If `a=hati+hatj+hatk,b=hati+hatj-hatk`, then vectors perpendicular to a and b is/are

Options

A. `lamda(hati+hatj)`
B. `lamda(hati+hatj+hatk)`
C. `lamda(hati+hatk)`
D. none of these

Answer

To find a vector that is perpendicular to both vectors \( \mathbf{a} \) and \( \mathbf{b} \), we can use the cross product. Let's go through the solution step by step.

### Step 1: Define the vectors
Given:
\[ \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{b} = \hat{i} + \hat{j} - \hat{k} \]

### Step 2: Set up the cross product
The cross product \( \mathbf{a} \times \mathbf{b} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of vectors \( \mathbf{a} \) and \( \mathbf{b} \).

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can expand it as follows:

\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2 \quad (\text{same as above}) \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 1 - 1 = 0 \]

Putting it all together:
\[ \mathbf{a} \times \mathbf{b} = -2\hat{i} + 2\hat{j} + 0\hat{k} \]
Thus,
\[ \mathbf{a} \times \mathbf{b} = -2\hat{i} + 2\hat{j} \]

### Step 4: Conclusion
The vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ \mathbf{n} = -2\hat{i} + 2\hat{j} \]
We can also express this as:
\[ \mathbf{n} = \lambda (-2\hat{i} + 2\hat{j}) \quad \text{for any scalar } \lambda \]