If `bara=hati+hatj+hatk, bara.barb=2 and bara times barb=2hati+hatj-3hatk` then `barb` is equal to

Options

A. `5hati-4hatj+2hatk`
B. `4hati-5hatj+hatk`
C. `2hati-hatj+hatk`
D. `hati-2hatj-3hatk`

Answer

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

### Step 1: Define the vectors
We are given:
\[ \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \]
Let \( \mathbf{b} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \).

### Step 2: Use the dot product
We know that:
\[ \mathbf{a} \cdot \mathbf{b} = 2 \]
Calculating the dot product:
\[ (\hat{i} + \hat{j} + \hat{k}) \cdot (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) = \alpha + \beta + \gamma = 2 \quad \text{(1)} \]

### Step 3: Use the cross product
We are also given:
\[ \mathbf{a} \times \mathbf{b} = 2\hat{i} + \hat{j} - 3\hat{k} \]
Calculating the cross product:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \alpha & \beta & \gamma \end{vmatrix} \]
This determinant expands to:
\[ \hat{i} \begin{vmatrix} 1 & 1 \\ \beta & \gamma \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ \alpha & \gamma \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ \alpha & \beta \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i} (1\gamma - 1\beta) - \hat{j} (1\gamma - 1\alpha) + \hat{k} (1\beta - 1\alpha) \]
\[ = (\gamma - \beta) \hat{i} - (\gamma - \alpha) \hat{j} + (\beta - \alpha) \hat{k} \]
Setting this equal to \( 2\hat{i} + \hat{j} - 3\hat{k} \):
\[ \gamma - \beta = 2 \quad \text{(2)} \]
\[ -(\gamma - \alpha) = 1 \quad \Rightarrow \quad \gamma - \alpha = -1 \quad \text{(3)} \]
\[ \beta - \alpha = -3 \quad \text{(4)} \]

### Step 4: Solve the system of equations
From equations (2), (3), and (4), we can express \( \gamma \), \( \beta \), and \( \alpha \):
1. From (2): \( \gamma = \beta + 2 \)
2. From (3): \( \alpha = \gamma + 1 \)
3. From (4): \( \beta = \alpha - 3 \)

Substituting \( \gamma \) from (2) into (3):
\[ \alpha = (\beta + 2) + 1 \quad \Rightarrow \quad \alpha = \beta + 3 \quad \text{(5)} \]

Now substitute (5) into (4):
\[ \beta = (\beta + 3) - 3 \quad \Rightarrow \quad \beta = \beta \quad \text{(no new information)} \]

Now substitute (5) into (1):
\[ (\beta + 3) + \beta + (\beta + 2) = 2 \]
\[ 3\beta + 5 = 2 \quad \Rightarrow \quad 3\beta = -3 \quad \Rightarrow \quad \beta = -1 \]

Now substituting \( \beta = -1 \) back into (5):
\[ \alpha = -1 + 3 = 2 \]
And substituting \( \beta = -1 \) into (2):
\[ \gamma = -1 + 2 = 1 \]

### Step 5: Write the vector \( \mathbf{b} \)
Thus, we have:
\[ \alpha = 2, \quad \beta = -1, \quad \gamma = 1 \]
So,
\[ \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} \]

### Final Answer
\[ \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} \]