Find a unit vector perpendicular to the plane of two vectors `a=hat(i)-hat(j)+2hat(k) and b=2hat(i)+3hat(j)-hat(k)`.

Answer

To find a unit vector perpendicular to the plane formed by the vectors \( \mathbf{a} = \hat{i} - \hat{j} + 2\hat{k} \) and \( \mathbf{b} = 2\hat{i} + 3\hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Calculate the Cross Product
The cross product \( \mathbf{C} = \mathbf{a} \times \mathbf{b} \) will give us a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).

Using the determinant method:
\[ \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix} \]

### Step 2: Expand the Determinant
Calculating the determinant:
\[ \mathbf{C} = \hat{i} \begin{vmatrix} -1 & 2 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -1 & 2 \\ 3 & -1 \end{vmatrix} = (-1)(-1) - (2)(3) = 1 - 6 = -5 \)
2. \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)
3. \( \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5 \)

Putting it all together:
\[ \mathbf{C} = -5\hat{i} + 5\hat{j} + 5\hat{k} \]

### Step 3: Write the Resulting Vector
Thus, we have:
\[ \mathbf{C} = -5\hat{i} + 5\hat{j} + 5\hat{k} \]

### Step 4: Find the Magnitude of \( \mathbf{C} \)
To find the unit vector, we first need the magnitude of \( \mathbf{C} \):
\[ |\mathbf{C}| = \sqrt{(-5)^2 + 5^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 5: Calculate the Unit Vector
Now, the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{C} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{-5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

### Final Answer
The unit vector perpendicular to the plane of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ \mathbf{u} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]