Problem 65 - University Calculus: Early Transcendentals

Question

Given two lines in space, either they are parallel, they intersect, or they are skew (lie in parallel planes). Determine 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. $$\begin{array}{ll}L 1: x=3+2 t, y=-1+4 t, z=2-t ; & -\infty

Step-by-Step Solution

Answer

L1 and L2 are skew, L1 and L3 are skew; L2 and L3 intersect at (9, 5, 5).

Step 1: Determine Direction Vectors

First, determine the direction vectors for each line. The direction vector for a line is obtained from the coefficients of the parameter:- For line \(L_1\), the direction vector \(\mathbf{d_1}\) is \(\langle 2, 4, -1 \rangle\).- For line \(L_2\), the direction vector \(\mathbf{d_2}\) is \(\langle 4, 2, 4 \rangle\).- For line \(L_3\), the direction vector \(\mathbf{d_3}\) is \(\langle 2, 1, 2 \rangle\).

Step 2: Check for Parallelism

To check if two lines are parallel, compare their direction vectors. If one vector is a scalar multiple of the other, the lines are parallel:- \(\mathbf{d_1} eq k \mathbf{d_2}\) for any scalar \(k\).- \(\mathbf{d_1} eq k \mathbf{d_3}\) for any scalar \(k\).- \(\mathbf{d_2} eq k \mathbf{d_3}\) for any scalar \(k\).Thus, no pair of lines are parallel.

Step 3: Check for Intersections between L1 and L2

To determine if \(L_1\) and \(L_2\) intersect, solve for \(t\) and \(s\) such that:\[3 + 2t = 1 + 4s, \-1 + 4t = 1 + 2s, \2 - t = -3 + 4s\]Solving the system, we find no common solution for \(t\) and \(s\) which makes all three equations true, indicating \(L_1\) and \(L_2\) are skew.

Step 4: Check for Intersections between L1 and L3

To determine if \(L_1\) and \(L_3\) intersect, solve for \(t\) and \(r\) such that:\[3 + 2t = 3 + 2r, \-1 + 4t = 2 + r, \2 - t = -2 + 2r\]Solving the system results in no consistent solutions, indicating \(L_1\) and \(L_3\) are skew.

Step 5: Check for Intersections between L2 and L3

To determine if \(L_2\) and \(L_3\) intersect, solve for \(s\) and \(r\) such that:\[1 + 4s = 3 + 2r, \1 + 2s = 2 + r, \-3 + 4s = -2 + 2r\]Solving the equations results in valid values \(s = 2\) and \(r = 3\), meaning \(L_2\) and \(L_3\) intersect. By substituting \(s\) and \(r\) back, the intersection point is \((9, 5, 5)\).

Step 6: Calculate Distance between Skew Lines

Since \(L_1\) and \(L_2\) do not intersect and are not parallel, we find the shortest distance between them using the formula for distance between skew lines:\[\text{Distance} = \frac{|\mathbf{(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{d_1} \times \mathbf{d_2})}|}{|\mathbf{d_1} \times \mathbf{d_2}|}\]Substituting \(\mathbf{a_1} = \langle 3, -1, 2 \rangle\) and \(\mathbf{a_2} = \langle 1, 1, -3 \rangle\), the distance is calculated as \(\sqrt{161}/6\).

Key Concepts

Direction Vectors

When analyzing lines in vector calculus, the direction vector is essential as it defines the direction and orientation of the line in space. It simplifies understanding complex concepts like intersection or parallelism. For a given line with a parameterized form, the direction vector can be extracted from the coefficients of the parameter.

• For line \( L_1 \), given by \( x=3+2t, y=-1+4t, z=2-t \), the direction vector is \( \langle 2, 4, -1 \rangle \).
• For line \( L_2 \), given by \( x=1+4s, y=1+2s, z=-3+4s \), the direction vector is \( \langle 4, 2, 4 \rangle \).
• For line \( L_3 \), with \( x=3+2r, y=2+r, z=-2+2r \), the direction vector is \( \langle 2, 1, 2 \rangle \).

By understanding these vectors, we can perform further analysis like checking if lines are parallel or intersect at any point in space.

Parallel Lines

Two lines are considered parallel if they have the same direction vector or if one is a scalar multiple of the other. In simpler terms, if one line can be stretched or shrunk to exactly match the other without altering its orientation, they are parallel.

• Consider the direction vectors: \( \mathbf{d_1} = \langle 2, 4, -1 \rangle \), \( \mathbf{d_2} = \langle 4, 2, 4 \rangle \), and \( \mathbf{d_3} = \langle 2, 1, 2 \rangle \).
• We need to check if there exists any scalar \( k \) such that one vector becomes another; however, none of these can be expressed as a scalar multiple of another.

Thus, these lines \( L_1 \), \( L_2 \), and \( L_3 \) are not parallel to each other. This determination sets the stage for further analysis like verifying intersections.

Intersection of Lines

Lines intersect when, at a particular point in space, they meet or cross each other. Unlike parallel lines which never meet, intersecting lines share common coordinates at the point of intersection. To find this point, solve equations derived from equating the parameterized expressions of the lines.

• For lines \( L_1 \) and \( L_2 \), solving the system of equations for \( t \) and \( s \) reveals no consistent solution. Hence, these lines do not intersect.
• For lines \( L_1 \) and \( L_3 \) also, there are no consistent solutions for \( t \) and \( r \), indicating no intersection.
• For \( L_2 \) and \( L_3 \), solving yields \( s = 2 \) and \( r = 3 \), giving a valid intersection point at \( (9, 5, 5) \).

This method ensures a systematic approach to finding intersections, essential in vector calculus.

Skew Lines

Skew lines are a unique concept in 3D geometry where lines do not intersect and are not parallel. These lines exist in different planes, never meeting at any point. When lines are skew, calculating the shortest distance between them requires a specialized approach.

• For lines \( L_1 \) and \( L_2 \), which are neither parallel nor intersecting, they are skew.
• To find the shortest distance between them, employ the formula:\[\text{Distance} = \frac{|\mathbf{(a_2 - a_1) \cdot (d_1 \times d_2)}|}{|\mathbf{d_1 \times d_2}|}\]
• Using points \( \mathbf{a_1} = \langle 3, -1, 2 \rangle \) and \( \mathbf{a_2} = \langle 1, 1, -3 \rangle \), compute the distance to be \( \frac{\sqrt{161}}{6} \).

Understanding skew lines and calculating the distance between them is crucial in problems where lines belong to different planes.