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 defined by the vectors \( \vec{u} = 5\vec{a} + 2\vec{b} \) and \( \vec{v} = \vec{a} - 3\vec{b} \), we will follow these steps:

### Step 1: Find the diagonals of the parallelogram
The diagonals of a parallelogram can be expressed as:
1. \( \vec{d_1} = \vec{u} + \vec{v} \)
2. \( \vec{d_2} = \vec{u} - \vec{v} \)

Calculating \( \vec{d_1} \):
\[ \vec{d_1} = (5\vec{a} + 2\vec{b}) + (\vec{a} - 3\vec{b}) = (5\vec{a} + \vec{a}) + (2\vec{b} - 3\vec{b}) = 6\vec{a} - \vec{b} \]

Calculating \( \vec{d_2} \):
\[ \vec{d_2} = (5\vec{a} + 2\vec{b}) - (\vec{a} - 3\vec{b}) = (5\vec{a} - \vec{a}) + (2\vec{b} + 3\vec{b}) = 4\vec{a} + 5\vec{b} \]

### Step 2: Calculate the lengths of the diagonals
We will calculate the lengths of both diagonals \( |\vec{d_1}| \) and \( |\vec{d_2}| \).

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

Given:
- \( |\vec{a}| = 2\sqrt{2} \)
- \( |\vec{b}| = 3 \)
- The angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \) (or \( 45^\circ \)), thus \( \cos\theta = \frac{1}{\sqrt{2}} \).

Calculating:
\[ |\vec{d_1}| = \sqrt{(6 \cdot 2\sqrt{2})^2 + (3)^2 - 2 \cdot (6 \cdot 2\sqrt{2}) \cdot (3) \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{(12\sqrt{2})^2 + 3^2 - 2 \cdot 12\sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{288 + 9 - 36} \]
\[ = \sqrt{261} \]

#### Length of \( \vec{d_2} \):
\[ |\vec{d_2}| = |4\vec{a} + 5\vec{b}| \]
Using the same formula:
\[ |\vec{d_2}| = \sqrt{|4\vec{a}|^2 + |5\vec{b}|^2 + 2|4\vec{a}||5\vec{b}|\cos\theta} \]
Calculating:
\[ |\vec{d_2}| = \sqrt{(4 \cdot 2\sqrt{2})^2 + (5 \cdot 3)^2 + 2 \cdot (4 \cdot 2\sqrt{2}) \cdot (5) \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{(8\sqrt{2})^2 + 15^2 + 2 \cdot 8\sqrt{2} \cdot 15 \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{128 + 225 + 240} \]
\[ = \sqrt{593} \]

### Step 3: Determine the length of the longer diagonal
Comparing the lengths:
- \( |\vec{d_1}| = \sqrt{261} \)
- \( |\vec{d_2}| = \sqrt{593} \)

Since \( \sqrt{593} > \sqrt{261} \), the longer diagonal is:
\[ \text{Length of the longer diagonal} = \sqrt{593} \]

### Final Answer
The length of the longer diagonal is \( \sqrt{593} \).

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