Show that the lines 
L1: x = 2 - t, y = 2t, z = 1 + t 

and 

L2: x = 1 + 2t, y = 3 - 4t, z = 5 - 2t 

are parallel and find the distance between them.

L1 and L2 are parallel because they are parallel to vectors v1 and v2 that satisfies: v1 = kv2

and what is D?

### Step 1: Parametric Equations of the Lines

The lines \(L_1\) and \(L_2\) are given in parametric form:

- For line \(L_1\):
  \[
  L_1: \left\{
  \begin{aligned}
  x &= 2 - t \\
  y &= 2t \\
  z &= 1 + t
  \end{aligned}
  \right.
  \]
  The direction vector \( \mathbf{v_1} \) is derived from the coefficients of \(t\):
  \[
  \mathbf{v_1} = \langle -1, 2, 1 \rangle
  \]

- For line \(L_2\):
  \[
  L_2: \left\{
  \begin{aligned}
  x &= 1 + 2t \\
  y &= 3 - 4t \\
  z &= 5 - 2t
  \end{aligned}
  \right.
  \]
  The direction vector \( \mathbf{v_2} \) is derived from the coefficients of \(t\):
  \[
  \mathbf{v_2} = \langle 2, -4, -2 \rangle
  \]

### Step 2: Check if the Lines are Parallel

To check if the lines are parallel, we need to determine if the direction vectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are proportional, i.e., if there exists a scalar \(k\) such that:
\[
\mathbf{v_1} = k \mathbf{v_2}
\]

Compare the components of \( \mathbf{v_1} = \langle -1, 2, 1 \rangle \) and \( \mathbf{v_2} = \langle 2, -4, -2 \rangle \):

\[
\frac{-1}{2} = \frac{2}{-4} = \frac{1}{-2} = -\frac{1}{2}
\]

Since all the ratios are equal, the vectors are proportional, meaning \( \mathbf{v_1} = -\frac{1}{2} \mathbf{v_2} \). Therefore, the lines \(L_1\) and \(L_2\) are parallel.

### Step 3: Distance Between the Parallel Lines

The formula for the distance between two parallel lines in 3D is:
\[
D = \frac{|\mathbf{v} \cdot (\mathbf{p_2} - \mathbf{p_1})|}{|\mathbf{v}|}
\]
where:
- \( \mathbf{v} \) is the direction vector of the lines (which is the same for both lines),
- \( \mathbf{p_1} \) is a point on \(L_1\),
- \( \mathbf{p_2} \) is a point on \(L_2\).

Let's take:
- A point on \(L_1\) at \(t = 0\): \( \mathbf{p_1} = (2, 0, 1) \),
- A point on \(L_2\) at \(t = 0\): \( \mathbf{p_2} = (1, 3, 5) \).

The vector \( \mathbf{p_2} - \mathbf{p_1} \) is:
\[
\mathbf{p_2} - \mathbf{p_1} = \langle 1 - 2, 3 - 0, 5 - 1 \rangle = \langle -1, 3, 4 \rangle
\]

The direction vector \( \mathbf{v_1} = \langle -1, 2, 1 \rangle \) can be used to calculate the distance. First, find the dot product \( \mathbf{v_1} \cdot (\mathbf{p_2} - \mathbf{p_1}) \):
\[
\mathbf{v_1} \cdot (\mathbf{p_2} - \mathbf{p_1}) = (-1)(-1) + (2)(3) + (1)(4) = 1 + 6 + 4 = 11
\]

Next, find the magnitude of \( \mathbf{v_1} \):
\[
|\mathbf{v_1}| = \sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
\]

Thus, the distance \(D\) is:
\[
D = \frac{|11|}{\sqrt{6}} = \frac{11}{\sqrt{6}} \approx 4.49
\]

### Conclusion:

- The lines \(L_1\) and \(L_2\) are parallel.
- The distance between the two lines is \( \frac{11}{\sqrt{6}} \) or approximately \(4.49\).

Would you like more details or have any questions?

### Related Questions:
1. How do you find the shortest distance between two skew lines?
2. What is the geometric interpretation of the direction vector of a line?
3. How can you determine if two vectors are perpendicular?
4. What is the formula for the distance between a point and a plane in 3D?
5. How do you find the point of intersection between two non-parallel lines?

### Tip: 
For parallel lines in space, the direction vectors are scalar multiples of each other, and the shortest distance is measured perpendicularly between any two points on the lines.

Show that the lines L1: x = 2 - t, y = 2t, z = 1 + t and L2: x = 1 + 2t, y = 3 - 4t, z = 5 - 2t are parallel and find the distance between them.

Learn how to show that two lines are parallel using vector algebra and calculate the distance between them. The problem involves 3D geometry concepts, parametric equations, and the distance formula for parallel lines. A step-by-step solution for advanced high school students.

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