Step 1: Verify that the lines are parallel

To determine whether the lines $l_1$ and $l_2$ are parallel, we must compare their direction vectors. A line in parametric form, given as:

$x = x_0 + a t, \quad y = y_0 + b t, \quad z = z_0 + c t$

has the direction vector $\vec{d} = \langle a, b, c \rangle$.

Direction Vector for $l_1$:

Given $l_1$ in parametric form: $x = 2t, \quad y = 3 + 4t, \quad z = 2 - 6t$ The direction vector $\vec{d_1}$ for $l_1$ is: $\vec{d_1} = \langle 2, 4, -6 \rangle$

Direction Vector for $l_2$:

Given $l_2$ in parametric form: $x = 1 + 3t, \quad y = 6t, \quad z = -9t$ The direction vector $\vec{d_2}$ for $l_2$ is: $\vec{d_2} = \langle 3, 6, -9 \rangle$

Checking Parallelism:

Two vectors are parallel if one is a scalar multiple of the other. Check if $\vec{d_2}$ is a scalar multiple of $\vec{d_1}$:

$\vec{d_2} = k \vec{d_1}$

We compare the corresponding components: $\frac{3}{2} = \frac{6}{4} = \frac{-9}{-6} = \frac{3}{2}$

Since all ratios are equal, $\vec{d_2} = \frac{3}{2} \vec{d_1}$, confirming that the lines $l_1$ and $l_2$ are parallel.

Step 2: Find the distance between the two parallel lines

The distance between two parallel lines is the perpendicular distance between any point on one line and the other line.

Find a point on $l_1$:

We can choose $t = 0$ to get the point $P_1$ on $l_1$: $P_1 = (2(0), 3 + 4(0), 2 - 6(0)) = (0, 3, 2)$

Parametric form of $l_2$:

The parametric form of $l_2$ is: $x = 1 + 3t, \quad y = 6t, \quad z = -9t$ We need to find the perpendicular distance from the point $P_1(0, 3, 2)$ to this line.

Vector from $P_1$ to any point on $l_2$:

A general point on $l_2$ is $Q_2(1 + 3t, 6t, -9t)$. The vector from $P_1$ to $Q_2$ is: $\overrightarrow{P_1 Q_2} = \langle (1 + 3t) - 0, (6t) - 3, (-9t) - 2 \rangle = \langle 1 + 3t, 6t - 3, -9t - 2 \rangle$

Find the cross product of the direction vectors:

We need to calculate the perpendicular distance using the formula: $d = \frac{|\overrightarrow{P_1 Q_2} \times \vec{d_2}|}{|\vec{d_2}|}$ First, calculate the cross product of $\overrightarrow{P_1 Q_2}$ and $\vec{d_2}$. Let me compute that.The distance between the two parallel lines $l_1$ and $l_2$ is approximately $3.73$ units.

Summary:

1. The lines $l_1$ and $l_2$ are parallel because their direction vectors are scalar multiples of each other.
2. The perpendicular distance between the two lines is approximately $3.73$ units.

Would you like more details on any step, or have any further questions?

Here are five related questions:

1. How do you find the angle between two lines in space?
2. How do you compute the distance between two skew lines?
3. What is the formula for the shortest distance from a point to a plane?
4. How can you determine if two lines are intersecting?
5. What is the general equation of a plane passing through a given point?

Tip: When working with vector cross products, the magnitude of the cross product gives the area of the parallelogram formed by two vectors.