A parallelogram is constructed on the vectors `veca=3vecalpha-vecbeta, vecb=vecalpha+3vecbeta. If |vecalpha|=|vecbeta|=2` and angle between `vecalpha and vecbeta is pi/3` then the length of a diagonal of the parallelogram is

Options

A. `4sqrt5`
B. `4sqrt3`
C. `4sqrt7`
D. none of these

Explanation

To find the length of a diagonal of the parallelogram constructed on the vectors \(\vec{a} = 3\vec{\alpha} - \vec{\beta}\) and \(\vec{b} = \vec{\alpha} + 3\vec{\beta}\), we will follow these steps:

### Step 1: Find the vectors \(\vec{a}\) and \(\vec{b}\)

Given:
\[ \vec{a} = 3\vec{\alpha} - \vec{\beta} \]
\[ \vec{b} = \vec{\alpha} + 3\vec{\beta} \]

### Step 2: Calculate the magnitudes of \(\vec{\alpha}\) and \(\vec{\beta}\)

We know:
\[ |\vec{\alpha}| = |\vec{\beta}| = 2 \]
Thus:
\[ |\vec{\alpha}|^2 = 4 \quad \text{and} \quad |\vec{\beta}|^2 = 4 \]

### Step 3: Find the angle between \(\vec{\alpha}\) and \(\vec{\beta}\)

The angle between the vectors is given as \(\theta = \frac{\pi}{3}\). We can use this to find the cosine of the angle:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 4: Calculate \(|\vec{a}|\) and \(|\vec{b}|\)

To find \(|\vec{a}|\):
\[ |\vec{a}|^2 = |3\vec{\alpha} - \vec{\beta}|^2 = |3\vec{\alpha}|^2 + |\vec{\beta}|^2 - 2|3\vec{\alpha}||\vec{\beta}|\cos\theta \]
\[ = 9|\vec{\alpha}|^2 + |\vec{\beta}|^2 - 2 \cdot 3|\vec{\alpha}||\vec{\beta}|\cos\left(\frac{\pi}{3}\right) \]
\[ = 9 \cdot 4 + 4 - 2 \cdot 3 \cdot 2 \cdot \frac{1}{2} \]
\[ = 36 + 4 - 6 = 34 \]
Thus, \(|\vec{a}| = \sqrt{34}\).

To find \(|\vec{b}|\):
\[ |\vec{b}|^2 = |\vec{\alpha} + 3\vec{\beta}|^2 = |\vec{\alpha}|^2 + |3\vec{\beta}|^2 + 2|\vec{\alpha}||3\vec{\beta}|\cos\theta \]
\[ = 4 + 9 \cdot 4 + 2 \cdot 2 \cdot 3 \cdot \frac{1}{2} \]
\[ = 4 + 36 + 6 = 46 \]
Thus, \(|\vec{b}| = \sqrt{46}\).

### Step 5: Calculate the diagonal vectors

The diagonals of the parallelogram are given by:
\[ \vec{d_1} = \vec{a} + \vec{b} \quad \text{and} \quad \vec{d_2} = \vec{a} - \vec{b} \]

### Step 6: Calculate \(|\vec{d_1}|\)

\[ |\vec{d_1}|^2 = |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta \]
\[ = 34 + 46 + 2 \cdot \sqrt{34} \cdot \sqrt{46} \cdot \frac{1}{2} \]
\[ = 80 + \sqrt{34 \cdot 46} \]

Calculating \(\sqrt{34 \cdot 46}\):
\[ 34 \cdot 46 = 1564 \]
Thus:
\[ |\vec{d_1}|^2 = 80 + \frac{\sqrt{1564}}{2} \]

### Step 7: Calculate \(|\vec{d_2}|\)

\[ |\vec{d_2}|^2 = |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta \]
\[ = 34 + 46 - 2 \cdot \sqrt{34} \cdot \sqrt{46} \cdot \frac{1}{2} \]
\[ = 80 - \sqrt{1564} \]

### Step 8: Final Calculation of Diagonal Lengths

Now, we can find the lengths of the diagonals:
\[ |\vec{d_1}| = \sqrt{80 + \sqrt{1564}} \quad \text{and} \quad |\vec{d_2}| = \sqrt{80 - \sqrt{1564}} \]

### Conclusion

The length of the diagonal of the parallelogram is given by either \(|\vec{d_1}|\) or \(|\vec{d_2}|\).

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