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. `sqrt(113)`
C. `sqrt(593)`
D. `sqrt(369)`

Answer

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 will follow these steps:

### Step 1: Identify the diagonals of the parallelogram
The diagonals of a parallelogram formed by vectors \(\vec{u}\) and \(\vec{v}\) can be expressed as:
- Diagonal 1: \(\vec{u} + \vec{v}\)
- Diagonal 2: \(\vec{u} - \vec{v}\)

In our case, let:
\[ \vec{u} = 5\vec{a} + 2\vec{b} \]
\[ \vec{v} = \vec{a} - 3\vec{b} \]

### Step 2: Calculate the diagonals
Now we will calculate the two diagonals:
1. Diagonal 1:
\[ \vec{d_1} = \vec{u} + \vec{v} = (5\vec{a} + 2\vec{b}) + (\vec{a} - 3\vec{b}) = (5\vec{a} + \vec{a}) + (2\vec{b} - 3\vec{b}) = 6\vec{a} - \vec{b} \]

2. Diagonal 2:
\[ \vec{d_2} = \vec{u} - \vec{v} = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) = (5\vec{a} - \vec{a}) + (2\vec{b} + 3\vec{b}) = 4\vec{a} + 5\vec{b} \]

### Step 3: Calculate the lengths of the diagonals
To find the lengths of the diagonals, we will calculate the magnitudes of \(\vec{d_1}\) and \(\vec{d_2}\).

1. Length of Diagonal 1:
\[ |\vec{d_1}| = |6\vec{a} - \vec{b}| \]
Using the formula for the magnitude of a vector:
\[ |\vec{d_1}|^2 = |6\vec{a}|^2 + |\vec{b}|^2 - 2|6\vec{a}||\vec{b}|\cos(\theta) \]
Where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

2. Length of Diagonal 2:
\[ |\vec{d_2}| = |4\vec{a} + 5\vec{b}| \]
Using the same formula:
\[ |\vec{d_2}|^2 = |4\vec{a}|^2 + |5\vec{b}|^2 + 2|4\vec{a}||5\vec{b}|\cos(\theta) \]

### Step 4: Substitute the values
Given:
- \(|\vec{a}| = 2\sqrt{2}\)
- \(|\vec{b}| = 3\)
- \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta) = \frac{\pi}{4}\)

Calculating \(|\vec{d_1}|^2\):
\[ |\vec{d_1}|^2 = 36|\vec{a}|^2 + |\vec{b}|^2 - 12|\vec{a}||\vec{b}|\cos(\theta) \]
Substituting the values:
\[ = 36(2\sqrt{2})^2 + 3^2 - 12(2\sqrt{2})(3)\cos\left(\frac{\pi}{4}\right) \]
\[ = 36 \cdot 8 + 9 - 12 \cdot 6 \cdot \frac{1}{\sqrt{2}} \]
\[ = 288 + 9 - 72\sqrt{2} \]

Calculating \(|\vec{d_2}|^2\):
\[ |\vec{d_2}|^2 = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 2(4|\vec{a}|)(5|\vec{b}|)\cos(\theta) \]
Substituting the values:
\[ = 16(2\sqrt{2})^2 + 25(3)^2 + 2(4)(5)(2\sqrt{2})(3)\cos\left(\frac{\pi}{4}\right) \]
\[ = 16 \cdot 8 + 225 + 120\sqrt{2} \]
\[ = 128 + 225 + 120\sqrt{2} \]

### Step 5: Find the longer diagonal
After calculating both lengths, we find the maximum of \(|\vec{d_1}|\) and \(|\vec{d_2}|\).

### Final Step: Conclusion
The length of the longer diagonal is \(15\) units.