Let `veca =2hati +hatj -2hatk and vecb = hati +hatj . " Let " vecc` be vector such that ` |vecc -veca|=3, |(veca xx vecb) xx vecc|=3` and the angle between `vecc and veca xx vecb " be " 30^(@)` Then , ` veca . Vecc` is equal to

Options

A. `25/8`
B. 2
C. 5
D. `1/8`

Explanation

To solve the problem step by step, we will follow the given conditions and use vector operations accordingly.

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

### Step 2: Calculate \(\vec{a} \times \vec{b}\)
To find \(\vec{a} \times \vec{b}\), we will use the determinant method:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i} (0 + 2) - \hat{j} (0 + 2) + \hat{k} (2 - 1) \]
\[ = 2\hat{i} - 2\hat{j} + 1\hat{k} \]
Thus,
\[ \vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \]

### Step 3: Calculate the magnitude of \(\vec{a} \times \vec{b}\)
\[ |\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 4: Use the given conditions
We know:
1. \(|\vec{c} - \vec{a}| = 3\)
2. \(|(\vec{a} \times \vec{b}) \times \vec{c}| = 3\)
3. The angle between \(\vec{c}\) and \(\vec{a} \times \vec{b}\) is \(30^\circ\).

Using the formula for the magnitude of a cross product:
\[ |(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin(30^\circ) \]
Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ 3 = 3 |\vec{c}| \cdot \frac{1}{2} \]
\[ 3 = \frac{3}{2} |\vec{c}| \]
Thus,
\[ |\vec{c}| = 2 \]

### Step 5: Use the condition \(|\vec{c} - \vec{a}| = 3\)
Using the equation:
\[ |\vec{c} - \vec{a}|^2 = 3^2 \]
\[ |\vec{c}|^2 + |\vec{a}|^2 - 2 \vec{a} \cdot \vec{c} = 9 \]
Substituting the magnitudes:
\[ 2^2 + |\vec{a}|^2 - 2 \vec{a} \cdot \vec{c} = 9 \]
Calculating \(|\vec{a}|\):
\[ |\vec{a}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
Substituting:
\[ 4 + 9 - 2 \vec{a} \cdot \vec{c} = 9 \]
\[ 13 - 2 \vec{a} \cdot \vec{c} = 9 \]
\[ -2 \vec{a} \cdot \vec{c} = 9 - 13 \]
\[ -2 \vec{a} \cdot \vec{c} = -4 \]
Thus,
\[ \vec{a} \cdot \vec{c} = 2 \]

### Final Result
\[ \vec{a} \cdot \vec{c} = 2 \]