A unit vector perpendicular to the vector
`4i-j+3k and -2i+j-2k` is

Options

A. `(1)/(3) (i-2j+2k)`
B. `(1)/(3) (-i+2j+2k)`
C. `(1)/(3) (2i+j+2k)`
D. `(1)/(3) (2i-2j+2k)`

Answer

To find a unit vector that is perpendicular to the vectors \( \mathbf{A} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k} \) and \( \mathbf{B} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \), we can use the cross product of the two vectors. The cross product will yield a vector that is perpendicular to both.

### Step-by-step Solution:

1. **Identify the vectors**:
\[ \mathbf{A} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k} \]
\[ \mathbf{B} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \]

2. **Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)**:
The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ = \mathbf{i} \begin{vmatrix} -1 & 3 \\ 1 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 4 & 3 \\ -2 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 4 & -1 \\ -2 & 1 \end{vmatrix} \]

- For \( \mathbf{i} \):
\[ = \mathbf{i}((-1)(-2) - (3)(1)) = \mathbf{i}(2 - 3) = -\mathbf{i} \]

- For \( \mathbf{j} \):
\[ = -\mathbf{j}((4)(-2) - (3)(-2)) = -\mathbf{j}(-8 + 6) = -\mathbf{j}(-2) = 2\mathbf{j} \]

- For \( \mathbf{k} \):
\[ = \mathbf{k}((4)(1) - (-1)(-2)) = \mathbf{k}(4 - 2) = 2\mathbf{k} \]

Combining these results, we have:
\[ \mathbf{A} \times \mathbf{B} = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \]

4. **Resulting vector**:
\[ \mathbf{C} = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \]

5. **Calculate the magnitude of \( \mathbf{C} \)**:
\[ |\mathbf{C}| = \sqrt{(-1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

6. **Find the unit vector**:
The unit vector \( \mathbf{U} \) in the direction of \( \mathbf{C} \) is given by:
\[ \mathbf{U} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} \]

Thus, the unit vector is:
\[ \mathbf{U} = -\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \]

### Final Answer:
The unit vector perpendicular to the given vectors is:
\[ \mathbf{U} = -\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \]