The valueof sine of the angle between the vectors `i -2j +3k` and `2i + j + k` is :

Options

A. `(5)/(21)`
B. `(5)/(sqrt(7))`
C. `(5)/(sqrt(14))`
D. `(5)/(2sqrt(7))`

Answer

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

### Step 1: Calculate the dot product of the vectors \( \mathbf{A} \) and \( \mathbf{B} \)

The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:

\[ \mathbf{A} \cdot \mathbf{B} = (1)(2) + (-2)(1) + (3)(1) = 2 - 2 + 3 = 3 \]

### Step 2: Calculate the magnitudes of the vectors \( \mathbf{A} \) and \( \mathbf{B} \)

The magnitude of vector \( \mathbf{A} \) is:

\[ |\mathbf{A}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

The magnitude of vector \( \mathbf{B} \) is:

\[ |\mathbf{B}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 3: Use the dot product to find \( \cos \theta \)

Using the formula for the dot product:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]

We can rearrange to find \( \cos \theta \):

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{3}{\sqrt{14} \cdot \sqrt{6}} = \frac{3}{\sqrt{84}} = \frac{3}{2\sqrt{21}} \]

### Step 4: Calculate \( \sin \theta \)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \sin^2 \theta = 1 - \cos^2 \theta \]

Calculating \( \cos^2 \theta \):

\[ \cos^2 \theta = \left(\frac{3}{2\sqrt{21}}\right)^2 = \frac{9}{4 \cdot 21} = \frac{9}{84} = \frac{3}{28} \]

Now substituting back to find \( \sin^2 \theta \):

\[ \sin^2 \theta = 1 - \frac{3}{28} = \frac{28 - 3}{28} = \frac{25}{28} \]

Taking the square root to find \( \sin \theta \):

\[ \sin \theta = \sqrt{\frac{25}{28}} = \frac{5}{\sqrt{28}} = \frac{5}{2\sqrt{7}} \]

### Final Result

Thus, the value of sine of the angle between the vectors is:

\[ \sin \theta = \frac{5}{2\sqrt{7}} \]