The number of vectors of unit length perpendicular to the vectors ` hata = hati +hatj and vecb = hatj + hatk` is

Options

A. 1
B. 2
C. 4
D. infinite

Explanation

To solve the problem, we need to find the number of unit vectors that are perpendicular to the given vectors \( \vec{A} = \hat{i} + \hat{j} \) and \( \vec{B} = \hat{j} + \hat{k} \).

### Step-by-step Solution:

1. **Identify the Vectors**:
We have:
\[ \vec{A} = \hat{i} + \hat{j} \]
\[ \vec{B} = \hat{j} + \hat{k} \]

2. **Find the Cross Product**:
The vector that is perpendicular to both \( \vec{A} \) and \( \vec{B} \) can be found using the cross product:
\[ \vec{C} = \vec{A} \times \vec{B} \]

To compute \( \vec{C} \), we can use the determinant of a matrix formed by the unit vectors:
\[ \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]

Expanding this determinant, we have:
\[ \vec{C} = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating the minors:
\[ \vec{C} = \hat{i}(1 \cdot 1 - 0 \cdot 1) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 1 - 0 \cdot 1) \]
\[ = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) \]
\[ = \hat{i} - \hat{j} + \hat{k} \]

3. **Magnitude of the Cross Product**:
Now, we need to find the magnitude of \( \vec{C} \):
\[ |\vec{C}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

4. **Unit Vector Calculation**:
The unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \hat{C} = \frac{\vec{C}}{|\vec{C}|} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]

5. **Number of Unit Vectors**:
Since the unit vector can point in two opposite directions (positive and negative), the number of unit vectors perpendicular to both \( \vec{A} \) and \( \vec{B} \) is:
\[ 2 \]

### Final Answer:
The number of unit vectors perpendicular to the vectors \( \vec{A} \) and \( \vec{B} \) is **2**.