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The problem you're asking to solve is about finding the distance between two skew lines (garis bersilangan) from the given parametric equations. Let's break it down step by step for each part.

### Part (a):
We are given the parametric equations for two lines:
1. \( L_1: \, x = 2 + 4t, \, y = 6 - 4t, \, z = 5t \)
2. \( L_2: \, x = 3 + 8t, \, y = 5 - 3t, \, z = 6 + t \)

### Part (b):
We have two other parametric lines:
1. \( L_1: \, x = 1 + 7t, \, y = 3 + t, \, z = 5 - 3t \)
2. \( L_2: \, x = 4 - t, \, y = 6, \, z = 7 + 2t \)

#### To find the distance between two skew lines:

1. **Direction vectors**: First, we calculate the direction vectors for both lines.
   - For \( L_1 \), the direction vector is found from the coefficients of \( t \).
   - For \( L_2 \), similarly from the \( t \)-coefficients.

2. **Vector between points**: Pick points on both lines (for example, \( t = 0 \)) and calculate the vector joining these points.

3. **Cross product**: Take the cross product of the two direction vectors to get a vector perpendicular to both.

4. **Distance formula**: The distance between two skew lines is given by the formula:
   \[
   D = \frac{|(\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}
   \]
   where \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) are points on the respective lines, and \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \) are their direction vectors.

I will calculate the distances for both parts (a) and (b). Let's proceed with the calculation.The distances between the two skew lines are:

1. **Part (a)**: The distance is approximately **2.23 units**.
2. **Part (b)**: The distance is approximately **2.23 units**.

These values represent the minimum distances between the two skew lines in each case.

Would you like more details on the calculation or further clarifications?

Here are 5 related questions to expand on this:
1. How is the cross product used to determine the perpendicular vector between two skew lines?
2. What happens when two lines are parallel? How would the distance calculation change?
3. Can skew lines exist in two-dimensional space, or only in three dimensions?
4. How does the direction vector of a line affect the calculation of distance?
5. Can you explain the geometric interpretation of the dot product in the distance formula?

**Tip**: When dealing with skew lines, remember that they never intersect but are not parallel, and the cross product of their direction vectors plays a key role in calculating the perpendicular distance.

Carilah jarak dua garis bersilangan berikut. a. x = 2 + 4t, y = 6 - 4t, z = 5t; x = 3 + 8t, y = 5 - 3t, z = 6 + t. b. x = 1 + 7t, y = 3 + t, z = 5 - 3t; x = 4 - t, y = 6, z = 7 + 2t.

Solve the problem of finding the distance between two skew lines using parametric equations. This solution involves vector calculus, including cross and dot products, suitable for advanced high school or university students.

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