A force `vec(F) = hat(i) + 3 hat(j) + 2 hat(k)` acts on a particle to displace it from the point A `(hat(i) + 2hat(j) - 3 hat(k))` to the point `B(3hat(i) - hat(j)+5hat(k))`. The work done by the force will be

Options

A. 5 units
B. 7 units
C. 9 units
D. 10 units

Explanation

To find the work done by the force on the particle as it moves from point A to point B, we can follow these steps:

### Step 1: Identify the Force and Points
The force acting on the particle is given as:
\[ \vec{F} = \hat{i} + 3\hat{j} + 2\hat{k} \]
The initial point A is:
\[ A = \hat{i} + 2\hat{j} - 3\hat{k} \]
And the final point B is:
\[ B = 3\hat{i} - \hat{j} + 5\hat{k} \]

### Step 2: Calculate the Displacement Vector \(\vec{AB}\)
The displacement vector \(\vec{AB}\) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Substituting the coordinates of A and B:
\[ \vec{AB} = (3\hat{i} - \hat{j} + 5\hat{k}) - (\hat{i} + 2\hat{j} - 3\hat{k}) \]
Calculating this gives:
\[ \vec{AB} = (3 - 1)\hat{i} + (-1 - 2)\hat{j} + (5 + 3)\hat{k} \]
\[ \vec{AB} = 2\hat{i} - 3\hat{j} + 8\hat{k} \]

### Step 3: Calculate the Work Done
The work done \(W\) by the force is given by the dot product of the force and the displacement:
\[ W = \vec{F} \cdot \vec{AB} \]
Substituting the values of \(\vec{F}\) and \(\vec{AB}\):
\[ W = (\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 3\hat{j} + 8\hat{k}) \]
Calculating the dot product:
\[ W = (1 \cdot 2) + (3 \cdot -3) + (2 \cdot 8) \]
\[ W = 2 - 9 + 16 \]
\[ W = 2 + 16 - 9 = 9 \]

### Final Answer
The work done by the force is:
\[ \boxed{9} \text{ units} \]