A force of magnitude 6 units acting parallel to `2i-2j+k` displaces the point of application from `A(1, 2, 3) ` to `B(5,3,7)`. Then the work done is

Options

A. 20
B. 30
C. 40
D. 50

Answer

To solve the problem, we need to calculate the work done by a force acting on an object during a displacement. The work done is given by the formula:

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

where \( \mathbf{F} \) is the force vector and \( \mathbf{d} \) is the displacement vector.

### Step 1: Determine the Force Vector

The force has a magnitude of 6 units and acts parallel to the vector \( \mathbf{v} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \).

First, we need to find the unit vector in the direction of \( \mathbf{v} \):

1. Calculate the magnitude of \( \mathbf{v} \):
\[ |\mathbf{v}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

2. Find the unit vector \( \mathbf{u} \):
\[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{3} = \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \]

3. Now, calculate the force vector \( \mathbf{F} \):
\[ \mathbf{F} = 6\mathbf{u} = 6 \left( \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \right) = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k} \]

### Step 2: Calculate the Displacement Vector

The displacement vector \( \mathbf{d} \) from point \( A(1, 2, 3) \) to point \( B(5, 3, 7) \) is calculated as follows:

\[ \mathbf{d} = \mathbf{B} - \mathbf{A} = (5 - 1)\mathbf{i} + (3 - 2)\mathbf{j} + (7 - 3)\mathbf{k} = 4\mathbf{i} + 1\mathbf{j} + 4\mathbf{k} \]

### Step 3: Calculate the Work Done

Now we can find the work done using the dot product:

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

Calculating the dot product:

\[ W = 4 \cdot 4 + (-4) \cdot 1 + 2 \cdot 4 = 16 - 4 + 8 = 20 \]

### Final Answer

The work done is:

\[ \boxed{20} \text{ units} \]