Consider The Two Lines L_{1}: X=-2 T, Y=1+2 T, Z=3 T And L_{2}: X=-9+5 S, Y=2+3 S, Z=4+2 S Find The Point

Consider The Two Lines L{1}: X=-2 T, Y=1+2 T, Z=3 T And L{2}: X=-9+5 S, Y=2+3 S, Z=4+2 S Find The Point

---

Introduction

In the realm of analytic geometry, understanding the relationships between lines in three-dimensional space is fundamental. Whether you're solving for the intersection point, analyzing whether two lines are skew, intersecting, or parallel, mastering the concepts of parametric equations and their applications is essential. This article provides a comprehensive guide to analyzing two lines given in parametric form, specifically focusing on the lines:

• Line L₁:

\( X = -2T \)
\( Y = 1 + 2T \)
\( Z = 3T \)

• Line L₂:

\( X = -9 + 5S \)
\( Y = 2 + 3S \)
\( Z = 4 + 2S \)

The goal is to find the point of intersection between these two lines, if it exists, or determine that they do not intersect.

---

Understanding the Parametric Equations of Lines

Parametric Equations in Geometry

Parametric equations describe a line in space by expressing each coordinate as a function of one or more parameters—in this case, \( T \) and \( S \). The general form of a line in 3D space:

\[
\begin{cases}
X = X_0 + aT \\
Y = Y_0 + bT \\
Z = Z_0 + cT
\end{cases}
\]

where \( (X0, Y0, Z_0) \) is a point on the line, and \( (a, b, c) \) is the direction vector.

In the given equations:

• For L₁:


• Point on the line when \( T=0 \): \( (0, 1, 0) \)


• Direction vector: \( \vec{d}_1 = (-2, 2, 3) \)


• For L₂:


• Point on the line when \( S=0 \): \( (-9, 2, 4) \)


• Direction vector: \( \vec{d}_2 = (5, 3, 2) \)

---

Steps to Find the Intersection Point

1. Set Up the Equations

To find the common point between the two lines, we need to find values of \( T \) and \( S \) such that:

\[
\begin{cases}
-2T = -9 + 5S \quad \text{(X-coordinates)} \\
1 + 2T = 2 + 3S \quad \text{(Y-coordinates)} \\
3T = 4 + 2S \quad \text{(Z-coordinates)}
\end{cases}
\]

This system ensures that at the intersection point, the coordinates from both lines are equal.

2. Formulate the System of Equations

Rearranging each:

• From \( X \):

\[
-2T = -9 + 5S \implies 2T + 5S = 9
\]

• From \( Y \):

\[
1 + 2T = 2 + 3S \implies 2T - 3S = 1
\]

• From \( Z \):

\[
3T = 4 + 2S \implies 3T - 2S = 4
\]

Now, we have a system of three equations:

\[
\begin{cases}
2T + 5S = 9 \quad (1) \\
2T - 3S = 1 \quad (2) \\
3T - 2S = 4 \quad (3)
\end{cases}
\]

---

Solving the System of Equations

1. Using Equations (1) and (2)

Subtract (2) from (1):

\[
(2T + 5S) - (2T - 3S) = 9 - 1
\]

\[
2T + 5S - 2T + 3S = 8
\]

\[
8S = 8 \implies S = 1
\]

2. Find \( T \) using \( S=1 \)

Plug \( S=1 \) into equation (2):

\[
2T - 3(1) = 1 \implies 2T - 3 = 1 \implies 2T = 4 \implies T = 2
\]

3. Verify with Equation (3)

Substitute \( T=2 \) and \( S=1 \):

\[
3(2) - 2(1) = 6 - 2 = 4
\]

which matches the right side of equation (3). This confirms the solution's consistency.

---

Finding the Intersection Point

Now that we have \( T=2 \) and \( S=1 \), substitute back into the parametric equations:

• For L₁:

\[
X = -2T = -2 \times 2 = -4
\]
\[
Y = 1 + 2T = 1 + 2 \times 2 = 1 + 4 = 5
\]
\[
Z = 3T = 3 \times 2 = 6
\]

• For L₂:

\[
X = -9 + 5S = -9 + 5 \times 1 = -9 + 5 = -4
\]
\[
Y = 2 + 3S = 2 + 3 \times 1 = 2 + 3 = 5
\]
\[
Z = 4 + 2S = 4 + 2 \times 1 = 4 + 2 = 6
\]

The coordinates match perfectly, confirming that the lines intersect at the point:

\[
\boxed{(-4, 5, 6)}
\]

---

Geometric Interpretation

The intersection point \( (-4, 5, 6) \) indicates that the two lines cross at this point in three-dimensional space. Since they intersect at exactly one point, they are neither parallel nor skew lines but are intersecting lines.

---

Additional Concepts Related to Lines in 3D Space

Parallel Lines

• Two lines are parallel if their direction vectors are scalar multiples.
• For example, if \( \vec{d}1 = k \times \vec{d}2 \) for some scalar \( k \), lines are parallel.

Skew Lines

• Lines that are neither parallel nor intersecting.
• They do not meet and are not coplanar.

Coincident Lines

• Lines that lie exactly on top of each other.
• Their parametric equations are scalar multiples of each other.

---

Summary

• Given two lines in parametric form, set their coordinate equations equal to find the intersection point.
• Formulate a system of equations based on the parametric forms.
• Solve the system for the parameters.
• Substitute the parameters back into the original equations to find the point of intersection.
• Confirm the solution's consistency across all equations.

---

Conclusion

The process of finding the intersection point between two lines in 3D space involves translating parametric equations into a system of algebraic equations and solving for the parameters. For the lines:

\[
L_{1}: X = -2T, \quad Y=1+2T, \quad Z=3T
\]
\[
L_{2}: X=-9+5S, \quad Y=2+3S, \quad Z=4+2S
\]

we determined that they intersect at the point (-4, 5, 6), with parameters \( T=2 \) and \( S=1 \). This comprehensive approach is fundamental in analytical geometry, aiding in various applications such as computer graphics, robotics, and physics.

---

Keywords: 3D lines intersection, parametric equations, analytic geometry, solving for intersection point, line equations in space, vector equations of lines, geometry in space, algebraic solutions in 3D, line relationships.