Find the unit vector perpendicular to the vectors `vecA=(hati+2hatj-3hatk)andvecB=(-hati+hatj-hatk)`

Explanation

To find the unit vector that is perpendicular to the vectors \(\vec{A} = \hat{i} + 2\hat{j} - 3\hat{k}\) and \(\vec{B} = -\hat{i} + \hat{j} - \hat{k}\), we will follow these steps:

### Step 1: Calculate the Cross Product of \(\vec{A}\) and \(\vec{B}\)

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

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

Calculating the determinant:

1. For \(\hat{i}\):
\[ \hat{i} \cdot \begin{vmatrix} 2 & -3 \\ 1 & -1 \end{vmatrix} = \hat{i} (2 \cdot -1 - (-3) \cdot 1) = \hat{i} (-2 + 3) = \hat{i} (1) \]

2. For \(-\hat{j}\):
\[ -\hat{j} \cdot \begin{vmatrix} 1 & -3 \\ -1 & -1 \end{vmatrix} = -\hat{j} (1 \cdot -1 - (-3) \cdot -1) = -\hat{j} (-1 - 3) = -\hat{j} (-4) = 4\hat{j} \]

3. For \(\hat{k}\):
\[ \hat{k} \cdot \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = \hat{k} (1 \cdot 1 - 2 \cdot -1) = \hat{k} (1 + 2) = 3\hat{k} \]

Combining these results, we have:
\[ \vec{A} \times \vec{B} = \hat{i} + 4\hat{j} + 3\hat{k} \]

### Step 2: Calculate the Magnitude of the Cross Product

The magnitude of \(\vec{A} \times \vec{B}\) is given by:
\[ |\vec{A} \times \vec{B}| = \sqrt{(1)^2 + (4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]

### Step 3: Find the Unit Vector

The unit vector \(\hat{n}\) that is perpendicular to both \(\vec{A}\) and \(\vec{B}\) is given by dividing the cross product by its magnitude:
\[ \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{\hat{i} + 4\hat{j} + 3\hat{k}}{\sqrt{26}} \]

Thus, the unit vector perpendicular to \(\vec{A}\) and \(\vec{B}\) is:
\[ \hat{n} = \frac{1}{\sqrt{26}} \hat{i} + \frac{4}{\sqrt{26}} \hat{j} + \frac{3}{\sqrt{26}} \hat{k} \]

### Final Answer
\[ \hat{n} = \frac{1}{\sqrt{26}} \hat{i} + \frac{4}{\sqrt{26}} \hat{j} + \frac{3}{\sqrt{26}} \hat{k} \]