A unit vector perpendicular to both the vectors `2hati-2hatj+hatk` and `3hati+4hatj-5hatk`, is

Options

A. `(1)/(sqrt(410))(6hati+13hatj+14hatk)`
B. `(1)/(sqrt(410))(6hati+13hatj+14hatk)`
C. `(1)/(sqrt(401))(6hati+13hatj+14hatk)`
D. `(1)/(sqrt(405))(6hati+13hatj+14hatk)`

Explanation

To find a unit vector that is perpendicular to both vectors \( \mathbf{A} = 2\hat{i} - 2\hat{j} + \hat{k} \) and \( \mathbf{B} = 3\hat{i} + 4\hat{j} - 5\hat{k} \), we can follow these steps:

### Step 1: Calculate the Cross Product
The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) will give us a vector that is perpendicular to both.

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 3 & 4 & -5 \end{vmatrix} \]

### Step 2: Evaluate the Determinant
Calculating the determinant, we have:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -2 & 1 \\ 4 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 3 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ (-2)(-5) - (1)(4) = 10 - 4 = 6 \]

2. For \( \hat{j} \):
\[ (2)(-5) - (1)(3) = -10 - 3 = -13 \]

3. For \( \hat{k} \):
\[ (2)(4) - (-2)(3) = 8 + 6 = 14 \]

Putting it all together, we get:
\[ \mathbf{A} \times \mathbf{B} = 6\hat{i} + 13\hat{j} + 14\hat{k} \]

### Step 3: Find the Magnitude of the Cross Product
Next, we need to find the magnitude of the resulting vector:

\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{6^2 + 13^2 + 14^2} \]
Calculating each term:
- \( 6^2 = 36 \)
- \( 13^2 = 169 \)
- \( 14^2 = 196 \)

Adding these together:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{36 + 169 + 196} = \sqrt{401} \]

### Step 4: Calculate the Unit Vector
To find the unit vector \( \mathbf{n} \) that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \), we divide the cross product by its magnitude:

\[ \mathbf{n} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|} = \frac{6\hat{i} + 13\hat{j} + 14\hat{k}}{\sqrt{401}} \]

### Final Answer
Thus, the unit vector perpendicular to both given vectors is:

\[ \mathbf{n} = \frac{6\hat{i} + 13\hat{j} + 14\hat{k}}{\sqrt{401}} \]

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