If `theta` is the angle between the vectors `2hati-2hatj+4hatk` and `3hati+hatj+2hatk`, then `sin theta` is equal to

Options

A. `(2)/(3)`
B. `(2)/(sqrt(7))`
C. `(sqrt(2))/(7)`
D. `sqrt((2)/(7))`

Answer

To find \( \sin \theta \) where \( \theta \) is the angle between the vectors \( \mathbf{A} = 2\hat{i} - 2\hat{j} + 4\hat{k} \) and \( \mathbf{B} = 3\hat{i} + \hat{j} + 2\hat{k} \), we will use the formula:

\[ \sin \theta = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 1: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)

The cross product can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 4 \\ 3 & 1 & 2 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -2 & 4 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -2 & 4 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (4)(1) = -4 - 4 = -8 \)
2. \( \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} = (2)(2) - (4)(3) = 4 - 12 = -8 \)
3. \( \begin{vmatrix} 2 & -2 \\ 3 & 1 \end{vmatrix} = (2)(1) - (-2)(3) = 2 + 6 = 8 \)

Putting it all together:

\[ \mathbf{A} \times \mathbf{B} = -8\hat{i} + 8\hat{j} + 8\hat{k} \]

### Step 2: Calculate the magnitude of the cross product

\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(-8)^2 + 8^2 + 8^2} = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3} \]

### Step 3: Calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \)

1. For \( \mathbf{A} \):

\[ |\mathbf{A}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

2. For \( \mathbf{B} \):

\[ |\mathbf{B}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 4: Substitute into the sine formula

Now substitute the values into the sine formula:

\[ \sin \theta = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}| |\mathbf{B}|} = \frac{8\sqrt{3}}{(2\sqrt{6})(\sqrt{14})} \]

Calculating the denominator:

\[ |\mathbf{A}| |\mathbf{B}| = 2\sqrt{6} \cdot \sqrt{14} = 2\sqrt{84} = 2 \cdot 2\sqrt{21} = 4\sqrt{21} \]

Thus,

\[ \sin \theta = \frac{8\sqrt{3}}{4\sqrt{21}} = \frac{2\sqrt{3}}{\sqrt{21}} \]

### Step 5: Rationalize the denominator

To rationalize the denominator:

\[ \sin \theta = \frac{2\sqrt{3}}{\sqrt{21}} \cdot \frac{\sqrt{21}}{\sqrt{21}} = \frac{2\sqrt{63}}{21} = \frac{2 \cdot 3\sqrt{7}}{21} = \frac{6\sqrt{7}}{21} = \frac{2\sqrt{7}}{7} \]

### Final Answer

Thus,

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