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, we need to find the value of \(\vec{c} \cdot \vec{b}\) given the conditions \(\vec{b} \times \vec{c} = \vec{b} \times \vec{a}\) and \(\vec{c} \cdot \vec{a} = 1\).

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

### Step 2: Use the vector cross product condition
From the condition \(\vec{b} \times \vec{c} = \vec{b} \times \vec{a}\), we can use the vector triple product identity:
\[ \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \implies \vec{c} \cdot \vec{a} = 1 \]

### Step 3: Apply the vector triple product identity
Using the identity:
\[ \vec{b} \times (\vec{c} - \vec{a}) = \vec{0} \]
This implies that \(\vec{c} - \vec{a}\) is parallel to \(\vec{b}\). Therefore, we can express \(\vec{c}\) as:
\[ \vec{c} = \vec{a} + k\vec{b} \]
for some scalar \(k\).

### Step 4: Use the dot product condition
Now, we use the condition \(\vec{c} \cdot \vec{a} = 1\):
\[ (\vec{a} + k\vec{b}) \cdot \vec{a} = 1 \]
Expanding this gives:
\[ \vec{a} \cdot \vec{a} + k(\vec{b} \cdot \vec{a}) = 1 \]

### Step 5: Calculate \(\vec{a} \cdot \vec{a}\) and \(\vec{b} \cdot \vec{a}\)
First, calculate \(\vec{a} \cdot \vec{a}\):
\[ \vec{a} \cdot \vec{a} = 1^2 + (-2)^2 + 2^2 = 1 + 4 + 4 = 9 \]

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

### Step 6: Substitute back into the equation
Substituting these values into the equation:
\[ 9 + 6k = 1 \]
Solving for \(k\):
\[ 6k = 1 - 9 \implies 6k = -8 \implies k = -\frac{4}{3} \]

### Step 7: Find \(\vec{c}\)
Now substitute \(k\) back into the expression for \(\vec{c}\):
\[ \vec{c} = \vec{a} - \frac{4}{3}\vec{b} \]

### Step 8: Calculate \(\vec{c} \cdot \vec{b}\)
Now we can find \(\vec{c} \cdot \vec{b}\):
\[ \vec{c} \cdot \vec{b} = \left(\vec{a} - \frac{4}{3}\vec{b}\right) \cdot \vec{b} \]
This expands to:
\[ \vec{a} \cdot \vec{b} - \frac{4}{3}(\vec{b} \cdot \vec{b}) \]

### Step 9: Substitute known values
We already calculated:
\[ \vec{a} \cdot \vec{b} = 6 \]
And we need \(\vec{b} \cdot \vec{b}\):
\[ \vec{b} \cdot \vec{b} = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6 \]

### Step 10: Final calculation
Substituting these values:
\[ \vec{c} \cdot \vec{b} = 6 - \frac{4}{3}(6) = 6 - 8 = -2 \]

Thus, the final answer is:
\[ \vec{c} \cdot \vec{b} = -2 \]