https://file.aiquickdraw.com/mathbot/file/1727070974073srr8vl0f.jpg

We are given two lines, \( L_1 \) and \( L_2 \), in parametric form:

\[
L_1: \begin{cases} 
x = -1 + 4t \\ 
y = 3 + t \\
z = 1
\end{cases}
\]

\[
L_2: \begin{cases} 
x = -13 + 12s \\ 
y = 1 + 6s \\
z = 2 + 3s
\end{cases}
\]

We need to determine whether these lines are parallel, skew, or intersecting. If they intersect, we will find the point of intersection.

### Step 1: Check if the lines are parallel
Two lines are parallel if their direction vectors are scalar multiples of each other. The direction vector for \( L_1 \) is \( \mathbf{d_1} = (4, 1, 0) \), and the direction vector for \( L_2 \) is \( \mathbf{d_2} = (12, 6, 3) \).

Let's check if \( \mathbf{d_1} \) is a scalar multiple of \( \mathbf{d_2} \). We can see that:

\[
\frac{12}{4} = 3, \quad \frac{6}{1} = 6, \quad \frac{3}{0}
\]

Since the ratios are not the same, the direction vectors are not scalar multiples of each other. Therefore, the lines are **not parallel**.

### Step 2: Check if the lines intersect
To check if the lines intersect, we need to solve for \( t \) and \( s \) such that the coordinates \( (x, y, z) \) of both lines are equal. This gives us the system of equations:

\[
-1 + 4t = -13 + 12s \quad \text{(for } x\text{-coordinates)}
\]
\[
3 + t = 1 + 6s \quad \text{(for } y\text{-coordinates)}
\]
\[
1 = 2 + 3s \quad \text{(for } z\text{-coordinates)}
\]

### Step 3: Solve the system of equations
Start by solving the third equation for \( s \):
\[
1 = 2 + 3s \quad \Rightarrow \quad 3s = -1 \quad \Rightarrow \quad s = -\frac{1}{3}
\]

Now substitute \( s = -\frac{1}{3} \) into the first two equations.

For the first equation:
\[
-1 + 4t = -13 + 12\left(-\frac{1}{3}\right) \quad \Rightarrow \quad -1 + 4t = -13 - 4 \quad \Rightarrow \quad -1 + 4t = -17 \quad \Rightarrow \quad 4t = -16 \quad \Rightarrow \quad t = -4
\]

For the second equation:
\[
3 + t = 1 + 6\left(-\frac{1}{3}\right) \quad \Rightarrow \quad 3 + t = 1 - 2 \quad \Rightarrow \quad 3 + t = -1 \quad \Rightarrow \quad t = -4
\]

### Step 4: Verify the solution
We found that \( t = -4 \) and \( s = -\frac{1}{3} \). Now substitute these values back into the parametric equations of both lines to verify that they yield the same point.

For \( L_1 \) with \( t = -4 \):
\[
x = -1 + 4(-4) = -1 - 16 = -17
\]
\[
y = 3 + (-4) = 3 - 4 = -1
\]
\[
z = 1
\]

For \( L_2 \) with \( s = -\frac{1}{3} \):
\[
x = -13 + 12\left(-\frac{1}{3}\right) = -13 - 4 = -17
\]
\[
y = 1 + 6\left(-\frac{1}{3}\right) = 1 - 2 = -1
\]
\[
z = 2 + 3\left(-\frac{1}{3}\right) = 2 - 1 = 1
\]

Both lines intersect at the point \( (-17, -1, 1) \).

### Conclusion
The lines are **intersecting**, and the point of intersection is \( (-17, -1, 1) \).

Would you like further details on any part of the solution, or do you have any questions?

---

Here are five related questions for further practice:

1. What conditions must hold for two lines in 3D to be skew?
2. Can two lines with different direction vectors ever be parallel?
3. How would you check if two planes are parallel or intersecting?
4. If one line lies on a plane and another line is perpendicular to the plane, how do you determine if they intersect?
5. What is the geometric interpretation of the dot product of two direction vectors?

**Tip**: When solving for the intersection of lines, always verify the solution by substituting values back into the parametric equations.

Determine whether the following lines are parallel, skewed or intersecting. If the lines are intersecting, find the point of intersection.

L1: x = -1 + 4t, y = 3 + t, z = 1

L2: x = -13 + 12s, y = 1 + 6s, z = 2 + 3s

Learn how to determine whether two lines in 3D space are parallel, skew, or intersecting. This problem involves parametric equations of lines, vector direction calculations, and solving a system of linear equations. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation