Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance between the skew (nonintersecting and nonparallel) lines \(x=2-t, y=3+4 t, z=2 t\) and \(x=-1+t\), \(y=2, z=-1+2 t\) by using the following steps. (a) Note by setting \(t=0\) that \((2,3,0)\) is on the first line. (b) Find the equation of the plane \(\pi\) through \((2,3,0)\) parallel to both given lines (i.e., with normal perpendicular to both). (c) Find a point \(Q\) on the second line. (d) Find the distance from \(Q\) to the plane \(\pi\). (See Example 10 of Section 11.3.) See Problem 32 for another way to do this problem.

Step-by-Step Solution

Answer

The distance between the lines is 3.2.

Step 1: Identify Point on the First Line

Using the first condition \(t=0\) for the line \(x=2-t, y=3+4t, z=2t\), we can find a point on the first line. Substituting \(t=0\) gives the point \((2, 3, 0)\).

Step 2: Determine Direction Vectors

Identify the direction vectors for the lines: the vector of the first line is \((-1, 4, 2)\) from \(x=2-t, y=3+4t, z=2t\) and for the second line is \((1, 0, 2)\) from \(x=-1+t, y=2, z=-1+2t\).

Step 3: Calculate Normal Vector for Plane

To find the normal vector \(\mathbf{N}\) for the plane \(\pi\), compute the cross product of the direction vectors \((-1, 4, 2)\) and \((1, 0, 2)\). The cross product \[\mathbf{N} = (-1, 4, 2) \times (1, 0, 2) = (8, 4, -4)\].

Step 4: Equation of Plane Through Point

Plug the normal vector \((8, 4, -4)\) and substitute the point \((2, 3, 0)\) into the plane equation \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\). Thus, \[8(x-2) + 4(y-3) - 4(z-0) = 0\] or \[8x + 4y - 4z = 32\].

Step 5: Find a Point on the Second Line

Select a convenient value for \(t\) in the second line equation \(x=-1+t, y=2, z=-1+2t\). Let \(t=0\), so the point \((-1, 2, -1)\) is on the second line.

Step 6: Compute Distance from Point to Plane

Use the point-plane distance formula \(d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\). Substitute \((x_1, y_1, z_1) = (-1, 2, -1)\) and plane equation \(8x + 4y - 4z - 32 = 0\). Thus, \(d = \frac{|8(-1) + 4(2) - 4(-1) - 32|}{\sqrt{8^2 + 4^2 + (-4)^2}}\), which simplifies to \(d = \frac{32}{10} = 3.2\).

Key Concepts

Distance Between Skew Lines

Skew lines are lines that do not intersect and are not parallel. They exist in three-dimensional space. Unlike parallel lines, they don't maintain a constant distance as they are positioned in different planes.

To find the distance between skew lines, we need to follow a methodical approach. First, identify points on each line. Then, compute the directional vectors of each line, and subsequently, find a plane through a point on one of the lines that is parallel to both lines. Finally, the distance calculation involves measuring how far a point on the second line is from this plane. This reveals the shortest distance between the skew lines.

Direction Vectors

Direction vectors are crucial in describing the direction and orientation of lines. For any line given in parametric form, the direction vector can be derived from the coefficients of the parameter.

For the line equations given in the exercise, the first line with equations:

• \(x = 2-t, y = 3 + 4t, z = 2t\)

provides the direction vector \((-1, 4, 2)\). The second line's equations are:

• \(x = -1 + t, y = 2, z = -1 + 2t\)

giving the direction vector \((1, 0, 2)\).

These direction vectors form the foundation for further calculations and help in determining the orientation of a plane that is parallel to both lines.

Normal Vector Calculation

Calculating the normal vector to the plane is crucial because it involves the orientation perpendicular to the direction vectors of the lines. The normal vector is found using the cross product of the direction vectors of the two lines.

Given the direction vectors \((-1, 4, 2)\) and \((1, 0, 2)\), the cross product is calculated as follows:\[\mathbf{N} = (-1, 4, 2) \times (1, 0, 2)\]This results in \[(8, 4, -4)\]

This normal vector \((8, 4, -4)\) is instrumental in forming the plane equation and serves as a guideline for perpendicularity, meaning it is orthogonal to both direction vectors involved.

Plane Equation

With the normal vector and a point on one of the lines, we can construct the equation of the plane that is parallel to both skew lines. The formula for the plane equation is derived from:\[a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\]Substituting the normal vector \((8, 4, -4)\) and the point \((2, 3, 0)\), we obtain:\[8(x-2) + 4(y-3) - 4(z-0) = 0\]Simplifying, the plane equation becomes:\[8x + 4y - 4z = 32\]

This plane equation is used to determine the distance from a point on the second line, thus enabling the calculation of the shortest distance between the skew lines.

Point-to-Plane Distance

Calculating the point-to-plane distance is the final step to find the shortest distance between the skew lines. We take a point from the second line, substitute it into the plane equation, and use the point-plane distance formula.

For the point \((-1, 2, -1)\) on the second line and plane equation:\[8x + 4y - 4z - 32 = 0\]Substitute these values into the distance formula:\[d = \frac{|8(-1) + 4(2) - 4(-1) - 32|}{\sqrt{8^2 + 4^2 + (-4)^2}}\]Simplifies to:\[d = \frac{32}{10} = 3.2\]

With this calculation, we effectively determine that the shortest distance between the two skew lines is 3.2 units. This approach ensures precision and understanding of spatial relationships between lines and planes.