Let `vec(a) = hat(i) - 2hat(j) + 2hat(k)` and `vec(b) = 2hat(i) - hat(j) + hat(k)` be two vectors. If `vec(c)` is a vector such that `vec(b) 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 follow the given information and derive the required value of \(\vec{c} \cdot \vec{b}\).

### Given:
- \(\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}\)
- \(\vec{b} = 2\hat{i} - \hat{j} + \hat{k}\)

### Step 1: Use the given condition \(\vec{b} \times \vec{c} = \vec{b} \times \vec{a}\)

From the equation \(\vec{b} \times \vec{c} = \vec{b} \times \vec{a}\), we can rearrange it as:
\[ \vec{b} \times \vec{c} - \vec{b} \times \vec{a} = 0 \]
This implies that:
\[ \vec{b} \times (\vec{c} - \vec{a}) = 0 \]
Since the cross product of two vectors is zero, it indicates that \(\vec{b}\) is parallel to \(\vec{c} - \vec{a}\). Therefore, we can express this relationship as:
\[ \vec{b} = \lambda (\vec{c} - \vec{a}) \]
for some scalar \(\lambda\).

### Step 2: Take the dot product with \(\vec{a}\)

Now, we take the dot product of both sides with \(\vec{a}\):
\[ \vec{a} \cdot \vec{b} = \lambda (\vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{a}) \]
Given that \(\vec{a} \cdot \vec{c} = 1\), we can substitute this into the equation:
\[ \vec{a} \cdot \vec{b} = \lambda (1 - \vec{a} \cdot \vec{a}) \]

### Step 3: Calculate \(\vec{a} \cdot \vec{b}\) and \(\vec{a} \cdot \vec{a}\)

First, we calculate \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = (1)(2) + (-2)(-1) + (2)(1) = 2 + 2 + 2 = 6 \]

Next, we calculate \(\vec{a} \cdot \vec{a}\):
\[ \vec{a} \cdot \vec{a} = (1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9 \]

### Step 4: Substitute back to find \(\lambda\)

Now substituting these values back into the equation:
\[ 6 = \lambda (1 - 9) \]
This simplifies to:
\[ 6 = \lambda (-8) \]
Thus, we find:
\[ \lambda = -\frac{3}{4} \]

### Step 5: Express \(\vec{c}\) in terms of \(\vec{b}\) and \(\vec{a}\)

Now we can substitute \(\lambda\) back into the expression for \(\vec{b}\):
\[ \vec{b} = -\frac{3}{4} (\vec{c} - \vec{a}) \]
Rearranging gives:
\[ \vec{c} = -\frac{4}{3} \vec{b} + \vec{a} \]

### Step 6: Find \(\vec{c} \cdot \vec{b}\)

Now we need to find \(\vec{c} \cdot \vec{b}\):
\[ \vec{c} \cdot \vec{b} = \left(-\frac{4}{3} \vec{b} + \vec{a}\right) \cdot \vec{b} \]
Distributing the dot product:
\[ \vec{c} \cdot \vec{b} = -\frac{4}{3} (\vec{b} \cdot \vec{b}) + \vec{a} \cdot \vec{b} \]

### Step 7: Calculate \(\vec{b} \cdot \vec{b}\)

Now we calculate \(\vec{b} \cdot \vec{b}\):
\[ \vec{b} \cdot \vec{b} = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6 \]

### Step 8: Substitute back to find \(\vec{c} \cdot \vec{b}\)

Now substituting back:
\[ \vec{c} \cdot \vec{b} = -\frac{4}{3} (6) + 6 \]
This simplifies to:
\[ \vec{c} \cdot \vec{b} = -8 + 6 = -2 \]

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