The work done by the force `4i-3j+2k` in moving a particle along a straight line from the point (3,2,-1 ) to (2,-1,4 ) is

Options

A. 0 units
B. 4 units
C. 15 units
D. 19 units

Answer

To find the work done by the force \( \mathbf{F} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) in moving a particle from the point \( (3, 2, -1) \) to the point \( (2, -1, 4) \), we will follow these steps:

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

\[ \mathbf{d} = (2 - 3)\mathbf{i} + (-1 - 2)\mathbf{j} + (4 - (-1))\mathbf{k} \]

Calculating each component:

- \( 2 - 3 = -1 \)
- \( -1 - 2 = -3 \)
- \( 4 - (-1) = 4 + 1 = 5 \)

Thus, the displacement vector is:

\[ \mathbf{d} = -1\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \]

### 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 of \( \mathbf{F} \) and \( \mathbf{d} \):

\[ W = (4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) \cdot (-1\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) \]

### Step 3: Calculate the Dot Product
Using the formula for the dot product:

\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \]

Calculating each term:

- \( 4 \cdot (-1) = -4 \)
- \( -3 \cdot (-3) = 9 \)
- \( 2 \cdot 5 = 10 \)

Adding these results together:

\[ W = -4 + 9 + 10 = 15 \]

### Final Result
The work done by the force in moving the particle is:

\[ \boxed{15} \]