Let's re-examine Part (b) and correct the mistake.

### Part (b) Revised

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 \( \mathbf{v} = (3, 1) \)

We express vector \( \mathbf{w} \) as:
\[
\mathbf{w} = (4 - 0, 2 - y) = (4, 2 - y)
\]
Since \( \mathbf{w} \) must be a scalar multiple of \( \mathbf{v} = (3, 1) \), we have:
\[
\mathbf{w} = k(3, 1) = (3k, k)
\]
This gives us the equations:
\[
4 = 3k \quad \text{and} \quad 2 - y = k
\]

#### Step 1: Solve for \( k \)
From the first equation:
\[
k = \frac{4}{3}
\]

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

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

I made an error in interpreting the vector components earlier. The correct interpretation should consider the vector \( \mathbf{w} \) itself. If we re-examine \( \mathbf{w} \), its component should be correctly expressed as:

\[
\mathbf{w} = \left(4, \frac{2}{3}\right)
\]

However, the calculation does give \( y = \frac{2}{3} \), but the direction wasn't what I expected. Let's reverse the procedure:

#### Step 3: Write \( y \) in the correct terminal point
\[
y = 2 - \frac{4}{3} = \frac{2}{3

Understand how to correct vector calculations with a detailed explanation. This problem involves vectors, scalar multiplication, and correcting errors in vector components. Suitable for advanced high school students or anyone learning vector mathematics.

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