk) N` acting on a particle takes it from the point
`r_(1) =(hati + hatk)` the point `r_(2) =(hati -hatj + 2hatk)` is .

Options

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

Explanation

To find the work done by the force \( \mathbf{F} = \hat{i} + 2\hat{j} + 3\hat{k} \) when it moves a particle from point \( \mathbf{r}_1 = \hat{i} + \hat{k} \) to point \( \mathbf{r}_2 = \hat{i} - \hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Calculate the Displacement Vector
The displacement vector \( \mathbf{d} \) can be calculated as:
\[ \mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 \]
Substituting the given points:
\[ \mathbf{d} = (\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} + \hat{k}) \]
Simplifying this gives:
\[ \mathbf{d} = \hat{i} - \hat{j} + 2\hat{k} - \hat{i} - \hat{k} = -\hat{j} + \hat{k} \]

### Step 2: Calculate the Work Done
The work done \( W \) by the force is given by the dot product of the force vector \( \mathbf{F} \) and the displacement vector \( \mathbf{d} \):
\[ W = \mathbf{F} \cdot \mathbf{d} \]
Substituting the values:
\[ W = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{j} + \hat{k}) \]
Calculating the dot product:
\[ W = \hat{i} \cdot (-\hat{j}) + 2\hat{j} \cdot (-\hat{j}) + 3\hat{k} \cdot \hat{k} \]
This results in:
\[ W = 0 - 2 + 3 = 1 \text{ joule} \]

### Final Answer
The work done is \( 1 \text{ joule} \).

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