The ratio of lengths of diagonals of the parallelogram constructed on the vectors `vec(a)=3vec(p)-vec(q),vec(b)=vec(p)+3vec(q)` is (given that `|vec(p)|=|vec(q)|=2` and angle between `vec(p)` and `vec(q)` is `pi//3`).

Options

A. `sqrt(7):sqrt(3)`
B. `sqrt(3):sqrt(7)`
C. `sqrt(5):sqrt(7)`
D. `sqrt(5):sqrt(3)`

Answer

To solve the problem, we need to find the ratio of the lengths of the diagonals of a parallelogram formed by the vectors \(\vec{a} = 3\vec{p} - \vec{q}\) and \(\vec{b} = \vec{p} + 3\vec{q}\). We are given that \(|\vec{p}| = |\vec{q}| = 2\) and the angle between \(\vec{p}\) and \(\vec{q}\) is \(\frac{\pi}{3}\).

### Step 1: Calculate the first diagonal \(\vec{d_1}\)

The first diagonal \(\vec{d_1}\) of the parallelogram is given by the sum of the vectors \(\vec{a}\) and \(\vec{b}\):
\[ \vec{d_1} = \vec{a} + \vec{b} = (3\vec{p} - \vec{q}) + (\vec{p} + 3\vec{q}) = 4\vec{p} + 2\vec{q} \]

### Step 2: Find the magnitude of \(\vec{d_1}\)

To find the length of diagonal \(\vec{d_1}\), we calculate its magnitude:
\[ |\vec{d_1}| = |4\vec{p} + 2\vec{q}| \]
Using the formula for the magnitude of a vector:
\[ |\vec{d_1}| = \sqrt{(4|\vec{p}|)^2 + (2|\vec{q}|)^2 + 2(4|\vec{p}|)(2|\vec{q}|)\cos(\theta)} \]
Substituting \(|\vec{p}| = 2\), \(|\vec{q}| = 2\), and \(\theta = \frac{\pi}{3}\) (where \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)):
\[ |\vec{d_1}| = \sqrt{(4 \cdot 2)^2 + (2 \cdot 2)^2 + 2(4 \cdot 2)(2 \cdot 2)\left(\frac{1}{2}\right)} \]
Calculating each term:
\[ = \sqrt{(8)^2 + (4)^2 + 2 \cdot 8 \cdot 4 \cdot \frac{1}{2}} \]
\[ = \sqrt{64 + 16 + 32} = \sqrt{112} \]

### Step 3: Calculate the second diagonal \(\vec{d_2}\)

The second diagonal \(\vec{d_2}\) of the parallelogram is given by the difference of the vectors \(\vec{a}\) and \(\vec{b}\):
\[ \vec{d_2} = \vec{a} - \vec{b} = (3\vec{p} - \vec{q}) - (\vec{p} + 3\vec{q}) = 2\vec{p} - 4\vec{q} \]

### Step 4: Find the magnitude of \(\vec{d_2}\)

To find the length of diagonal \(\vec{d_2}\), we calculate its magnitude:
\[ |\vec{d_2}| = |2\vec{p} - 4\vec{q}| \]
Using the formula for the magnitude of a vector:
\[ |\vec{d_2}| = \sqrt{(2|\vec{p}|)^2 + (-4|\vec{q}|)^2 + 2(2|\vec{p})(-4|\vec{q}|)\cos(\theta)} \]
Substituting the values:
\[ = \sqrt{(2 \cdot 2)^2 + (-4 \cdot 2)^2 + 2(2 \cdot 2)(-4 \cdot 2)\left(\frac{1}{2}\right)} \]
Calculating each term:
\[ = \sqrt{(4)^2 + (-8)^2 + 2 \cdot 4 \cdot -8 \cdot \frac{1}{2}} \]
\[ = \sqrt{16 + 64 - 32} = \sqrt{48} \]

### Step 5: Find the ratio of the lengths of the diagonals

Now we can find the ratio of the lengths of the diagonals:
\[ \text{Ratio} = \frac{|\vec{d_1}|}{|\vec{d_2}|} = \frac{\sqrt{112}}{\sqrt{48}} = \frac{\sqrt{112/48}} = \frac{\sqrt{7/3}} = \frac{\sqrt{7}}{\sqrt{3}} \]

### Final Answer

Thus, the ratio of the lengths of the diagonals of the parallelogram is:
\[ \frac{\sqrt{7}}{\sqrt{3}} \]