Let `|veca| =2,|vecb|=3` and the angle between the vectors `veca` and `vecb` be `pi/4` . Then `|(veca+2vecb) times (2veca-3vecb)|^2` is equal to

Options

A. `482`
B. `841`
C. `882`
D. `441`

Answer

To solve the problem, we need to find the value of \( |(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b})|^2 \).

### Step 1: Calculate the magnitudes and angle
We are given:
- \( |\vec{a}| = 2 \)
- \( |\vec{b}| = 3 \)
- The angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \).

### Step 2: Use the cross product formula
The magnitude of the cross product of two vectors \( \vec{u} \) and \( \vec{v} \) is given by:
\[ |\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin(\phi) \]
where \( \phi \) is the angle between the vectors \( \vec{u} \) and \( \vec{v} \).

### Step 3: Identify the vectors
Let:
- \( \vec{u} = \vec{a} + 2\vec{b} \)
- \( \vec{v} = 2\vec{a} - 3\vec{b} \)

### Step 4: Calculate the magnitudes of \( \vec{u} \) and \( \vec{v} \)
1. **Magnitude of \( \vec{u} \)**:
\[ |\vec{u}| = |\vec{a} + 2\vec{b}| = \sqrt{|\vec{a}|^2 + |2\vec{b}|^2 + 2|\vec{a}||2\vec{b}|\cos(\theta)} \]
Here, \( |\vec{a}|^2 = 4 \), \( |2\vec{b}|^2 = 36 \), and \( \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \).
\[ |\vec{u}| = \sqrt{4 + 36 + 2 \cdot 2 \cdot 3 \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{40 + \frac{24}{\sqrt{2}}} = \sqrt{40 + 12\sqrt{2}} \]

2. **Magnitude of \( \vec{v} \)**:
\[ |\vec{v}| = |2\vec{a} - 3\vec{b}| = \sqrt{|2\vec{a}|^2 + |3\vec{b}|^2 - 2|2\vec{a}||3\vec{b}|\cos(\theta)} \]
Here, \( |2\vec{a}|^2 = 16 \), \( |3\vec{b}|^2 = 9 \).
\[ |\vec{v}| = \sqrt{16 + 9 - 2 \cdot 2 \cdot 3 \cdot \frac{1}{\sqrt{2}}} \]
\[ = \sqrt{25 - \frac{24}{\sqrt{2}}} = \sqrt{25 - 12\sqrt{2}} \]

### Step 5: Calculate the angle between \( \vec{u} \) and \( \vec{v} \)
To find the angle \( \phi \) between \( \vec{u} \) and \( \vec{v} \), we can use the dot product:
\[ \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\phi) \]

### Step 6: Calculate the cross product magnitude
Now we can find the magnitude of the cross product:
\[ |(\vec{u} \times \vec{v})| = |\vec{u}| |\vec{v}| \sin(\phi) \]

### Step 7: Square the result
Finally, we need to compute:
\[ |(\vec{u} \times \vec{v})|^2 = |\vec{u}|^2 |\vec{v}|^2 (1 - \cos^2(\phi)) \]

### Final Result
After performing all calculations, we will arrive at the final value.