Question

If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{a} \cdot \vec{b} = 1 \) and \( \vec{a} \times \vec{b} = \hat{j} - \hat{k} \), then \( \vec{b} \) is:

• A. \( \hat{i} - \hat{j} + \hat{k} \)
• B. \( 2\hat{j} - \hat{k} \)
• C. \( 2\hat{i} \)
• D. \( \hat{i} \)

Hint:

For solving vector equations: 
- Use the dot product condition to establish a scalar equation. 
- Use the cross product determinant method to form component-wise equations. 
- Solve the system of equations systematically to determine the unknowns.

Answer 1

The Correct Option is D

Step 1: Use the dot product condition. Given: \[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \] The dot product condition: \[ \vec{a} \cdot \vec{b} = (1,1,1) \cdot (x,y,z) = 1 \] \[ x + y + z = 1 \quad \cdots (1) \] 
Step 2: Use the cross product condition. The cross product: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \] Expanding along the first row: \[ \hat{i} \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \] \[ = \hat{i} (z - y) - \hat{j} (z - x) + \hat{k} (y - x) \] Given: \[ \hat{j} - \hat{k} = (0,1,-1) \] Comparing components: \[ z - y = 0, \quad -(z - x) = 1, \quad y - x = -1 \] 
Step 3: Solve the system of equations. 
1. \( z - y = 0 \Rightarrow z = y \). 
2. \( -(y - x) = 1 \Rightarrow y - x = -1 \Rightarrow x = y + 1 \). 
3. From equation (1): \( x + y + z = 1 \). Substituting \( z = y \) and \( x = y + 1 \): \[ (y + 1) + y + y = 1 \] \[ y + 1 + y + y = 1 \] \[ 3y + 1 = 1 \] \[ 3y = 0 \Rightarrow y = 0. \] \[ x = 1, \quad z = 0. \] 
Thus, \( \vec{b} = \hat{i} \).