A force `F=(2hati-hatj+4hatk)N` displaces a particle upto `d=(3hati+2hatj+hat k)m.` Work done by the force is

Options

A. zero
B. 8 J
C. 4 J
D. 12 J

Explanation

To find the work done by the force \( F \) on the particle during its displacement \( d \), we can use the formula for work done, which is given by the dot product of the force vector and the displacement vector:

\[ W = \vec{F} \cdot \vec{d} \]

### Step 1: Identify the Force and Displacement Vectors
The force vector \( \vec{F} \) is given as:
\[ \vec{F} = 2\hat{i} - \hat{j} + 4\hat{k} \, \text{N} \]
The displacement vector \( \vec{d} \) is given as:
\[ \vec{d} = 3\hat{i} + 2\hat{j} + \hat{k} \, \text{m} \]

### Step 2: Calculate the Dot Product
To calculate the work done, we need to perform the dot product \( \vec{F} \cdot \vec{d} \):
\[ \vec{F} \cdot \vec{d} = (2\hat{i} - \hat{j} + 4\hat{k}) \cdot (3\hat{i} + 2\hat{j} + \hat{k}) \]

Using the properties of the dot product:
\[ \vec{F} \cdot \vec{d} = (2 \cdot 3) + (-1 \cdot 2) + (4 \cdot 1) \]

### Step 3: Perform the Multiplications
Now, calculate each term:
- \( 2 \cdot 3 = 6 \)
- \( -1 \cdot 2 = -2 \)
- \( 4 \cdot 1 = 4 \)

### Step 4: Sum the Results
Now, add these results together:
\[ W = 6 - 2 + 4 \]
\[ W = 8 \, \text{J} \]

### Conclusion
The work done by the force is:
\[ \boxed{8 \, \text{J}} \]