Problem 62 - Thomas Calculus

Question

Given two lines in space, either they are parallel, they intersect, or they are skew (lie in parallel planes). In Exercises 61 and \(62,\) deter- mine whether the lines, taken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersection. Otherwise, find the distance between the two lines. $$L 1 : x=1+2 t y=-1-t z=3 t -\infty < t < \infty$$ $$L 2 : x=2-s y=3 s z=1+s -\infty < s < \infty $$ $$L 3 : x=5+2 r y=1-r z=8+3 r -\infty < r < \infty$$

Step-by-Step Solution

Answer

Lines L1 and L2 intersect at \((\frac{9}{5}, -\frac{7}{5}, \frac{6}{5})\); Lines L1 and L3 are parallel; Lines L2 and L3 are skew.

Step 1: Analyzing Line Equations

We are given three line equations in parametric form. Each line is expressed as a function of a parameter: \( L_1 = (x=1+2t, y=-1-t, z=3t) \), \( L_2 = (x=2-s, y=3s, z=1+s) \), and \( L_3 = (x=5+2r, y=1-r, z=8+3r) \). We need to compare two lines at a time.

Step 2: Comparing Line L1 and L2

Check if the direction vectors of the lines are parallel. For \( L_1 \), the direction vector is \((2, -1, 3)\), and for \( L_2 \), it is \((-1, 3, 1)\). Since these vectors are not scalar multiples of each other, \( L_1 \) and \( L_2 \) are not parallel. Next, we check if they intersect by equating their parametric equations and solving for \( t \) and \( s \).

Step 3: Solving for Intersection of L1 and L2

Set the parametric equations equal: \(1 + 2t = 2 - s\), \(-1 - t = 3s\), \(3t = 1 + s\). Solving these, we first isolate \(s\) in terms of \(t\) from the first and third equations: \(s = 1 - 2t\), \(s = 3t - 1\). Equate and solve: \(1 - 2t = 3t - 1\) leading to \(t = \frac{2}{5}\). Substitute \(t\) back to find \(s = \frac{1}{5}\).

Step 4: Verifying Intersection Point for L1 and L2

Plug \(t = \frac{2}{5}\) into \(L_1\): \(x = \frac{9}{5}, y = -\frac{7}{5}, z = \frac{6}{5}\). Plug \(s = \frac{1}{5}\) into \(L_2\): \(x = \frac{9}{5}, y = -\frac{7}{5}, z = \frac{6}{5}\). Since both give the same point, the lines intersect at \( (\frac{9}{5}, -\frac{7}{5}, \frac{6}{5}) \).

Step 5: Comparing Line L1 and L3

For \( L_1 \), the direction vector is \((2, -1, 3)\) and for \( L_3 \), it's \((2, -1, 3)\). These vectors are parallel because they are scalar multiples. We check the vectors joining points from each line. Using the point \((1, -1, 0)\) from both lines, we get a direction vector between the lines \((4, 2, 8)\). Solving for a common point yields no results, meaning they are skew.

Step 6: Comparing Line L2 and L3

For \( L_2 \), the direction vector is \((-1, 3, 1)\) and for \( L_3 \), it's \((2, -1, 3)\). These vectors are not parallel. Attempting to set the lines equal \(x, y, z\) yields no common solution, indicating that \(L_2\) and \(L_3\) do not intersect and are skew lines.

Step 7: Finding Distance Between Skew Lines L2 and L3

Distance between skew lines is calculated using the formula: \( \text{Distance} = \frac{|d \cdot (b \times c)|}{|b \times c|} \) where \(d\) is the vector between the lines, \(b = (-1, 3, 1)\) from \(L_2\) and \(c = (2, -1, 3)\) from \(L_3\). Perform cross product and dot product, simplify to find the distance.

Key Concepts

Parametric Equations

When working with 3D geometry, parametric equations become incredibly useful to describe lines in space.
A parametric equation expresses the coordinates of the points that make up a geometrical object as functions of a number of variables, called parameters.
For lines, we typically have three functions, one for each coordinate (x, y, z), each of which depends on a single parameter.Understanding how these parametric equations are structured is essential for other tasks, such as finding intersections or distances between lines.
Suppose we have two lines with parametric equations:

• Line 1: \( x_1 = a_1 + m_1 t \), \( y_1 = b_1 + n_1 t \), \( z_1 = c_1 + p_1 t \)
• Line 2: \( x_2 = a_2 + m_2 s \), \( y_2 = b_2 + n_2 s \), \( z_2 = c_2 + p_2 s \)

Here, the coefficients \( m_1, n_1, p_1 \) and \( m_2, n_2, p_2 \) are the direction vectors of the respective lines.
The parameters \( t \) and \( s \) are variables that can take on any real number, effectively "tracing out" each line in three dimensions.

Line Intersection

Checking if two lines intersect involves finding a common point on both lines.
In terms of parametric equations, this requires solving a system of equations to find parameters where:

• The x-coordinates are equal: \( x_1(t) = x_2(s) \)
• The y-coordinates are equal: \( y_1(t) = y_2(s) \)
• The z-coordinates are equal: \( z_1(t) = z_2(s) \)

The key is to determine values for the parameters \( t \) and \( s \) such that all three coordinate equations are simultaneously satisfied.
If such a pair \( (t, s) \) can be found, the lines intersect, and the coordinates can be plugged back into the parametric equations to find the exact intersection point.In our exercise, we found that lines \( L_1 \) and \( L_2 \) intersect at the point \( \left( \frac{9}{5}, -\frac{7}{5}, \frac{6}{5} \right) \), after solving the system of equations for \( t = \frac{2}{5} \) and \( s = \frac{1}{5} \).

Skew Lines

Skew lines are an interesting concept in 3D geometry.
Unlike parallel lines, which might never meet because they run alongside each other, skew lines do not meet either, but for a different reason.
They lie in different planes and thus have no intersection point.To determine if two lines are skew, check that:

• The lines are not parallel (their direction vectors are not scalar multiples).
• No value of parameters makes all coordinates align (no point of intersection).

For lines in our problem, \( L_2 \) and \( L_3 \) are skew because their direction vectors differ, and there is no solution to the coordinate equations that makes both lines meet.
This means that although they extend infinitely, they will never cross or touch each other.

Distance Between Lines

In 3D geometry, finding the distance between skew lines is a common challenge.
The shortest distance is along the line perpendicular to both lines.To calculate this distance, you can use the vector form:

• Find vector \( d \) as the difference in position vectors on each line: \( d = (a_2 - a_1, b_2 - b_1, c_2 - c_1) \).
• Compute the cross product of direction vectors \( b \) and \( c \).
• Using the formula \( \frac{|d \cdot (b \times c)|}{|b \times c|} \) gives the minimum distance.

The numerator is the absolute value of the dot product of vector \( d \) with the cross product of the direction vectors, while the denominator is the magnitude of the cross product.
This formula ensures the solution lies along a perpendicular, providing the shortest separation between the skew lines.