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We are tasked with proving that vectors \( \vec{CA} \) and \( \vec{CB} \) are perpendicular using the dot product. Let's break down the problem step by step.

### Given:
- \( AB \) is the diameter of the circle.
- \( O \) is the center of the circle.
- \( C \) is a point on the circle.
- Vectors \( \vec{OA} = -\vec{u} \), \( \vec{OB} = \vec{u} \), and \( \vec{OC} = \vec{v} \).

### Objective:
To show that vectors \( \vec{CA} \) and \( \vec{CB} \) are perpendicular by using the dot product.

### Steps:

1. **Express the vectors \( \vec{CA} \) and \( \vec{CB} \):**
   \[
   \vec{CA} = \vec{OA} - \vec{OC} = -\vec{u} - \vec{v}
   \]
   \[
   \vec{CB} = \vec{OB} - \vec{OC} = \vec{u} - \vec{v}
   \]

2. **Dot product of \( \vec{CA} \) and \( \vec{CB} \):**
   The dot product of \( \vec{CA} \) and \( \vec{CB} \) is:
   \[
   \vec{CA} \cdot \vec{CB} = (-\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v})
   \]
   Expanding this:
   \[
   = (-\vec{u}) \cdot \vec{u} + (-\vec{u}) \cdot (-\vec{v}) + (-\vec{v}) \cdot \vec{u} + (-\vec{v}) \cdot (-\vec{v})
   \]
   Simplifying each term:
   \[
   = -\vec{u} \cdot \vec{u} + \vec{u} \cdot \vec{v} + (-\vec{v}) \cdot \vec{u} + \vec{v} \cdot \vec{v}
   \]
   Since \( \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \), we have:
   \[
   = -|\vec{u}|^2 + \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{v} + |\vec{v}|^2
   \]
   Which simplifies to:
   \[
   = -|\vec{u}|^2 + 2\vec{u} \cdot \vec{v} + |\vec{v}|^2
   \]

3. **Using the property of a circle:**
   Since \( AB \) is the diameter of the circle, the vector \( \vec{OA} \) is perpendicular to the vector \( \vec{OB} \). Thus, by the geometric properties of the circle, \( \vec{u} \cdot \vec{v} = 0 \).

   Substituting \( \vec{u} \cdot \vec{v} = 0 \) into the dot product equation:
   \[
   \vec{CA} \cdot \vec{CB} = -|\vec{u}|^2 + |\vec{v}|^2
   \]
   Now, because \( A \), \( B \), and \( C \) all lie on the circle, \( |\vec{u}| = |\vec{v}| \). Therefore:
   \[
   \vec{CA} \cdot \vec{CB} = -|\vec{u}|^2 + |\vec{u}|^2 = 0
   \]

4. **Conclusion:**
   Since \( \vec{CA} \cdot \vec{CB} = 0 \), the vectors \( \vec{CA} \) and \( \vec{CB} \) are perpendicular.

Thus, we have proven using the dot product that \( \vec{CA} \) and \( \vec{CB} \) are perpendicular.

---

Would you like more details or have any questions?

Here are some related questions to explore:
1. What are the properties of the dot product in vector mathematics?
2. How does the geometric property of a circle help in this kind of vector problem?
3. Why does \( \vec{u} \cdot \vec{v} = 0 \) in this specific scenario?
4. Can this perpendicularity be proven using another method such as trigonometry?
5. How does the length of vectors \( \vec{OA} \), \( \vec{OB} \), and \( \vec{OC} \) affect this proof?

**Tip:** Always remember that when two vectors are perpendicular, their dot product is zero. This is a quick way to test for perpendicularity.

Using dot product, show that the vectors CA and CB are perpendicular given that AB is the diameter of a circle with center O and C is a point on the circle.

In this problem, you are tasked to prove that the vectors CA and CB are perpendicular by using the dot product. This problem involves understanding vector algebra, the dot product operation, and the geometry of circles, where AB is the diameter. The proof applies key concepts in vector mathematics and circle properties, ideal for advanced high school or early college students.

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