If `vec A= 3i + j +2k` and `vec B = 2i - 2j +4k` and `theta` is the angle between the two vectors, then `sin theta` is equal to

Options

A. `2/3`
B. `2/(sqrt(3))`
C. `2/(sqrt(7))`
D. `2/(sqrt(13))`

Explanation

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

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

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

The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

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

Calculating this determinant, we expand it as follows:

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

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ 1 \cdot 4 - 2 \cdot (-2) = 4 + 4 = 8 \]

2. For \( \hat{j} \):
\[ 3 \cdot 4 - 2 \cdot 2 = 12 - 4 = 8 \]
(Note that we subtract this term, so it becomes \(-8\hat{j}\))

3. For \( \hat{k} \):
\[ 3 \cdot (-2) - 1 \cdot 2 = -6 - 2 = -8 \]

Putting it all together, we have:

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

### Step 2: Calculate the magnitude of \( \vec{A} \times \vec{B} \)

The magnitude is given by:

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

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

1. For \( |\vec{A}| \):
\[ |\vec{A}| = \sqrt{(3)^2 + (1)^2 + (2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

2. For \( |\vec{B}| \):
\[ |\vec{B}| = \sqrt{(2)^2 + (-2)^2 + (4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

### Step 4: Substitute into the formula for \( \sin \theta \)

Now we can substitute these values into the formula for \( \sin \theta \):

\[ \sin \theta = \frac{|\vec{A} \times \vec{B}|}{|\vec{A}| |\vec{B}|} = \frac{8\sqrt{3}}{\sqrt{14} \cdot 2\sqrt{6}} \]

Calculating the denominator:

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

Thus,

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

### Final Answer

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