Let `vecp2hati+hatj-2hatk, vecq=hati+hatj.` If `vecr` is a vector such that `vecp. Vecr=|vecr|, |vecr-vecp|=2sqrt2` and the angle between `vecp xx vecq and vecr` is `(pi)/(6)`, then the value of `|(vecpxxvecq)xx vecr|` is equal to

Options

A. `(3)/(2)`
B. `(1)/(2)`
C. `(3sqrt3)/(2)`
D. 3

Answer

To solve the problem step by step, we will follow the given conditions and calculate the required value.

### Step 1: Define the vectors
Given:
\[ \vec{p} = 2\hat{i} + \hat{j} - 2\hat{k}, \quad \vec{q} = \hat{i} + \hat{j} \]

### Step 2: Calculate \(\vec{p} \times \vec{q}\)
To find \(\vec{p} \times \vec{q}\), we will use the determinant of a matrix formed by the unit vectors and the components of \(\vec{p}\) and \(\vec{q}\).

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

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

Calculating each of the 2x2 determinants:
\[ = \hat{i} (1 \cdot 0 - 1 \cdot (-2)) - \hat{j} (2 \cdot 0 - 1 \cdot (-2)) + \hat{k} (2 \cdot 1 - 1 \cdot 1) \]
\[ = \hat{i} (0 + 2) - \hat{j} (0 + 2) + \hat{k} (2 - 1) \]
\[ = 2\hat{i} - 2\hat{j} + 1\hat{k} \]
Thus,
\[ \vec{p} \times \vec{q} = 2\hat{i} - 2\hat{j} + \hat{k} \]

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

### Step 4: Use the given conditions
We know:
1. \(\vec{p} \cdot \vec{r} = |\vec{r}|\)
2. \(|\vec{r} - \vec{p}| = 2\sqrt{2}\)

From the second condition, squaring both sides gives:
\[ |\vec{r} - \vec{p}|^2 = (2\sqrt{2})^2 = 8 \]
This expands to:
\[ |\vec{r}|^2 - 2\vec{r} \cdot \vec{p} + |\vec{p}|^2 = 8 \]

### Step 5: Calculate \(|\vec{p}|\)
\[ |\vec{p}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 6: Substitute into the equation
Substituting \(\vec{p} \cdot \vec{r} = |\vec{r}|\) into the equation:
\[ |\vec{r}|^2 - 2|\vec{r}| + 9 - 8 = 0 \]
This simplifies to:
\[ |\vec{r}|^2 - 2|\vec{r}| + 1 = 0 \]
Factoring gives:
\[ (|\vec{r}| - 1)^2 = 0 \implies |\vec{r}| = 1 \]

### Step 7: Find \(|(\vec{p} \times \vec{q}) \times \vec{r}|\)
Using the formula:
\[ |(\vec{p} \times \vec{q}) \times \vec{r}| = |\vec{p} \times \vec{q}| \cdot |\vec{r}| \cdot \sin(\theta) \]
Where \(\theta = \frac{\pi}{6}\):
\[ |(\vec{p} \times \vec{q}) \times \vec{r}| = 3 \cdot 1 \cdot \sin\left(\frac{\pi}{6}\right) = 3 \cdot 1 \cdot \frac{1}{2} = \frac{3}{2} \]

### Final Answer
\[ |(\vec{p} \times \vec{q}) \times \vec{r}| = \frac{3}{2} \]

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