If `veca= hati +hatj, vecb= hatj + hatk, vecc = hatk + hati ` then in the reciprocal system of vectors `veca, vecb, vecc` reciprocal `veca` of vector `veca` is

Options

A. `(hati + hatj + hatk)/2`
B. `(hati - hatj + hatk)/2`
C. `(-hati - hatj + hatk)/2`
D. `(hati + hatj - hatk)/2`

Explanation

To find the reciprocal of the vector \(\vec{a}\) in the reciprocal system of vectors \(\vec{a}, \vec{b}, \vec{c}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j}, \quad \vec{b} = \hat{j} + \hat{k}, \quad \vec{c} = \hat{k} + \hat{i} \]

### Step 2: Find the cross product \(\vec{b} \times \vec{c}\)
To find \(\vec{b} \times \vec{c}\), we will set up the determinant:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \]

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

Putting this together:
\[ \vec{b} \times \vec{c} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(-1) = \hat{i} + \hat{j} - \hat{k} \]

### Step 3: Find the determinant of \(\vec{a}, \vec{b}, \vec{c}\)
Now we calculate the determinant of the matrix formed by \(\vec{a}, \vec{b}, \vec{c}\):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]

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

Putting this together:
\[ = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \hat{i} - \hat{j} + \hat{k} \]

### Step 4: Find the reciprocal of \(\vec{a}\)
Using the formula for the reciprocal of vector \(\vec{a}\):
\[ \text{Reciprocal of } \vec{a} = \frac{\vec{b} \times \vec{c}}{\text{det}(\vec{a}, \vec{b}, \vec{c})} \]
Substituting the values we found:
\[ = \frac{\hat{i} + \hat{j} - \hat{k}}{2} \]

### Final Answer
The reciprocal of vector \(\vec{a}\) is:
\[ \frac{1}{2}(\hat{i} + \hat{j} - \hat{k}) \]