The length of the longer diagonal of the parallelogram constructed on ` 5veca + 2vecb and veca - 3vecb, ` if it is given that ` |veca|=2sqrt2, |vecb|=3 and veca. Vecb= pi/4` is

Options

A. 15
B. `sqrt3`
C. `sqrt593`
D. `sqrt369`

Explanation

To find the length of the longer diagonal of the parallelogram constructed on the vectors \( 5\vec{a} + 2\vec{b} \) and \( \vec{a} - 3\vec{b} \), we can follow these steps:

### Step 1: Identify the diagonals of the parallelogram
The diagonals of the parallelogram are given by:
- \( \alpha = (5\vec{a} + 2\vec{b}) + (\vec{a} - 3\vec{b}) \)
- \( \beta = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) \)

### Step 2: Simplify the expressions for the diagonals
Calculating \( \alpha \):
\[ \alpha = 5\vec{a} + 2\vec{b} + \vec{a} - 3\vec{b} = (5\vec{a} + \vec{a}) + (2\vec{b} - 3\vec{b}) = 6\vec{a} - \vec{b} \]

Calculating \( \beta \):
\[ \beta = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) = (5\vec{a} - \vec{a}) + (2\vec{b} + 3\vec{b}) = 4\vec{a} + 5\vec{b} \]

### Step 3: Find the magnitudes of the diagonals
To find the lengths of the diagonals, we need to calculate \( |\alpha| \) and \( |\beta| \).

#### For \( |\alpha| \):
\[ |\alpha| = |6\vec{a} - \vec{b}| = \sqrt{(6\vec{a})^2 + (-\vec{b})^2 - 2(6\vec{a}) \cdot (-\vec{b})} \]
Using the formula \( |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \cdot \vec{v} \):
\[ |\alpha|^2 = 36|\vec{a}|^2 + |\vec{b}|^2 - 12\vec{a} \cdot \vec{b} \]

#### Substitute the values:
Given \( |\vec{a}| = 2\sqrt{2} \), \( |\vec{b}| = 3 \), and \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta) = 2\sqrt{2} \cdot 3 \cdot \frac{\pi}{4} \):
\[ |\vec{a}|^2 = (2\sqrt{2})^2 = 8, \quad |\vec{b}|^2 = 3^2 = 9 \]
\[ \vec{a} \cdot \vec{b} = 2\sqrt{2} \cdot 3 \cdot \frac{\pi}{4} = \frac{6\sqrt{2}\pi}{4} = \frac{3\sqrt{2}\pi}{2} \]

Now substituting into the equation for \( |\alpha|^2 \):
\[ |\alpha|^2 = 36 \cdot 8 + 9 - 12 \cdot \frac{3\sqrt{2}\pi}{2} \]
\[ |\alpha|^2 = 288 + 9 - 18\sqrt{2}\pi \]
\[ |\alpha|^2 = 297 - 18\sqrt{2}\pi \]

#### For \( |\beta| \):
\[ |\beta| = |4\vec{a} + 5\vec{b}| = \sqrt{(4\vec{a})^2 + (5\vec{b})^2 + 2(4\vec{a}) \cdot (5\vec{b})} \]
\[ |\beta|^2 = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40\vec{a} \cdot \vec{b} \]
Substituting the values:
\[ |\beta|^2 = 16 \cdot 8 + 25 \cdot 9 + 40 \cdot \frac{3\sqrt{2}\pi}{2} \]
\[ |\beta|^2 = 128 + 225 + 60\sqrt{2}\pi \]
\[ |\beta|^2 = 353 + 60\sqrt{2}\pi \]

### Step 4: Determine the longer diagonal
To find the longer diagonal, we compare \( |\alpha| \) and \( |\beta| \):
- \( |\alpha| = \sqrt{297 - 18\sqrt{2}\pi} \)
- \( |\beta| = \sqrt{353 + 60\sqrt{2}\pi} \)

Since \( 353 + 60\sqrt{2}\pi > 297 - 18\sqrt{2}\pi \), we conclude that \( |\beta| \) is the longer diagonal.

### Final Answer
Thus, the length of the longer diagonal is:
\[ |\beta| = \sqrt{353 + 60\sqrt{2}\pi} \]