(a) Find the components of a vector  u
 which has initial point  (−5,−1)
 , terminal point  (x,0)
 for some  x∈R
 and which points in the same direction as  (1,3)
 
Use exact values as vector entries, so use fractions rather than decimal approximations, for example (1/7,6).
Preview 

(b) Find the components of a vector  w
 which has initial point  (0,y)
 for some  y∈R
 , terminal point  (4,2)
 and which points in the same direction as  (3,1)
 .

Let's solve each part of the problem step by step.

### Part (a)

Given:
- Initial point of vector \( \mathbf{u} \) is \( (-5, -1) \)
- Terminal point of vector \( \mathbf{u} \) is \( (x, 0) \)
- The vector \( \mathbf{u} \) points in the same direction as \( \mathbf{v} = (1, 3) \)

To find the components of vector \( \mathbf{u} \), we first express it as:
\[
\mathbf{u} = (x - (-5), 0 - (-1)) = (x + 5, 1)
\]
This vector must be a scalar multiple of \( \mathbf{v} = (1, 3) \), so:
\[
\mathbf{u} = k(1, 3) = (k, 3k)
\]
Therefore, we have the equations:
\[
x + 5 = k \quad \text{and} \quad 1 = 3k
\]

Solving for \( k \) in the second equation:
\[
k = \frac{1}{3}
\]

Now, substitute \( k = \frac{1}{3} \) into the first equation:
\[
x + 5 = \frac{1}{3}
\]

Solving for \( x \):
\[
x = \frac{1}{3} - 5 = \frac{1}{3} - \frac{15}{3} = \frac{-14}{3}
\]

So the vector \( \mathbf{u} \) is:
\[
\mathbf{u} = \left(\frac{1}{3}, 1\right)
\]

### Part (b)

Given:
- Initial point of vector \( \mathbf{w} \) is \( (0, y) \)
- Terminal point of vector \( \mathbf{w} \) is \( (4, 2) \)
- The vector \( \mathbf{w} \) points in the same direction as \( \mathbf{v} = (3, 1) \)

To find the components of vector \( \mathbf{w} \), we express it as:
\[
\mathbf{w} = (4 - 0, 2 - y) = (4, 2 - y)
\]
This vector must be a scalar multiple of \( \mathbf{v} = (3, 1) \), so:
\[
\mathbf{w} = k(3, 1) = (3k, k)
\]
Therefore, we have the equations:
\[
4 = 3k \quad \text{and} \quad 2 - y = k
\]

Solving for \( k \) in the first equation:
\[
k = \frac{4}{3}
\]

Substitute \( k = \frac{4}{3} \) into the second equation:
\[
2 - y = \frac{4}{3}
\]

Solving for \( y \):
\[
y = 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}
\]

So the vector \( \mathbf{w} \) is:
\[
\mathbf{w} = \left(4, \frac{2}{3}\right)
\]

### Summary of Results
- **Part (a):** The components of vector \( \mathbf{u} \) are \( \left(\frac{1}{3}, 1\right) \).
- **Part (b):** The components of vector \( \mathbf{w} \) are \( \left(4, \frac{2}{3}\right) \).

Would you like more details or have any questions? Here are some related questions to consider:

1. How do you determine the direction of a vector?
2. What is the significance of scalar multiplication in vectors?
3. How would you find the magnitude of the vectors \( \mathbf{u} \) and \( \mathbf{w} \)?
4. What happens if the initial or terminal points of the vector are modified?
5. How do you normalize a vector, and why might it be useful?

**Tip:** When solving vector problems, always check if the direction and magnitude make sense in the context of the problem.

Learn how to find vector components given initial and terminal points, with a detailed solution. This problem involves vectors pointing in the same direction as specified vectors (1, 3) and (3, 1). Suitable for high school students studying vector algebra.

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