Let `x=2i+j-2k" and "y=i+j`. If z is a vector such that `x.z=abs(z), abs(z-x)=2sqrt(2)` and the angle between `x times y` and z is `30^(@)`, then the magnitude of the vector `(x timesy) timesz` is:

Options

A. `sqrt(3)/2`
B. `3/2`
C. `1/2`
D. `(3sqrt(3))/2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
Given:
- \( \mathbf{x} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \)
- \( \mathbf{y} = \mathbf{i} + \mathbf{j} \)

### Step 2: Calculate the cross product \( \mathbf{x} \times \mathbf{y} \)
We can calculate the cross product using the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{x} \) and \( \mathbf{y} \):

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

Calculating the determinant:

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

Calculating each of these 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. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 1 \)

Thus,

\[ \mathbf{x} \times \mathbf{y} = 2\mathbf{i} - 2\mathbf{j} + 1\mathbf{k} \]

### Step 3: Calculate the magnitude of \( \mathbf{x} \times \mathbf{y} \)
The magnitude is given by:

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

### Step 4: Use the given conditions to find \( |\mathbf{z}| \)
We have two conditions:
1. \( \mathbf{x} \cdot \mathbf{z} = |\mathbf{z}| \)
2. \( |\mathbf{z} - \mathbf{x}| = 2\sqrt{2} \)

From the second condition, we can express it as:

\[ |\mathbf{z} - \mathbf{x}|^2 = (2\sqrt{2})^2 = 8 \]

Expanding this using the dot product:

\[ |\mathbf{z}|^2 - 2\mathbf{z} \cdot \mathbf{x} + |\mathbf{x}|^2 = 8 \]

Let \( |\mathbf{z}| = r \). Then, substituting \( \mathbf{x} \cdot \mathbf{z} = r \):

\[ r^2 - 2r + |\mathbf{x}|^2 = 8 \]

Now, calculate \( |\mathbf{x}| \):

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

Substituting back:

\[ r^2 - 2r + 9 = 8 \implies r^2 - 2r + 1 = 0 \implies (r - 1)^2 = 0 \implies r = 1 \]

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

### Step 5: Calculate the magnitude of \( (\mathbf{x} \times \mathbf{y}) \times \mathbf{z} \)
Using the formula for the magnitude of the cross product:

\[ |(\mathbf{x} \times \mathbf{y}) \times \mathbf{z}| = |\mathbf{x} \times \mathbf{y}| \cdot |\mathbf{z}| \cdot \sin(\theta) \]

Where \( \theta = 30^\circ \) and \( \sin(30^\circ) = \frac{1}{2} \):

\[ |(\mathbf{x} \times \mathbf{y}) \times \mathbf{z}| = 3 \cdot 1 \cdot \frac{1}{2} = \frac{3}{2} \]

### Final Answer
The magnitude of the vector \( (\mathbf{x} \times \mathbf{y}) \times \mathbf{z} \) is \( \frac{3}{2} \).

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