A body is displaced from position `(hati+hatj+hatk)` m to position `(4hati+5hatj+6hatk)` m under the action of force `(5hati+4hatj-3hatk)` N. The angle between the force and displcement vector is

Options

A. `cos^-1(frac{8}{25}`
B. `sin^-1(frac{8}{25})`
C. `cos^-1(frac{11}{25})`
D. `cos^-1(frac{3}{25})`

Answer

To find the angle between the force vector and the displacement vector, we can follow these steps:

### Step 1: Determine the Displacement Vector
The displacement vector \( \vec{d} \) can be calculated by subtracting the initial position vector from the final position vector.

Given:
- Initial position vector \( \vec{P_1} = \hat{i} + \hat{j} + \hat{k} \)
- Final position vector \( \vec{P_2} = 4\hat{i} + 5\hat{j} + 6\hat{k} \)

The displacement vector \( \vec{d} \) is:
\[ \vec{d} = \vec{P_2} - \vec{P_1} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) \]
\[ \vec{d} = (4 - 1)\hat{i} + (5 - 1)\hat{j} + (6 - 1)\hat{k} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

### Step 2: Identify the Force Vector
The force vector \( \vec{F} \) is given as:
\[ \vec{F} = 5\hat{i} + 4\hat{j} - 3\hat{k} \]

### Step 3: Calculate the Dot Product of the Vectors
The dot product \( \vec{d} \cdot \vec{F} \) is calculated as follows:
\[ \vec{d} \cdot \vec{F} = (3\hat{i} + 4\hat{j} + 5\hat{k}) \cdot (5\hat{i} + 4\hat{j} - 3\hat{k}) \]
\[ = 3 \cdot 5 + 4 \cdot 4 + 5 \cdot (-3) = 15 + 16 - 15 = 16 \]

### Step 4: Calculate the Magnitudes of the Vectors
Now we need to find the magnitudes of \( \vec{d} \) and \( \vec{F} \).

Magnitude of \( \vec{d} \):
\[ |\vec{d}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]

Magnitude of \( \vec{F} \):
\[ |\vec{F}| = \sqrt{5^2 + 4^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} \]

### Step 5: Calculate the Cosine of the Angle
Using the dot product and magnitudes, we can find the cosine of the angle \( \theta \):
\[ \cos \theta = \frac{\vec{d} \cdot \vec{F}}{|\vec{d}| |\vec{F}|} = \frac{16}{\sqrt{50} \cdot \sqrt{50}} = \frac{16}{50} = \frac{8}{25} \]

### Step 6: Find the Angle
Finally, we find the angle \( \theta \) by taking the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{8}{25}\right) \]

### Summary
The angle between the force vector and the displacement vector is \( \theta = \cos^{-1}\left(\frac{8}{25}\right) \).

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