Consider a parallelogram constructed as `5bara+2barb and bara-3barb` where `|a|=2sqrt2 and |b|=3` the angle between `bara and barb` is `pi//4` then the length of the longer diagonal is

Options

A. `sqrt473`
B. `sqrt593`
C. `sqrt474`
D. `sqrt594`

Answer

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

### Step 1: Identify the diagonals
The diagonals of the parallelogram can be expressed as:
- \(D_1 = (5\mathbf{a} + 2\mathbf{b}) + (\mathbf{a} - 3\mathbf{b})\)
- \(D_2 = (5\mathbf{a} + 2\mathbf{b}) - (\mathbf{a} - 3\mathbf{b})\)

### Step 2: Simplify the expressions for the diagonals
Calculating \(D_1\):
\[ D_1 = (5\mathbf{a} + 2\mathbf{b}) + (\mathbf{a} - 3\mathbf{b}) = 6\mathbf{a} - \mathbf{b} \]

Calculating \(D_2\):
\[ D_2 = (5\mathbf{a} + 2\mathbf{b}) - (\mathbf{a} - 3\mathbf{b}) = 4\mathbf{a} + 5\mathbf{b} \]

### Step 3: Calculate the lengths of the diagonals
To find the lengths of the diagonals, we need to compute \(|D_1|\) and \(|D_2|\).

#### Length of \(D_1\):
\[ |D_1| = |6\mathbf{a} - \mathbf{b}| \]
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:
- \(|\mathbf{a}| = 2\sqrt{2}\)
- \(|\mathbf{b}| = 3\)
- \(\theta = \frac{\pi}{4}\) (so \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\))

Calculating \(|D_1|\):
\[ |D_1| = \sqrt{(6 \cdot 2\sqrt{2})^2 + 3^2 - 2(6 \cdot 2\sqrt{2})(3)\left(\frac{1}{\sqrt{2}}\right)} \]
\[ = \sqrt{(12\sqrt{2})^2 + 3^2 - 2(12\sqrt{2})(3)\left(\frac{1}{\sqrt{2}}\right)} \]
\[ = \sqrt{288 + 9 - 36} = \sqrt{261} \]

#### Length of \(D_2\):
\[ |D_2| = |4\mathbf{a} + 5\mathbf{b}| \]
Using the same formula:
\[ |D_2| = \sqrt{|4\mathbf{a}|^2 + |5\mathbf{b}|^2 + 2|4\mathbf{a}||5\mathbf{b}|\cos(\theta)} \]
Calculating \(|D_2|\):
\[ |D_2| = \sqrt{(4 \cdot 2\sqrt{2})^2 + (5 \cdot 3)^2 + 2(4 \cdot 2\sqrt{2})(5)(\frac{1}{\sqrt{2}})} \]
\[ = \sqrt{(8\sqrt{2})^2 + 15^2 + 40\sqrt{2}} \]
\[ = \sqrt{128 + 225 + 40\sqrt{2}} = \sqrt{353 + 40\sqrt{2}} \]

### Step 4: Compare the lengths
To find the longer diagonal, we compare \(|D_1|\) and \(|D_2|\).

### Final Result
After calculating both lengths, we find that:
- \(|D_1| = \sqrt{261}\)
- \(|D_2| = \sqrt{353 + 40\sqrt{2}}\)

Since \(|D_2|\) is greater than \(|D_1|\), the length of the longer diagonal is:
\[ \text{Length of the longer diagonal} = |D_2| = \sqrt{353 + 40\sqrt{2}} \]