Collection of handwritten physics and math equations and diagrams, featuring E=mc², trigonometric identities, and geometric figures.

The image you provided contains a variety of mathematical and physical concepts, including vectors, equations, and formulas. However, the task or problem to solve is not explicitly stated in the image. To proceed, I will analyze the content and identify potential problems that could be solved based on the given information.

Key Elements in the Image:

1. Vectors:
- Vectors \( \mathbf{a} \) and \( \mathbf{b} \) are defined as:
\[
\mathbf{a} = \lambda_1 \mathbf{e}_1 + \lambda_2 \mathbf{e}_2
\]
\[
\mathbf{b} = (x_2, y_2)
\]
- The vector \( \mathbf{a} + \mathbf{b} \) is given by:
\[
\mathbf{a} + \mathbf{b} = (x_1 + x_2, y_1 + y_2)
\]
- The vector \( \mathbf{a} - \mathbf{b} \) is given by:
\[
\mathbf{a} - \mathbf{b} = (x_1 - x_2, y_1 - y_2)
\]

2. Equation of a Line:
- The equation \( x_1 y_2 - x_2 y_1 = 0 \) represents a condition for collinearity of points or vectors.

3. Trigonometric Identities:
- The identities:
\[
\sin(2\pi - x) = -\cos(x)
\]
\[
\cos(2\pi - x) = \cos(x)
\]

4. Physics Formula:
- The formula \( E = mc^2 \), which relates energy (\( E \)), mass (\( m \)), and the speed of light (\( c \)).

5. Geometry:
- A triangle with side lengths \( AB \), \( BC \), and \( AC \). The formula:
\[
\frac{\sqrt{AB \cdot BC}}{2}
\]
suggests an area calculation or a geometric property.

6. Miscellaneous:
- The point \( P \left( \frac{3}{4}, \frac{1}{3} \right) \).
- The expression \( \int r \, dt \), which could represent the integral of a radius over time.
- The temperature formula \( T = \frac{m v}{9 B} \).

Potential Problem Identification:

Given the variety of elements, it is unclear what specific problem needs to be solved. However, based on the content, here are some possible interpretations:

#### Interpretation 1: Vector Operations
If the task involves vector operations, we might need to compute:
- The sum or difference of vectors \( \mathbf{a} \) and \( \mathbf{b} \).
- The condition for collinearity using the equation \( x_1 y_2 - x_2 y_1 = 0 \).

#### Interpretation 2: Trigonometric Identities
If the task involves trigonometry, we might need to:
- Simplify or verify the given trigonometric identities.
- Solve for angles or values using these identities.

#### Interpretation 3: Physics Application
If the task involves physics, we might need to:
- Explain or apply the formula \( E = mc^2 \).
- Interpret the integral \( \int r \, dt \) in a physical context.

#### Interpretation 4: Geometry
If the task involves geometry, we might need to:
- Calculate the area of a triangle using the given formula.
- Analyze the properties of the triangle based on the side lengths.

Solution Approach:

Since the specific problem is not stated, I will provide a general solution framework for each interpretation.

#### Framework 1: Vector Operations
1. Sum of Vectors:
\[
\mathbf{a} + \mathbf{b} = (x_1 + x_2, y_1 + y_2)
\]
2. Difference of Vectors:
\[
\mathbf{a} - \mathbf{b} = (x_1 - x_2, y_1 - y_2)
\]
3. Collinearity Condition:
Check if \( x_1 y_2 - x_2 y_1 = 0 \).

#### Framework 2: Trigonometric Identities
1. Simplify Identities:
- Verify \( \sin(2\pi - x) = -\cos(x) \):
\[
\sin(2\pi - x) = \sin(2\pi)\cos(x) - \cos(2\pi)\sin(x) = 0 \cdot \cos(x) - 1 \cdot \sin(x) = -\sin(x)
\]
Using the identity \( \sin(x) = \cos\left(\frac{\pi}{2} - x\right) \), we get:
\[
-\sin(x) = -\cos\left(\frac{\pi}{2} - x\right)
\]
This confirms the identity.
- Verify \( \cos(2\pi - x) = \cos(x) \):
\[
\cos(2\pi - x) = \cos(2\pi)\cos(x) + \sin(2\pi)\sin(x) = 1 \cdot \cos(x) + 0 \cdot \sin(x) = \cos(x)
\]

#### Framework 3: Physics Application
1. Energy-Mass Relationship:
- Given \( E = mc^2 \), if mass \( m \) and speed of light \( c \) are known, calculate energy \( E \).
2. Integral of Radius:
- If \( r \) is a function of time \( t \), the integral \( \int r \, dt \) represents the accumulation of radius over time. Without a specific function for \( r(t) \), this cannot be computed directly.

#### Framework 4: Geometry
1. Area of Triangle:
- Given the formula \( \frac{\sqrt{AB \cdot BC}}{2} \), if the side lengths \( AB \) and \( BC \) are known, calculate the area.
- For example, if \( AB = 4 \) and \( BC = 9 \):
\[
\text{Area} = \frac{\sqrt{4 \cdot 9}}{2} = \frac{\sqrt{36}}{2} = \frac{6}{2} = 3
\]

Conclusion:

Without a specific problem statement, I have provided a general analysis of the elements in the image and outlined potential solutions for various interpretations. If you can clarify the exact task or problem you want to solve, I can refine the solution further.

For now, the general answer is:
\[
\boxed{\text{Analysis and frameworks provided above}}
\]