The cross product of vectors Given : `vecA=hati+hatjandvecB=hati+hatk`.

Options

A. `hati+hatj+hatk`
B. `hati-hatj+hatk`
C. `hati+hatj-hatk`
D. `hati-hatj-hatk`

Answer

To find the cross product of the vectors \(\vec{A}\) and \(\vec{B}\), we can use the determinant method. Let's go through the solution step by step.

### Step 1: Write the vectors
Given:
\[ \vec{A} = \hat{i} + \hat{j} \]
\[ \vec{B} = \hat{i} + \hat{k} \]

### Step 2: Set up the determinant
We will set up a 3x3 determinant where the first row contains the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\). The second row will contain the coefficients of \(\vec{A}\) and the third row will contain the coefficients of \(\vec{B}\).

The coefficients for \(\vec{A}\) are:
- For \(\hat{i}\): 1
- For \(\hat{j}\): 1
- For \(\hat{k}\): 0

The coefficients for \(\vec{B}\) are:
- For \(\hat{i}\): 1
- For \(\hat{j}\): 0
- For \(\hat{k}\): 1

Thus, the determinant can be set up as follows:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we will use the formula for a 3x3 determinant:
\[ \text{det} = \hat{i} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1\)
2. \(\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 0\)
3. \(\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(1) = -1\)

Putting it all together:
\[ \text{det} = \hat{i}(1) - \hat{j}(0) + \hat{k}(-1) \]
This simplifies to:
\[ \vec{A} \times \vec{B} = \hat{i} - 0\hat{j} - \hat{k} = \hat{i} - \hat{k} \]

### Step 4: Final result
Thus, the cross product \(\vec{A} \times \vec{B}\) is:
\[ \vec{A} \times \vec{B} = \hat{i} - \hat{j} - \hat{k} \]