Step 1: Express \(\vec{b}\) in Terms of Components Let \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\). Step 2: Use the Dot Product Condition The dot product \(\vec{a} \cdot \vec{b}\) is: \[ \vec{a} \cdot \vec{b} = (1)(b_1) + (1)(b_2) + (1)(b_3) = b_1 + b_2 + b_3 = 1 \quad \text{(1)} \] Step 3: Use the Cross Product Condition The cross product \(\vec{a} \times \vec{b}\) is: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2) \hat{i} - (b_3 - b_1) \hat{j} + (b_2 - b_1) \hat{k} \] Given \(\vec{a} \times \vec{b} = \hat{j} - \hat{k}\), we equate the components: \[ b_3 - b_2 = 0 \quad \text{(2)} \] \[ -(b_3 - b_1) = 1 \quad \text{(3)} \] \[ b_2 - b_1 = -1 \quad \text{(4)} \] Step 4: Solve the System of Equations From equation (2): \[ b_3 = b_2 \quad \text{(5)} \] From equation (3): \[ b_3 - b_1 = -1 \quad \text{(6)} \] From equation (4): \[ b_2 - b_1 = -1 \quad \text{(7)} \] From equation (5) and (7): \[ b_3 - b_1 = -1 \quad \text{(6)} \] \[ b_2 - b_1 = -1 \quad \text{(7)} \] Since \(b_3 = b_2\), equations (6) and (7) are consistent. From equation (1): \[ b_1 + b_2 + b_3 = 1 \] Substitute \(b_3 = b_2\): \[ b_1 + 2b_2 = 1 \quad \text{(8)} \] From equation (7): \[ b_2 = b_1 - 1 \quad \text{(9)} \] Substitute equation (9) into equation (8): \[ b_1 + 2(b_1 - 1) = 1 \] \[ b_1 + 2b_1 - 2 = 1 \] \[ 3b_1 = 3 \Rightarrow b_1 = 1 \] From equation (9): \[ b_2 = 1 - 1 = 0 \] From equation (5): \[ b_3 = 0 \] Step 5: Write \(\vec{b}\) Thus, the vector \(\vec{b}\) is: \[ \vec{b} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i} \] Step 6: Conclusion The vector \(\vec{b}\) is: \[ \boxed{\hat{i}} \] Correct Answer: \(\boxed{C}\)