Step 1: Parametric Equations of the Lines

The lines $L_1$ and $L_2$ are given in parametric form:

• For line $L_1$: \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$$
• For line $L_2$: \begin{aligned} x &= 1 + 2t \\ y &= 3 - 4t \\ z &= 5 - 2t \end{aligned} \right.$$ 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.