https://file.aiquickdraw.com/mathbot/file/1729234452289i8rnzp4w.jpg

The given problem is written in Russian and translates as follows:

**Problem:**
Calculate \( 4 \left| \vec{a} \times \vec{b} \right| - 2 (\vec{a} \cdot \vec{b}) + \left| \vec{a} + \vec{b} \right|^2 \), where \( |\vec{a}| = |\vec{b}| = 1 \) and the angle between vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{6} \).

### Step-by-step solution:

1. **Cross product \( \left| \vec{a} \times \vec{b} \right| \):**
   The magnitude of the cross product is given by:
   \[
   \left| \vec{a} \times \vec{b} \right| = |\vec{a}| |\vec{b}| \sin \theta
   \]
   Here, \( |\vec{a}| = |\vec{b}| = 1 \) and \( \theta = \frac{\pi}{6} \), so:
   \[
   \left| \vec{a} \times \vec{b} \right| = 1 \cdot 1 \cdot \sin \frac{\pi}{6} = \frac{1}{2}
   \]

2. **Dot product \( \vec{a} \cdot \vec{b} \):**
   The dot product is given by:
   \[
   \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta
   \]
   With \( |\vec{a}| = |\vec{b}| = 1 \) and \( \theta = \frac{\pi}{6} \):
   \[
   \vec{a} \cdot \vec{b} = 1 \cdot 1 \cdot \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}
   \]

3. **Magnitude of \( \left| \vec{a} + \vec{b} \right|^2 \):**
   Using the formula for the square of the magnitude of the sum of two vectors:
   \[
   \left| \vec{a} + \vec{b} \right|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 (\vec{a} \cdot \vec{b})
   \]
   Since \( |\vec{a}| = |\vec{b}| = 1 \):
   \[
   \left| \vec{a} + \vec{b} \right|^2 = 1^2 + 1^2 + 2 \cdot \frac{\sqrt{3}}{2} = 2 + \sqrt{3}
   \]

4. **Substitute into the original expression:**
   Now we can substitute the computed values into the original expression:
   \[
   4 \left| \vec{a} \times \vec{b} \right| - 2 (\vec{a} \cdot \vec{b}) + \left| \vec{a} + \vec{b} \right|^2
   \]
   This becomes:
   \[
   4 \cdot \frac{1}{2} - 2 \cdot \frac{\sqrt{3}}{2} + (2 + \sqrt{3})
   \]
   Simplifying:
   \[
   2 - \sqrt{3} + 2 + \sqrt{3} = 4
   \]

### Final Answer:
The result is \( 4 \).

Would you like more details or have any questions?

### Related Questions:
1. What is the difference between a dot product and a cross product geometrically?
2. How does the magnitude of a vector affect the results of the dot and cross products?
3. How can we generalize this problem for vectors of different magnitudes?
4. Can we solve similar problems with vectors in three-dimensional space?
5. How would the result change if the angle between the vectors were different?

### Tip:
Always remember that the magnitude of a cross product gives the area of the parallelogram formed by two vectors, while the dot product gives the scalar projection of one vector onto another.

Вычислить 4|a × b| - 2(a · b) + |a + b|^2, где |a| = |b| = 1, угол(a, b) = π/6

Step-by-step solution to calculate 4|a × b| - 2(a · b) + |a + b|^2, where |a| = |b| = 1 and the angle between vectors is π/6. Covers vector calculus concepts, including dot and cross products, and trigonometric identities. Suitable for undergraduate students studying calculus and vector analysis.

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