The length of longer diagonal of the parallelogram constructed on `5a + 2b` and `a-3b` . If it is given that `|a| = 2sqrt(2),|b|` = 3 and angle between a and b is `(pi)/(4)` is

Options

A. 15
B. `sqrt(113)`
C. `sqrt(593)`
D. `sqrt(369)`

Explanation

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

### Step 1: Determine the diagonals of the parallelogram
The diagonals \(d_1\) and \(d_2\) of the parallelogram can be expressed as:
\[ d_1 = (5\mathbf{a} + 2\mathbf{b}) + (\mathbf{a} - 3\mathbf{b}) = 6\mathbf{a} - \mathbf{b} \]
\[ d_2 = (5\mathbf{a} + 2\mathbf{b}) - (\mathbf{a} - 3\mathbf{b}) = 4\mathbf{a} + 5\mathbf{b} \]

### Step 2: Calculate the magnitudes of the diagonals
To find the lengths of the diagonals, we need to calculate \(|d_1|\) and \(|d_2|\).

#### For \(d_1\):
\[ |d_1| = |6\mathbf{a} - \mathbf{b}| \]
Using the formula for the magnitude of a vector:
\[ |d_1| = \sqrt{(6\mathbf{a})^2 + (-\mathbf{b})^2 + 2(6\mathbf{a})(-\mathbf{b})\cos(\theta)} \]
where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

Given:
- \(|\mathbf{a}| = 2\sqrt{2}\)
- \(|\mathbf{b}| = 3\)
- \(\theta = \frac{\pi}{4}\)

Substituting these values:
\[ |d_1| = \sqrt{(6 \cdot 2\sqrt{2})^2 + (-3)^2 + 2(6 \cdot 2\sqrt{2})(-3)\cos\left(\frac{\pi}{4}\right)} \]

Calculating each term:
- \((6 \cdot 2\sqrt{2})^2 = 72\)
- \((-3)^2 = 9\)
- \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)

Putting it all together:
\[ |d_1| = \sqrt{72 + 9 - 36\sqrt{2}} \]
\[ |d_1| = \sqrt{81 - 36\sqrt{2}} \]

#### For \(d_2\):
\[ |d_2| = |4\mathbf{a} + 5\mathbf{b}| \]
Using the same formula:
\[ |d_2| = \sqrt{(4\mathbf{a})^2 + (5\mathbf{b})^2 + 2(4\mathbf{a})(5\mathbf{b})\cos(\theta)} \]
Substituting the values:
\[ |d_2| = \sqrt{(4 \cdot 2\sqrt{2})^2 + (5 \cdot 3)^2 + 2(4 \cdot 2\sqrt{2})(5)(3)\cos\left(\frac{\pi}{4}\right)} \]

Calculating each term:
- \((4 \cdot 2\sqrt{2})^2 = 64\)
- \((5 \cdot 3)^2 = 225\)

Putting it all together:
\[ |d_2| = \sqrt{64 + 225 + 60\sqrt{2}} \]
\[ |d_2| = \sqrt{289 + 60\sqrt{2}} \]

### Step 3: Compare the lengths of the diagonals
Now we need to determine which diagonal is longer. We will compare \(|d_1|\) and \(|d_2|\).

From our calculations:
- \(|d_1| = \sqrt{81 - 36\sqrt{2}}\)
- \(|d_2| = \sqrt{289 + 60\sqrt{2}}\)

Since \(289 + 60\sqrt{2}\) is greater than \(81 - 36\sqrt{2}\), we conclude that:
\[ |d_2| > |d_1| \]

### Final Result
The length of the longer diagonal of the parallelogram is:
\[ |d_2| = \sqrt{289 + 60\sqrt{2}} \]