The sine of the angle between the vector `a=3hati+hatj+hatk` and `b=2hati-2hatj+hatk` is

Options

A. `sqrt((74)/(99))`
B. `sqrt((55)/(99))`
C. `sqrt((37)/(99))`
D. `(5)/(sqrt(41))`

Answer

To find the sine of the angle between the vectors \( \mathbf{a} = 3\hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{b} = 2\hat{i} - 2\hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Calculate the magnitudes of the vectors
The magnitude of a vector \( \mathbf{v} = x\hat{i} + y\hat{j} + z\hat{k} \) is given by:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \]

For vector \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]

For vector \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 2: Calculate the dot product of the vectors
The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (3\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k}) \]
\[ = 3 \cdot 2 + 1 \cdot (-2) + 1 \cdot 1 = 6 - 2 + 1 = 5 \]

### Step 3: Use the dot product to find \( \cos \theta \)
The relationship between the dot product and the angle \( \theta \) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
Substituting the known values:
\[ 5 = \sqrt{11} \cdot 3 \cdot \cos \theta \]
\[ \cos \theta = \frac{5}{3\sqrt{11}} \]

### Step 4: Find \( \sin \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting for \( \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \left(\frac{5}{3\sqrt{11}}\right)^2 \]
\[ = 1 - \frac{25}{9 \cdot 11} = 1 - \frac{25}{99} = \frac{99 - 25}{99} = \frac{74}{99} \]
Thus,
\[ \sin \theta = \sqrt{\frac{74}{99}} = \frac{\sqrt{74}}{3\sqrt{11}} \]

### Final Answer
The sine of the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ \sin \theta = \frac{\sqrt{74}}{99} \]