Let `veca=2hati+hatj-2hatk and vecb=hati+hatj`. Let `vecc` be a vector such that `|vecc-veca|=3,|(vecaxxvecb)xxvecc|` = 3 and the angle between `vecc and veca xx vecb ` is `30^@` Find `veca.vecc`

Answer

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

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

### Step 1: Find \(\vec{a} \times \vec{b}\)

Using the determinant method for the cross product:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(-2) = 2\)
2. \(\begin{vmatrix} 2 & -2 \\ 1 & 0 \end{vmatrix} = (2)(0) - (1)(-2) = 2\)
3. \(\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 1\)

Putting it all together:

\[ \vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + 1\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k} \]

### Step 2: Find 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 3: Use the conditions given

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 \(30^\circ\).

From the second condition, we can express it as:

\[ |\vec{a} \times \vec{b}| \cdot |\vec{c}| \cdot \sin(30^\circ) = 3 \]

Since \(|\vec{a} \times \vec{b}| = 3\) and \(\sin(30^\circ) = \frac{1}{2}\):

\[ 3 \cdot |\vec{c}| \cdot \frac{1}{2} = 3 \implies |\vec{c}| = 2 \]

### Step 4: Use the first condition

We have:

\[ |\vec{c} - \vec{a}| = 3 \]

Squaring both sides:

\[ |\vec{c} - \vec{a}|^2 = 9 \implies (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 9 \]

Expanding this:

\[ \vec{c} \cdot \vec{c} - 2\vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{a} = 9 \]

Substituting \(|\vec{c}|^2 = 4\) and \(|\vec{a}|^2 = 9\):

\[ 4 - 2\vec{a} \cdot \vec{c} + 9 = 9 \]

This simplifies to:

\[ -2\vec{a} \cdot \vec{c} + 13 = 9 \implies -2\vec{a} \cdot \vec{c} = -4 \implies \vec{a} \cdot \vec{c} = 2 \]

### Final Answer

\(\vec{a} \cdot \vec{c} = 2\)