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resuelve este problema de manera humana y ponlo como a y b

Let's break down and solve the problem step by step.

### Problem Statement:

1. **Part a**: 
   - It mentions the effect of multiplying a vector by a scalar. Specifically, how this operation changes the magnitude but not the direction of the vector.
   - **Question**: If a vector's magnitude is 6, what would be the magnitude of that vector if it's multiplied by 10? Also, is the resulting vector a unit vector?

2. **Part b**: 
   - You need to normalize each of the given vectors to create unit vectors.

### Solution:

#### Part a:

- **Concept**: When you multiply a vector by a scalar, the direction of the vector remains unchanged, but its magnitude is scaled by the absolute value of the scalar. A unit vector has a magnitude of 1.

- **Given**: 
  - The magnitude of the original vector, \( |v| = 6 \).
  - The scalar to multiply by, \( k = 10 \).

- **Magnitude after multiplication**:
  \[
  |v'| = |k \cdot v| = |k| \cdot |v| = 10 \cdot 6 = 60
  \]
  So, the magnitude of the resulting vector is 60.

- **Is it a unit vector?**: 
  No, because a unit vector has a magnitude of 1, and the resulting vector has a magnitude of 60.

#### Part b:

- **Concept**: Normalizing a vector means scaling it to have a magnitude of 1. To normalize a vector \( \vec{v} = (x, y, z) \), you divide each component by the vector's magnitude.

- **Given Vectors**:
  1. \( \vec{v}_1 = (1, -2, 2) \)
  2. \( \vec{v}_2 = (-6, 8, -10) \)
  3. \( \vec{v}_3 = (6, 5, 9) \)

- **Steps to normalize each vector**:

  1. **Vector \( \vec{v}_1 = (1, -2, 2) \)**:
     \[
     \text{Magnitude} = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
     \]
     \[
     \text{Unit Vector} = \left(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}\right)
     \]

  2. **Vector \( \vec{v}_2 = (-6, 8, -10) \)**:
     \[
     \text{Magnitude} = \sqrt{(-6)^2 + 8^2 + (-10)^2} = \sqrt{36 + 64 + 100} = \sqrt{200} \approx 14.14
     \]
     \[
     \text{Unit Vector} = \left(\frac{-6}{14.14}, \frac{8}{14.14}, \frac{-10}{14.14}\right) \approx \left(-0.424, 0.566, -0.707\right)
     \]

  3. **Vector \( \vec{v}_3 = (6, 5, 9) \)**:
     \[
     \text{Magnitude} = \sqrt{6^2 + 5^2 + 9^2} = \sqrt{36 + 25 + 81} = \sqrt{142} \approx 11.92
     \]
     \[
     \text{Unit Vector} = \left(\frac{6}{11.92}, \frac{5}{11.92}, \frac{9}{11.92}\right) \approx \left(0.503, 0.419, 0.755\right)
     \]

### Final Answers:

**a)**
- The magnitude of the vector after being multiplied by 10 is **60**.
- The resulting vector is **not a unit vector** because its magnitude is not 1.

**b)**
- Normalized vectors (unit vectors):
  1. \( \left(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}\right) \)
  2. \( \left(-0.424, 0.566, -0.707\right) \)
  3. \( \left(0.503, 0.419, 0.755\right) \)

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Do you want any further details or have any questions?

Here are five related questions to further your understanding:

1. What is the significance of a unit vector in physics or engineering?
2. How would multiplying a vector by a negative scalar affect its direction?
3. If a vector has a magnitude of 0, what can be said about that vector?
4. How would you add two vectors together and normalize the result?
5. How does the process of normalization apply to higher dimensions or more components?

**Tip**: Always remember that the direction of a vector is preserved when multiplying by a positive scalar, but the magnitude scales according to the absolute value of the scalar.

Explore the effects of scalar multiplication on vector magnitudes and the process of normalizing vectors to unit vectors. Solve specific problems related to vector operations and understand their applications. Suitable for Grades 10-12.

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