Problem 56 - Essential Calculus Early Transcendentals

Question

Find the distance between the skew lines with parametric equations \(x=1+t, y=1+6 t, z=2 t,\) and \(x=1+2 s\) \(y=5+15 s, z=-2+6 s\)

Step-by-Step Solution

Answer

The shortest distance between the skew lines is \( \frac{2}{7} \).

Step 1: Understand the Problem

We are given two parametric equations representing skew lines and asked to find the shortest distance between them. Skew lines are lines that do not intersect and are not parallel.

Step 2: Write Parametric Equations in Vector Form

For the first line, the vector form is \( \mathbf{r_1} = \langle 1, 1, 0 \rangle + t \langle 1, 6, 2 \rangle \). For the second line, the vector form is \( \mathbf{r_2} = \langle 1, 5, -2 \rangle + s \langle 2, 15, 6 \rangle \).

Step 3: Identify Direction Vectors

The direction vector for the first line is \( \mathbf{d_1} = \langle 1, 6, 2 \rangle \) and for the second line is \( \mathbf{d_2} = \langle 2, 15, 6 \rangle \). These vectors indicate the direction each line follows.

Step 4: Compute the Cross Product of Direction Vectors

Calculate the cross product \( \mathbf{d_1} \times \mathbf{d_2} \). This cross product will give a vector that is perpendicular to both lines. \[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 6 & 2 \ 2 & 15 & 6 \end{vmatrix} = \langle 6, -2, -3 \rangle \].

Step 5: Find the Vector Between a Point on Each Line

Identify any point from each line, say \( P_1 = (1, 1, 0) \) from the first and \( P_2 = (1, 5, -2) \) from the second. Compute the vector \( \mathbf{n} = \overrightarrow{P_1P_2} = \langle 0, 4, -2 \rangle \).

Step 6: Calculate the Distance Formula

Use the formula for the distance between skew lines: \( \text{Distance} = \frac{| \mathbf{n} \cdot (\mathbf{d_1} \times \mathbf{d_2}) |}{\|\mathbf{d_1} \times \mathbf{d_2}\|} \).

Step 7: Compute the Dot Product

Calculate the dot product \( \mathbf{n} \cdot (\mathbf{d_1} \times \mathbf{d_2}) = \langle 0, 4, -2 \rangle \cdot \langle 6, -2, -3 \rangle = 0(6) + 4(-2) - 2(-3) = -8 + 6 = -2 \).

Step 8: Compute the Magnitude of the Cross Product

Find the magnitude of \( \mathbf{d_1} \times \mathbf{d_2} \): \( \| \mathbf{d_1} \times \mathbf{d_2} \| = \sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \).

Step 9: Calculate the Distance

Substitute the dot product and magnitude into the distance formula: \( \text{Distance} = \frac{| -2 |}{7} = \frac{2}{7} \).

Key Concepts

Parametric Equations

In mathematics, parametric equations are equations that express coordinates of points on a geometrical object, such as a line or a curve, as functions of a variable, often denoted as 't' or 's'. Using these formulas, we can represent complicated geometrical shapes succinctly. For skew lines, parametric equations show the trajectory of each line in three dimensions.
For example, in the given problem, we have two parametric equations:

• For the first line as: \( (x, y, z) = (1+t, 1+6t, 2t) \)
• For the second line as: \( (x, y, z) = (1+2s, 5+15s, -2+6s) \)

Here, 't' and 's' are parameters. As these parameters vary, each equation describes a line moving through three-dimensional space. Such representation helps us identify and work with 3D structures effectively without having to visualize every single point explicitly.

Vector Cross Product

The vector cross product is a fundamental concept in vector algebra that combines two vectors to produce a third vector perpendicular to the plane containing the original vectors. It is symbolically represented by \( \times \).
Mathematically, the cross product \( \mathbf{a} \times \mathbf{b} \) for vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by the determinant:

• \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)
• Calculating this involves steps similar to finding a determinant of a matrix.

In our problem, we calculate the cross product of direction vectors \( \mathbf{d_1} = \langle 1, 6, 2 \rangle \) and \( \mathbf{d_2} = \langle 2, 15, 6 \rangle \), yielding \( \langle 6, -2, -3 \rangle \), a vector perpendicular to both skew lines. This perpendicularity helps in finding the shortest distance between the skew lines.

Dot Product

The dot product is another key operation in vector algebra that measures the cosine of the angle between two vectors and scales it with their magnitudes. It results in a scalar rather than a vector.
The dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), where \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
• This represents the product of the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \) and the cosine of the angle between them.

In the exercise, we find the dot product of vector \( \mathbf{n} = \langle 0, 4, -2 \rangle \) and the previously found vector \( \mathbf{d_1} \times \mathbf{d_2} = \langle 6, -2, -3 \rangle \). This gives a scalar value of -2, which plays a crucial role in calculating the distance formula for skew lines.

Vector Magnitude

The magnitude of a vector is a measure of its length in space. Similar to finding the length of a segment in 2D, vector magnitude extends this to three dimensions.
For a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude is calculated using:

• \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)
• This aligns with Pythagoras' theorem, generalizing it to three dimensions.

In our methodology, the vector magnitude of the cross product \( \mathbf{d_1} \times \mathbf{d_2} \) gives us \( \sqrt{49} = 7\). This value helps in dividing the absolute dot product of \( \mathbf{n} \) by the magnitude to find the minimum distance between the skew lines.

Skew Lines

Skew lines are lines in three-dimensional space that do not intersect and are not parallel. They exist uniquely in 3D as they don't share a plane.
This differs from parallel lines which never meet but exist within the same plane, or intersecting lines that cross at a point. Skew lines can be visualized as two lines threading through each other without touching.
The challenge with skew lines lies in finding the minimum distance separating them. This is achieved by calculating a common perpendicular bisector, using vector algebra, between the two lines as demonstrated in this exercise. This shortest perpendicular gives the distance, in this case, \( \frac{2}{7} \), between the given skew lines.