why vector a = length of vector a times the unit vector of vector a? Proof in coordinate system 

To prove that a vector \( \mathbf{a} \) is equal to the product of its length (magnitude) and its unit vector, we will work in a coordinate system and use the definition of vector magnitude and unit vector.

### Let’s break down the components:

1. **Vector \( \mathbf{a} \):**
   - In a 3D coordinate system, a vector \( \mathbf{a} \) can be written as:
     \[
     \mathbf{a} = \langle a_x, a_y, a_z \rangle
     \]
     where \( a_x \), \( a_y \), and \( a_z \) are the components of the vector along the \(x\)-axis, \(y\)-axis, and \(z\)-axis, respectively.

2. **Magnitude of \( \mathbf{a} \):**
   - The magnitude (or length) of \( \mathbf{a} \), denoted \( |\mathbf{a}| \), is given by the formula:
     \[
     |\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}
     \]
     This is the Euclidean norm of the vector.

3. **Unit vector \( \hat{\mathbf{a}} \):**
   - The unit vector \( \hat{\mathbf{a}} \) in the direction of \( \mathbf{a} \) is defined as the vector with the same direction as \( \mathbf{a} \) but with a magnitude of 1. It is given by:
     \[
     \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}
     \]
     This normalizes the vector by dividing each component by the magnitude of the vector.

### Proof: 
We want to show that \( \mathbf{a} = |\mathbf{a}| \hat{\mathbf{a}} \).

Substitute the expression for \( \hat{\mathbf{a}} \) into \( |\mathbf{a}| \hat{\mathbf{a}} \):
\[
|\mathbf{a}| \hat{\mathbf{a}} = |\mathbf{a}| \cdot \frac{\mathbf{a}}{|\mathbf{a}|}
\]

Since \( |\mathbf{a}| \) is a scalar, it cancels out with the denominator:
\[
|\mathbf{a}| \cdot \frac{\mathbf{a}}{|\mathbf{a}|} = \mathbf{a}
\]

Thus, we have shown that:
\[
\mathbf{a} = |\mathbf{a}| \hat{\mathbf{a}}
\]

### Explanation:
- \( \hat{\mathbf{a}} \) is a unit vector that has the same direction as \( \mathbf{a} \) but a magnitude of 1.
- By multiplying the unit vector by the magnitude \( |\mathbf{a}| \), you recover the original vector \( \mathbf{a} \), since the magnitude of \( \mathbf{a} \) scales the unit vector back to the correct length.

### Conclusion:
We have proven that \( \mathbf{a} = |\mathbf{a}| \hat{\mathbf{a}} \), meaning that any vector can be expressed as the product of its magnitude and its corresponding unit vector in a coordinate system.

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Would you like more details on any of the steps, or have any questions? Here are five related questions to expand on this concept:

1. How do you compute the unit vector for any arbitrary vector in 3D?
2. What is the geometric interpretation of the unit vector?
3. How do vector components change when converting to polar or spherical coordinates?
4. Can the same concept of magnitude and direction be applied to higher-dimensional vectors?
5. How would you compute the magnitude of a vector in 2D instead of 3D?

**Tip:** Normalizing vectors (i.e., converting them to unit vectors) is often useful when you need to work with direction independently of magnitude.

why vector a = length of vector a times the unit vector of vector a? Proof in coordinate system

Explore the proof that any vector a can be expressed as the product of its magnitude and its unit vector in a 3D coordinate system. This solution covers vector magnitude, unit vector definitions, and key formulas, suitable for students in Grades 10-12.

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