Unit vector perpnicular to vector `A=-3hati-2hatj -3hatk and 2hati + 4hatj +6hatk` is

Explanation

To find the unit vector that is perpendicular to the vectors \( \mathbf{A} = -3\hat{i} - 2\hat{j} - 3\hat{k} \) and \( \mathbf{B} = 2\hat{i} + 4\hat{j} + 6\hat{k} \), we will use the cross product of the two vectors. Here’s a step-by-step solution:

### Step 1: Write down the vectors
We have:
\[ \mathbf{A} = -3\hat{i} - 2\hat{j} - 3\hat{k} \]
\[ \mathbf{B} = 2\hat{i} + 4\hat{j} + 6\hat{k} \]

### Step 2: Set up 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 \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \mathbf{A} \) and \( \mathbf{B} \):
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -2 & -3 \\ 2 & 4 & 6 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we expand it:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -2 & -3 \\ 4 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & -3 \\ 2 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & -2 \\ 2 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -2 & -3 \\ 4 & 6 \end{vmatrix} = (-2)(6) - (-3)(4) = -12 + 12 = 0 \]

2. For \( -\hat{j} \):
\[ \begin{vmatrix} -3 & -3 \\ 2 & 6 \end{vmatrix} = (-3)(6) - (-3)(2) = -18 + 6 = -12 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} -3 & -2 \\ 2 & 4 \end{vmatrix} = (-3)(4) - (-2)(2) = -12 + 4 = -8 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 0\hat{i} + 12\hat{j} - 8\hat{k} = 12\hat{j} - 8\hat{k} \]

### Step 4: Write the resulting vector
Thus,
\[ \mathbf{R} = 12\hat{j} - 8\hat{k} \]

### Step 5: Calculate the magnitude of \( \mathbf{R} \)
The magnitude of \( \mathbf{R} \) is given by:
\[ |\mathbf{R}| = \sqrt{(0)^2 + (12)^2 + (-8)^2} = \sqrt{0 + 144 + 64} = \sqrt{208} = 4\sqrt{13} \]

### Step 6: Find the unit vector
The unit vector \( \hat{u} \) in the direction of \( \mathbf{R} \) is:
\[ \hat{u} = \frac{\mathbf{R}}{|\mathbf{R}|} = \frac{12\hat{j} - 8\hat{k}}{4\sqrt{13}} = \frac{12}{4\sqrt{13}}\hat{j} - \frac{8}{4\sqrt{13}}\hat{k} = \frac{3}{\sqrt{13}}\hat{j} - \frac{2}{\sqrt{13}}\hat{k} \]

### Final Answer
The unit vector perpendicular to the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is:
\[ \hat{u} = \frac{3}{\sqrt{13}} \hat{j} - \frac{2}{\sqrt{13}} \hat{k} \]