Find a unit vector perpendicular to the vector `4hati + 3hatj + hatk` and `2hati-hatj+2hatk` Determine the sine angle between these two vectors.

Answer

To find a unit vector perpendicular to the vectors \( \mathbf{A} = 4\hat{i} + 3\hat{j} + \hat{k} \) and \( \mathbf{B} = 2\hat{i} - \hat{j} + 2\hat{k} \), and to determine the sine of the angle between these two vectors, we can follow these steps:

### Step 1: Calculate the Cross Product
The unit vector perpendicular to both vectors can be found using the cross product \( \mathbf{A} \times \mathbf{B} \).

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

Calculating the determinant:

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

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ 3 \cdot 2 - (-1) \cdot 1 = 6 + 1 = 7 \]

2. For \( \hat{j} \):
\[ 4 \cdot 2 - 1 \cdot 2 = 8 - 2 = 6 \]

3. For \( \hat{k} \):
\[ 4 \cdot (-1) - 3 \cdot 2 = -4 - 6 = -10 \]

Thus, we have:
\[ \mathbf{A} \times \mathbf{B} = 7\hat{i} - 6\hat{j} - 10\hat{k} \]

### Step 2: Calculate the Magnitude of the Cross Product
Next, we find the magnitude of \( \mathbf{A} \times \mathbf{B} \):

\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{7^2 + (-6)^2 + (-10)^2} = \sqrt{49 + 36 + 100} = \sqrt{185} \]

### Step 3: Calculate the Unit Vector
The unit vector \( \mathbf{N} \) perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \) is given by:

\[ \mathbf{N} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{7\hat{i} - 6\hat{j} - 10\hat{k}}{\sqrt{185}} = \frac{7}{\sqrt{185}}\hat{i} - \frac{6}{\sqrt{185}}\hat{j} - \frac{10}{\sqrt{185}}\hat{k} \]

### Step 4: Calculate the Magnitudes of Vectors A and B
Now, we calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \):

\[ |\mathbf{A}| = \sqrt{4^2 + 3^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \]

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

### Step 5: Calculate the Sine of the Angle Between the Vectors
Using the formula for the sine of the angle \( \theta \) between two vectors:

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

Substituting the values we have:

\[ \sqrt{185} = \sqrt{26} \cdot 3 \cdot \sin \theta \]

Thus, we can solve for \( \sin \theta \):

\[ \sin \theta = \frac{\sqrt{185}}{3\sqrt{26}} \]

### Final Answers
1. The unit vector perpendicular to both vectors is:
\[ \mathbf{N} = \frac{7}{\sqrt{185}}\hat{i} - \frac{6}{\sqrt{185}}\hat{j} - \frac{10}{\sqrt{185}}\hat{k} \]

2. The sine of the angle between the two vectors is:
\[ \sin \theta = \frac{\sqrt{185}}{3\sqrt{26}} \]