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

Options

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

Answer

To solve the problem step by step, we will analyze the given vectors and their relationships.

### Given:
1. \( \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \)
2. \( \mathbf{a} \cdot \mathbf{b} = 2 \)
3. \( \mathbf{a} \times \mathbf{b} = 2\hat{i} + \hat{j} - 3\hat{k} \)

### Step 1: Express the vector \( \mathbf{b} \)
Let’s assume \( \mathbf{b} = x\hat{i} + y\hat{j} + z\hat{k} \). We need to find \( x, y, z \).

### Step 2: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)
Using the dot product formula:
\[ \mathbf{a} \cdot \mathbf{b} = (1)(x) + (1)(y) + (1)(z) = x + y + z \]
Setting this equal to 2:
\[ x + y + z = 2 \quad \text{(1)} \]

### Step 3: Calculate the cross product \( \mathbf{a} \times \mathbf{b} \)
Using the determinant for the cross product:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating this determinant:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i}(1z - 1y) - \hat{j}(1z - 1x) + \hat{k}(1y - 1x) \]
\[ = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \]
Setting this equal to \( 2\hat{i} + \hat{j} - 3\hat{k} \):
\[ (z - y) = 2 \quad \text{(2)} \]
\[ -(z - x) = 1 \quad \Rightarrow \quad z - x = -1 \quad \text{(3)} \]
\[ (y - x) = -3 \quad \text{(4)} \]

### Step 4: Solve the system of equations
From equations (2), (3), and (4), we can express \( z \) in terms of \( y \) and \( x \):
1. From (2): \( z = y + 2 \)
2. From (3): \( z = x - 1 \)
3. From (4): \( y = x - 3 \)

Substituting \( y \) from (4) into (2):
\[ z = (x - 3) + 2 = x - 1 \]
Now we have:
\[ z = x - 1 \quad \text{and} \quad z = x - 1 \quad \text{(consistent)} \]

Now substituting \( z \) into (1):
\[ x + (x - 3) + (x - 1) = 2 \]
\[ 3x - 4 = 2 \]
\[ 3x = 6 \quad \Rightarrow \quad x = 2 \]
Now substituting \( x \) back to find \( y \) and \( z \):
\[ y = 2 - 3 = -1 \]
\[ z = 2 - 1 = 1 \]

### Final Result:
Thus, \( \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} \).