Let `a =2i+j-2k and b=i+j`. If c is a vector such that `a.c = |c|,|c-a| =2sqrt(2)` and the angle between `(a xxb)` and c. is `30^(@)`, then `|( axx b) xx c| =`

Options

A. `i-j+k`
B. `2j-k`
C. `3/2`
D. 2i

Answer

To solve the problem step by step, we will follow the given conditions and compute the required values systematically.

### Given:
- \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \)
- \( \mathbf{b} = \mathbf{i} + \mathbf{j} \)
- Conditions:
1. \( \mathbf{a} \cdot \mathbf{c} = |\mathbf{c}| \)
2. \( |\mathbf{c} - \mathbf{a}| = 2\sqrt{2} \)
3. The angle between \( \mathbf{a} \times \mathbf{b} \) and \( \mathbf{c} \) is \( 30^\circ \)

### Step 1: Calculate \( \mathbf{a} \times \mathbf{b} \)

To find \( \mathbf{a} \times \mathbf{b} \), we use the determinant method:

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

Calculating the determinant:

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

Calculating the minors:

- For \( \mathbf{i} \): \( 1 \cdot 0 - (-2) \cdot 1 = 2 \)
- For \( \mathbf{j} \): \( 2 \cdot 0 - (-2) \cdot 1 = 2 \)
- For \( \mathbf{k} \): \( 2 \cdot 1 - 1 \cdot 1 = 1 \)

Thus,

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

### Step 2: Calculate the magnitude of \( \mathbf{a} \times \mathbf{b} \)

\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 3: Find \( |\mathbf{c}| \)

From the condition \( \mathbf{a} \cdot \mathbf{c} = |\mathbf{c}| \), we can denote \( |\mathbf{c}| = k \). Thus, we have:

\[ \mathbf{a} \cdot \mathbf{c} = k \]

### Step 4: Use the second condition

From the condition \( |\mathbf{c} - \mathbf{a}| = 2\sqrt{2} \):

\[ |\mathbf{c} - \mathbf{a}|^2 = (2\sqrt{2})^2 = 8 \]

Expanding \( |\mathbf{c} - \mathbf{a}|^2 \):

\[ |\mathbf{c}|^2 + |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{c} = 8 \]

Substituting \( |\mathbf{c}| = k \) and \( |\mathbf{a}|^2 = 9 \):

\[ k^2 + 9 - 2k = 8 \]

Rearranging gives:

\[ k^2 - 2k + 1 = 0 \implies (k - 1)^2 = 0 \implies k = 1 \]

Thus, \( |\mathbf{c}| = 1 \).

### Step 5: Use the angle condition

The angle between \( \mathbf{a} \times \mathbf{b} \) and \( \mathbf{c} \) is \( 30^\circ \). Therefore,

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

Substituting the known values:

\[ 3 \cdot 1 \cdot \frac{1}{2} = |\mathbf{a} \times \mathbf{b} \times \mathbf{c}| \]

Thus,

\[ |\mathbf{a} \times \mathbf{b} \times \mathbf{c}| = \frac{3}{2} \]

### Final Answer:

\[ |\mathbf{(a \times b) \times c}| = \frac{3}{2} \]