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

Explanation

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

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

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

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

### Step 2: Expand the Determinant

Using the determinant formula, we can expand it as follows:

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

Calculating each of these 2x2 determinants:

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

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

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

Putting it all together, we have:

\[ \vec{A} \times \vec{B} = 1\hat{i} + 4\hat{j} + 3\hat{k} = \hat{i} + 4\hat{j} + 3\hat{k} \]

### Step 3: Calculate the Magnitude of \(\vec{A} \times \vec{B}\)

Now, we need to find the magnitude of the cross product:

\[ |\vec{A} \times \vec{B}| = \sqrt{(1)^2 + (4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]

### Step 4: Find the Unit Vector

The unit vector \(\hat{n}\) in the direction of \(\vec{A} \times \vec{B}\) is given by:

\[ \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{\hat{i} + 4\hat{j} + 3\hat{k}}{\sqrt{26}} \]

### Final Answer

Thus, the unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\) is:

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