Determine whether the lines L
_
(
1
)
 ﻿and L
_
(
2
)
 ﻿are parallel, skew, or intersecting. L
_
(
1
)
:
(
x
-
3
)
/
(
1
)
=
(
y
-
1
)
/
(
-
2
)
=
(
z
-
1
2
)
/
(
-
3
)
 ﻿L
_
(
2
)
:
(
x
-
4
)
/
(
1
)
=
(
y
+
6
)
/
(
3
)
=
(
z
-
1
3
)
/
(
-
7
)
 ﻿parallel skew intersecting If they intersect, find the point of intersection. 
(
If an answer does not exist, enter DNE.
)
 
(
x
,
y
,
z
)
=
(
,
x
)

We are given two lines \( L_1 \) and \( L_2 \) in parametric form. Our goal is to determine whether they are parallel, skew, or intersecting, and if they intersect, find the point of intersection.

### Step 1: Express the lines in parametric form

The equations for the lines are given as:

\[
L_1: \frac{x - 3}{1} = \frac{y - 1}{-2} = \frac{z - 12}{-3}
\]
\[
L_2: \frac{x - 4}{1} = \frac{y + 6}{3} = \frac{z - 13}{-7}
\]

We can express each line in parametric form. Let \( t_1 \) be the parameter for \( L_1 \) and \( t_2 \) for \( L_2 \).

For \( L_1 \):
\[
x = 3 + t_1
\]
\[
y = 1 - 2t_1
\]
\[
z = 12 - 3t_1
\]

For \( L_2 \):
\[
x = 4 + t_2
\]
\[
y = -6 + 3t_2
\]
\[
z = 13 - 7t_2
\]

### Step 2: Check if the lines are parallel

To check if the lines are parallel, we need to compare their direction vectors. 

- The direction vector of \( L_1 \) is \( \langle 1, -2, -3 \rangle \).
- The direction vector of \( L_2 \) is \( \langle 1, 3, -7 \rangle \).

For the lines to be parallel, their direction vectors must be scalar multiples of each other. Let's compare the ratios of the corresponding components:

\[
\frac{1}{1}, \quad \frac{-2}{3}, \quad \frac{-3}{-7}
\]

Since these ratios are not equal, the lines are **not parallel**.

### Step 3: Check if the lines intersect

To check if the lines intersect, we need to find values of \( t_1 \) and \( t_2 \) such that the parametric equations for \( L_1 \) and \( L_2 \) yield the same point \( (x, y, z) \).

From the \( x \)-coordinates:
\[
3 + t_1 = 4 + t_2 \quad \Rightarrow \quad t_1 = t_2 + 1
\]

From the \( y \)-coordinates:
\[
1 - 2t_1 = -6 + 3t_2
\]
Substitute \( t_1 = t_2 + 1 \) into this equation:
\[
1 - 2(t_2 + 1) = -6 + 3t_2
\]
\[
1 - 2t_2 - 2 = -6 + 3t_2
\]
\[
-1 - 2t_2 = -6 + 3t_2
\]
\[
5 = 5t_2 \quad \Rightarrow \quad t_2 = 1
\]

Now that we have \( t_2 = 1 \), substitute this into \( t_1 = t_2 + 1 \):
\[
t_1 = 2
\]

### Step 4: Verify the intersection point

Substitute \( t_1 = 2 \) into the parametric equations for \( L_1 \):
\[
x = 3 + 2 = 5, \quad y = 1 - 2(2) = -3, \quad z = 12 - 3(2) = 6
\]

Substitute \( t_2 = 1 \) into the parametric equations for \( L_2 \):
\[
x = 4 + 1 = 5, \quad y = -6 + 3(1) = -3, \quad z = 13 - 7(1) = 6
\]

Since both lines give the same point \( (5, -3, 6) \), the lines **intersect** at the point \( (5, -3, 6) \).

### Final Answer:
The lines intersect at the point \( (5, -3, 6) \).

---

Do you want more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you find the parametric equations of a line given two points?
2. What is the geometric interpretation of skew lines?
3. How can you determine the angle between two intersecting lines in space?
4. What conditions must be satisfied for two lines in 3D to be parallel?
5. How do you find the shortest distance between two skew lines?

**Tip**: Always double-check the direction vectors when comparing two lines for parallelism.

Determine whether the lines L_1 and L_2 are parallel, skew, or intersecting. L_1: (x - 3) / 1 = (y - 1) / -2 = (z - 12) / -3. L_2: (x - 4) / 1 = (y + 6) / 3 = (z - 13) / -7. If they intersect, find the point of intersection.

Learn how to determine if two lines in 3D space are parallel, skew, or intersecting using vector algebra. This problem walks through the step-by-step process of expressing lines in parametric form and solving for their intersection point. Suitable for advanced high school geometry students.

Math Problem Statement

Determine whether the lines L _ ( 1 ) and L _ ( 2 ) are parallel, skew, or intersecting. L _ ( 1 ) : ( x

3 ) / ( 1 )

( y

1 ) / (

2 )

( z

1 2 ) / (

3 ) L _ ( 2 ) : ( x

4 ) / ( 1 )

( y + 6 ) / ( 3 )

( z

1 3 ) / (

( x , y , z )

Solution

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