A groove is in the form of a broken line ABC and the position vectors fo the three points are respectively ` 2hati -3hatj +2hatk , 3hati -hatk , hati +hatj +hatk`, A force of magnitude ` 24sqrt3` acts on a particle of unit mass kept at the point A and moves it angle the groove to the point C. If the line of action of the force is parallel to the vector ` hati +2hatj +hatk` all along, the number of units of work done by the force is

Options

A. `144sqrt2`
B. `144sqrt3`
C. `72sqrt2`
D. `72sqrt3`

Explanation

To solve the problem step by step, we will follow the outlined approach to calculate the work done by the force acting on the particle as it moves from point A to point C along the groove.

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{A} = 2\hat{i} - 3\hat{j} + 2\hat{k} \)
- \( \vec{B} = 3\hat{i} - \hat{k} \)
- \( \vec{C} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Calculate the displacement vector \( \vec{AC} \)
The displacement vector from point A to point C is given by:
\[ \vec{AC} = \vec{C} - \vec{A} \]
Substituting the position vectors:
\[ \vec{AC} = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - 3\hat{j} + 2\hat{k}) \]
\[ \vec{AC} = \hat{i} + \hat{j} + \hat{k} - 2\hat{i} + 3\hat{j} - 2\hat{k} \]
\[ \vec{AC} = (-1\hat{i} + 4\hat{j} - \hat{k}) \]

### Step 3: Determine the force vector
The force has a magnitude of \( 24\sqrt{3} \) and is parallel to the vector \( \hat{i} + 2\hat{j} + \hat{k} \). First, we need to find the unit vector in the direction of the force:
\[ \text{Magnitude of } \hat{i} + 2\hat{j} + \hat{k} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6} \]
Thus, the unit vector in the direction of the force is:
\[ \hat{u} = \frac{\hat{i} + 2\hat{j} + \hat{k}}{\sqrt{6}} \]
Now, the force vector \( \vec{F} \) can be expressed as:
\[ \vec{F} = 24\sqrt{3} \cdot \hat{u} = 24\sqrt{3} \cdot \frac{\hat{i} + 2\hat{j} + \hat{k}}{\sqrt{6}} = \frac{24\sqrt{3}}{\sqrt{6}} (\hat{i} + 2\hat{j} + \hat{k}) \]
\[ \vec{F} = 24\sqrt{2} (\hat{i} + 2\hat{j} + \hat{k}) \]

### Step 4: Calculate the work done
The work done \( W \) by the force is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{AC} \]
Substituting the vectors:
\[ W = (24\sqrt{2} (\hat{i} + 2\hat{j} + \hat{k})) \cdot (-\hat{i} + 4\hat{j} - \hat{k}) \]
Calculating the dot product:
\[ W = 24\sqrt{2} \left( (-1) + (2 \cdot 4) + (-1) \right) \]
\[ W = 24\sqrt{2} ( -1 + 8 - 1) = 24\sqrt{2} \cdot 6 \]
\[ W = 144\sqrt{2} \]

### Final Answer
The work done by the force is \( 144\sqrt{2} \) units.

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