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?

Hint: The line in 3-space that passes through the point Psub0 (xsub0, ysub0, zsub0) and is parallel to the nonzero vector v = (a, b, c) = ai + bj + ck has parametric equations:

x = xsub0 + at
y = ysub0 + bt
z = zsub0 + ct

To show that the lines \( L_1 \) and \( L_2 \) are parallel and find the distance between them, we'll follow these steps:

### Step 1: Represent the Lines in Parametric Form
The given parametric equations are:
- Line \( L_1 \): \( x = 2 - t \), \( y = 2t \), \( z = 1 + t \)
- Line \( L_2 \): \( x = 1 + 2t \), \( y = 3 - 4t \), \( z = 5 - 2t \)

### Step 2: Identify the Direction Vectors
For line \( L_1 \), the direction vector \( \mathbf{v}_1 \) can be found from the coefficients of \( t \):
\[
\mathbf{v}_1 = (-1, 2, 1)
\]
For line \( L_2 \), the direction vector \( \mathbf{v}_2 \) is:
\[
\mathbf{v}_2 = (2, -4, -2)
\]

### Step 3: Check if the Vectors are Parallel
Two vectors are parallel if one is a scalar multiple of the other, i.e., \( \mathbf{v}_1 = k \mathbf{v}_2 \) for some scalar \( k \).

Let's check if \( \mathbf{v}_1 \) is a scalar multiple of \( \mathbf{v}_2 \):
\[
(-1, 2, 1) = k(2, -4, -2)
\]
Comparing each component:
1. \( -1 = 2k \) implies \( k = -\frac{1}{2} \)
2. \( 2 = -4k \) also gives \( k = -\frac{1}{2} \)
3. \( 1 = -2k \) again gives \( k = -\frac{1}{2} \)

Since all components give the same value for \( k \), the vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are indeed parallel. Thus, the lines \( L_1 \) and \( L_2 \) are parallel.

### Step 4: Find the Distance Between the Lines
To find the distance \( D \) between two parallel lines in space, use the formula:
\[
D = \frac{| (\mathbf{P_1} - \mathbf{P_2}) \cdot (\mathbf{v}_1 \times \mathbf{v}_2) |}{|\mathbf{v}_1 \times \mathbf{v}_2|}
\]
where:
- \( \mathbf{P_1} = (2, 0, 1) \) is a point on line \( L_1 \)
- \( \mathbf{P_2} = (1, 3, 5) \) is a point on line \( L_2 \)
- \( \mathbf{v}_1 \times \mathbf{v}_2 \) is the cross product of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \)

#### Step 4.1: Compute \( \mathbf{P_1} - \mathbf{P_2} \)
\[
\mathbf{P_1} - \mathbf{P_2} = (2 - 1, 0 - 3, 1 - 5) = (1, -3, -4)
\]

#### Step 4.2: Compute \( \mathbf{v}_1 \times \mathbf{v}_2 \)
The cross product \( \mathbf{v}_1 \times \mathbf{v}_2 \) is calculated as:
\[
\mathbf{v}_1 \times \mathbf{v}_2 = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 2 & 1 \\
2 & -4 & -2 \\
\end{vmatrix}
= \hat{i}(2(-2) - 1(-4)) - \hat{j}((-1)(-2) - 1(2)) + \hat{k}((-1)(-4) - 2(2))
\]
\[
= \hat{i}(-4 + 4) - \hat{j}(2 - 2) + \hat{k}(4 - 4) = (0, 0, 0)
\]

Since the cross product results in the zero vector, the lines are not only parallel but also lie in the same plane. Hence, the distance between them is zero.



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 the lines L1: x = 2 - t, y = 2t, z = 1 + t and L2: x = 1 + 2t, y = 3 - 4t, z = 5 - 2t are parallel by analyzing their direction vectors, and calculate the distance between them using the formula for parallel lines. This solution covers key concepts like vectors, parametric equations, and cross products.

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