If `veca = (hati + hatj +hatk), veca. vecb= 1 and vecaxxvecb = hatj -hatk , " then " vecb` is (a)`hati - hatj + hatk` (b) `2hati - hatk` (c) `hati` (d) `2hati`

Options

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

Explanation

To solve the problem, we need to find the vector \(\vec{b}\) given the conditions involving vectors \(\vec{a}\) and \(\vec{b}\).

1. **Define the vectors**:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
Let \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\).

2. **Use the dot product condition**:
We know that:
\[ \vec{a} \cdot \vec{b} = 1 \]
This can be expanded as:
\[ (\hat{i} + \hat{j} + \hat{k}) \cdot (b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}) = b_1 + b_2 + b_3 = 1 \quad \text{(Equation 1)} \]

3. **Use the cross product condition**:
We are given:
\[ \vec{a} \times \vec{b} = \hat{j} - \hat{k} \]
We can calculate the cross product using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
This expands to:
\[ \hat{i} \begin{vmatrix} 1 & 1 \\ b_2 & b_3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ b_1 & b_3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ b_1 & b_2 \end{vmatrix} \]
Calculating the determinants:
- For \(\hat{i}\): \(1 \cdot b_3 - 1 \cdot b_2 = b_3 - b_2\)
- For \(\hat{j}\): \(1 \cdot b_3 - 1 \cdot b_1 = b_3 - b_1\)
- For \(\hat{k}\): \(1 \cdot b_2 - 1 \cdot b_1 = b_2 - b_1\)

Thus:
\[ \vec{a} \times \vec{b} = (b_3 - b_2) \hat{i} - (b_3 - b_1) \hat{j} + (b_2 - b_1) \hat{k} \]
Setting this equal to \(\hat{j} - \hat{k}\), we get:
\[ b_3 - b_2 = 0 \quad \text{(Equation 2)} \]
\[ -(b_3 - b_1) = 1 \quad \Rightarrow \quad b_3 - b_1 = -1 \quad \text{(Equation 3)} \]
\[ b_2 - b_1 = -1 \quad \text{(Equation 4)} \]

4. **Solve the equations**:
From Equation 2:
\[ b_2 = b_3 \]
Substitute \(b_2\) in Equation 4:
\[ b_3 - b_1 = -1 \quad \Rightarrow \quad b_1 = b_3 + 1 \quad \text{(from Equation 3)} \]
Substitute \(b_1\) and \(b_2\) in Equation 1:
\[ (b_3 + 1) + b_3 + b_3 = 1 \]
\[ 3b_3 + 1 = 1 \quad \Rightarrow \quad 3b_3 = 0 \quad \Rightarrow \quad b_3 = 0 \]
Thus, from \(b_3 = 0\):
\[ b_2 = 0 \quad \text{and} \quad b_1 = 1 \]

5. **Final vector**:
Therefore, the vector \(\vec{b}\) is:
\[ \vec{b} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i} \]

6. **Conclusion**:
The correct option is:
\[ \text{(c) } \hat{i} \]