Let `veca = hati - 2 hatj + 2hatk` and `vec b = 2 hati - hatj + hatk` be two vectors. If `vec c` is a vector such that `vecb xx vec c = vec b xx vec a` and `vec c vec a = 1` , then `vec c vec b` is equal to :

Options

A. `1`
B. `-1`
C. `2`
D. `-2`

Explanation

To solve the problem step by step, we will start by defining the vectors and using the given conditions to find the required value of \( \vec{c} \cdot \vec{b} \).

### Step 1: Define the vectors
Let:
\[ \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \]
\[ \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \]

### Step 2: Express vector \( \vec{c} \)
Assume:
\[ \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 3: Use the first condition \( \vec{c} \cdot \vec{a} = 1 \)
Calculating the dot product:
\[ \vec{c} \cdot \vec{a} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + 2\hat{k}) = x - 2y + 2z = 1 \]
This gives us our first equation:
\[ x - 2y + 2z = 1 \quad \text{(1)} \]

### Step 4: Use the second condition \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \)
Calculate \( \vec{b} \times \vec{c} \):
\[ \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-1)z - 1y) - \hat{j}(2z - 1x) + \hat{k}(2y - (-1)x) \]
\[ = (-z - y)\hat{i} - (2z - x)\hat{j} + (2y + x)\hat{k} \]

Now calculate \( \vec{b} \times \vec{a} \):
\[ \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \]
\[ \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & -2 & 2 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i}((-1)2 - 1(-2)) - \hat{j}(2(2) - 1(1)) + \hat{k}(2(-2) - (-1)(1)) \]
\[ = \hat{i}(-2 + 2) - \hat{j}(4 - 1) + \hat{k}(-4 + 1) \]
\[ = 0\hat{i} - 3\hat{j} - 3\hat{k} \]

### Step 5: Set the two cross products equal
From \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \):
\[ (-z - y)\hat{i} - (2z - x)\hat{j} + (2y + x)\hat{k} = 0\hat{i} - 3\hat{j} - 3\hat{k} \]
This gives us the following equations:
1. \(-z - y = 0 \quad \text{(2)}\)
2. \(2z - x = 3 \quad \text{(3)}\)
3. \(2y + x = -3 \quad \text{(4)}\)

### Step 6: Solve the system of equations
From equation (2):
\[ y = -z \]

Substituting \(y = -z\) into equation (3):
\[ 2z - x = 3 \implies x = 2z - 3 \quad \text{(5)} \]

Substituting \(y = -z\) into equation (4):
\[ 2(-z) + x = -3 \implies x - 2z = -3 \implies x = 2z - 3 \quad \text{(6)} \]
Both equations (5) and (6) are consistent.

### Step 7: Substitute into equation (1)
Now substitute \(y = -z\) and \(x = 2z - 3\) into equation (1):
\[ (2z - 3) - 2(-z) + 2z = 1 \]
\[ 2z - 3 + 2z + 2z = 1 \]
\[ 6z - 3 = 1 \implies 6z = 4 \implies z = \frac{2}{3} \]

### Step 8: Find \(y\) and \(x\)
Using \(z = \frac{2}{3}\):
\[ y = -z = -\frac{2}{3} \]
\[ x = 2z - 3 = 2 \cdot \frac{2}{3} - 3 = \frac{4}{3} - 3 = \frac{4}{3} - \frac{9}{3} = -\frac{5}{3} \]

### Step 9: Write \( \vec{c} \)
Thus,
\[ \vec{c} = -\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Step 10: Calculate \( \vec{c} \cdot \vec{b} \)
Now we calculate \( \vec{c} \cdot \vec{b} \):
\[ \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \]
\[ \vec{c} \cdot \vec{b} = \left(-\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \cdot (2\hat{i} - \hat{j} + \hat{k}) \]
Calculating the dot product:
\[ = -\frac{5}{3} \cdot 2 + -\frac{2}{3} \cdot (-1) + \frac{2}{3} \cdot 1 \]
\[ = -\frac{10}{3} + \frac{2}{3} + \frac{2}{3} = -\frac{10}{3} + \frac{4}{3} = -\frac{6}{3} = -2 \]

### Final Answer
Thus, \( \vec{c} \cdot \vec{b} = -2 \).