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We are given two vectors that define the diagonals of a parallelogram:

\[
\vec{a} = \vec{p} - 3 \vec{q} \quad \text{and} \quad \vec{b} = 2 \vec{p} - 4 \vec{q}
\]

Additionally, we know the magnitudes of the vectors \(\vec{p}\) and \(\vec{q}\):

\[
|\vec{p}| = |\vec{q}| = 1
\]

The angle between the vectors \(\vec{p}\) and \(\vec{q}\) is given as:

\[
\cos(\theta) = \cos(60^\circ) = \frac{1}{2}
\]

### Step 1: Calculate the dot product between the diagonals

We want to find the cosine of the angle between the diagonals \(\vec{a}\) and \(\vec{b}\). To do this, we need the dot product \(\vec{a} \cdot \vec{b}\):

\[
\vec{a} \cdot \vec{b} = (\vec{p} - 3 \vec{q}) \cdot (2 \vec{p} - 4 \vec{q})
\]

Expanding this:

\[
\vec{a} \cdot \vec{b} = 2 (\vec{p} \cdot \vec{p}) - 4 (\vec{p} \cdot \vec{q}) - 6 (\vec{q} \cdot \vec{p}) + 12 (\vec{q} \cdot \vec{q})
\]

Since \(\vec{p} \cdot \vec{p} = |\vec{p}|^2 = 1\) and \(\vec{q} \cdot \vec{q} = |\vec{q}|^2 = 1\), and \(\vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{p} = \frac{1}{2}\), we substitute these values:

\[
\vec{a} \cdot \vec{b} = 2(1) - 4\left(\frac{1}{2}\right) - 6\left(\frac{1}{2}\right) + 12(1)
\]

\[
\vec{a} \cdot \vec{b} = 2 - 2 - 3 + 12 = 9
\]

### Step 2: Calculate the magnitudes of the diagonals

Now we compute the magnitudes of \(\vec{a}\) and \(\vec{b}\). The magnitude of a vector \(\vec{v}\) is given by \(\sqrt{\vec{v} \cdot \vec{v}}\).

First, for \(\vec{a}\):

\[
\vec{a} \cdot \vec{a} = (\vec{p} - 3 \vec{q}) \cdot (\vec{p} - 3 \vec{q})
\]
\[
\vec{a} \cdot \vec{a} = (\vec{p} \cdot \vec{p}) - 6 (\vec{p} \cdot \vec{q}) + 9 (\vec{q} \cdot \vec{q})
\]
\[
\vec{a} \cdot \vec{a} = 1 - 6 \left(\frac{1}{2}\right) + 9 = 1 - 3 + 9 = 7
\]
Thus:
\[
|\vec{a}| = \sqrt{7}
\]

Next, for \(\vec{b}\):

\[
\vec{b} \cdot \vec{b} = (2 \vec{p} - 4 \vec{q}) \cdot (2 \vec{p} - 4 \vec{q})
\]
\[
\vec{b} \cdot \vec{b} = 4 (\vec{p} \cdot \vec{p}) - 16 (\vec{p} \cdot \vec{q}) + 16 (\vec{q} \cdot \vec{q})
\]
\[
\vec{b} \cdot \vec{b} = 4 - 16 \left(\frac{1}{2}\right) + 16 = 4 - 8 + 16 = 12
\]
Thus:
\[
|\vec{b}| = \sqrt{12} = 2\sqrt{3}
\]

### Step 3: Find the cosine of the angle between the diagonals

Now that we have the dot product and magnitudes, we can calculate the cosine of the angle \(\varphi\) between the diagonals using the formula:

\[
\cos(\varphi) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}
\]

Substitute the values we found:

\[
\cos(\varphi) = \frac{9}{\sqrt{7} \cdot 2\sqrt{3}} = \frac{9}{2\sqrt{21}}
\]

Now simplify:

\[
\cos(\varphi) = \frac{9}{2\sqrt{21}} \cdot \frac{\sqrt{21}}{\sqrt{21}} = \frac{9\sqrt{21}}{42} = \frac{\sqrt{21}}{7}
\]

Thus, the cosine of the angle between the diagonals is \(\frac{\sqrt{21}}{7}\).

Would you like further details or have any questions?

### Related Questions:
1. How do dot products help in finding angles between vectors?
2. What is the geometrical significance of the angle between the diagonals in a parallelogram?
3. How can you derive the magnitude of a vector in a general form?
4. What role does the angle between two vectors play in the projection of one onto the other?
5. Can you explain how vector addition or subtraction affects the shape of the parallelogram?

**Tip:** The dot product is a useful tool for determining the angle between two vectors, as it directly relates to the cosine of the angle between them.

Знайти кут між діагоналями паралелограма, побудованого на векторах \( \vec{a} = \vec{p} - 3 \vec{q} \) та \( \vec{b} = 2\vec{p} - 4 \vec{q} \), якщо \( |\vec{p}| = |\vec{q}| = 1 \), \( (\vec{p}, \vec{q}) = 60^\circ \).

Learn how to find the angle between the diagonals of a parallelogram constructed from vectors \( \vec{p} \) and \( \vec{q} \). This problem involves vector algebra, dot product, and cosine of the angle, using given magnitudes and angles between vectors. Detailed step-by-step solution provided.

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