What is the sine of the angle between the vectors `hat(i) + 2hat(j) + 3hat(k) and -hat(i) +2hat(j) + 3hat(k)`?

Options

A. `sqrt((13)/(7))`
B. `(sqrt13)/(7)`
C. `(13)/(sqrt7)`
D. none of these

Answer

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

### Step 1: Define the Vectors
Let:
\[ \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \mathbf{B} = -\hat{i} + 2\hat{j} + 3\hat{k} \]

### Step 2: Compute the Cross Product
The cross product \(\mathbf{A} \times \mathbf{B}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\mathbf{A}\) and \(\mathbf{B}\):

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

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ -1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 2 = 0\)
2. \(\begin{vmatrix} 1 & 3 \\ -1 & 3 \end{vmatrix} = 1 \cdot 3 - 3 \cdot (-1) = 3 + 3 = 6\)
3. \(\begin{vmatrix} 1 & 2 \\ -1 & 2 \end{vmatrix} = 1 \cdot 2 - 2 \cdot (-1) = 2 + 2 = 4\)

Thus, we have:
\[ \mathbf{A} \times \mathbf{B} = 0\hat{i} - 6\hat{j} + 4\hat{k} = -6\hat{j} + 4\hat{k} \]

### Step 3: Compute the Magnitude of the Cross Product
The magnitude of \(\mathbf{A} \times \mathbf{B}\) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]

### Step 4: Compute the Magnitudes of the Vectors
Now, we need to compute the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\):
\[ |\mathbf{A}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
\[ |\mathbf{B}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 5: Use the Formula for Sine of the Angle
The sine of the angle \(\theta\) between the two vectors can be found using the formula:
\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \]
Substituting the values we have:
\[ 2\sqrt{13} = \sqrt{14} \cdot \sqrt{14} \cdot \sin \theta \]
\[ 2\sqrt{13} = 14 \sin \theta \]
\[ \sin \theta = \frac{2\sqrt{13}}{14} = \frac{\sqrt{13}}{7} \]

### Final Answer
Thus, the sine of the angle between the vectors is:
\[ \sin \theta = \frac{\sqrt{13}}{7} \]