The unit vector `bot` to each of the vector `2i-j+k` and `3i+4j-k` is

Options

A. `-3i+4j+11k`
B. `(-3i+5j-11k)//sqrt(155)`
C. `(-3i+5j+11k)//sqrt((155))`
D. none of these

Answer

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

### Step 1: Define the vectors
Let:
\[ \mathbf{A} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \]
\[ \mathbf{B} = 3\mathbf{i} + 4\mathbf{j} - \mathbf{k} \]

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
To find a vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \), we compute the cross product \( \mathbf{A} \times \mathbf{B} \).

The formula for the cross product in determinant form is:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} -1 & 1 \\ 4 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -1 \\ 3 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \mathbf{i} \):
\[ (-1)(-1) - (1)(4) = 1 - 4 = -3 \]
2. For \( \mathbf{j} \):
\[ (2)(-1) - (1)(3) = -2 - 3 = -5 \quad \text{(remember to change the sign)} \Rightarrow +5 \]
3. For \( \mathbf{k} \):
\[ (2)(4) - (-1)(3) = 8 + 3 = 11 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = -3\mathbf{i} + 5\mathbf{j} + 11\mathbf{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude \( |\mathbf{C}| \) of the vector \( \mathbf{C} = -3\mathbf{i} + 5\mathbf{j} + 11\mathbf{k} \) is calculated as follows:
\[ |\mathbf{C}| = \sqrt{(-3)^2 + 5^2 + 11^2} = \sqrt{9 + 25 + 121} = \sqrt{155} \]

### Step 5: Calculate the unit vector
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{C} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{-3\mathbf{i} + 5\mathbf{j} + 11\mathbf{k}}{\sqrt{155}} \]

### Final Answer
Thus, the unit vector perpendicular to both vectors is:
\[ \mathbf{u} = \frac{-3}{\sqrt{155}} \mathbf{i} + \frac{5}{\sqrt{155}} \mathbf{j} + \frac{11}{\sqrt{155}} \mathbf{k} \]