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Chapter 1: Problem 36

Find the distance between the parallel lines $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right]+\cdot\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \text { and }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right]+t\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$

Short Answer

Expert verified
The distance between the two parallel lines is \( \frac{\sqrt{42}}{3} \).

Step by step solution

01

Understand the Line Equations

The given lines are vector equations of parallel lines since they have the same direction vector \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \). These lines can be interpreted as: \( L_1 : \mathbf{r} = \begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} + s \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \) and \( L_2 : \mathbf{r} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} + t \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \), where \( s \) and \( t \) are parameters.
02

Find a Vector Connecting the Lines

To find the perpendicular distance between the lines, find a vector connecting a point from \( L_1 \) to a point on \( L_2 \). Taking the points without parameter intervention gives: \( \mathbf{A} = \begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} \) from \( L_1 \) and \( \mathbf{B} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \) from \( L_2 \). Thus, the connecting vector is \( \mathbf{AB} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} - \begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} = \begin{bmatrix} -1 \ 1 \ 2 \end{bmatrix} \).
03

Projection of Connecting Vector onto Direction Vector

The parallel lines have the same direction vector \( \mathbf{d} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \). Find the projection of the vector \( \mathbf{AB} \) onto \( \mathbf{d} \) using the formula: \[ \text{proj}_{\mathbf{d}} \mathbf{AB} = \frac{\mathbf{AB} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} \] where \( \mathbf{AB} \cdot \mathbf{d} = -1 + 1 + 2 = 2 \) and \( \mathbf{d} \cdot \mathbf{d} = 3 \). Thus, \[ \text{proj}_{\mathbf{d}} \mathbf{AB} = \frac{2}{3} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \].
04

Find the Perpendicular Component

Subtract the projection from \( \mathbf{AB} \) to obtain the perpendicular component: \[ \mathbf{AB}_\perp = \mathbf{AB} - \text{proj}_{\mathbf{d}} \mathbf{AB} = \begin{bmatrix} -1 \ 1 \ 2 \end{bmatrix} - \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \] which simplifies to \[ \mathbf{AB}_\perp = \begin{bmatrix} -\frac{5}{3} \ \frac{1}{3} \ \frac{4}{3} \end{bmatrix} \].
05

Calculate the Distance

The distance between the two parallel lines is the magnitude of the perpendicular vector \( \mathbf{AB}_\perp \). Calculate this as: \[ |\mathbf{AB}_\perp| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{42}{9}} = \frac{\sqrt{42}}{3} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Equation
In the context of three-dimensional geometry, a vector equation represents a line. It is a mathematical expression that describes a line using direction vectors and points. For instance, the lines given in the exercise have vector equations:
  • Line 1: \[ \begin{bmatrix} \ x\ y\ z\end{bmatrix} = \begin{bmatrix} \ 1 \\ 0 \\ -1 \end{bmatrix} + s \begin{bmatrix} \ 1 \\ 1 \\ 1 \end{bmatrix} \]
  • Line 2: \[ \begin{bmatrix} \ x\ y\ z\end{bmatrix} = \begin{bmatrix} \ 0 \\ 1 \\ 1 \end{bmatrix} + t \begin{bmatrix} \ 1 \\ 1 \\ 1 \end{bmatrix} \]
Both lines share the same direction vector \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \), indicating that they are parallel. The vector equation can be broken down into a position vector, which is a specific point on the line, and a direction vector that determines the line鈥檚 path. Understanding vector equations is crucial for finding the characteristics of the line, such as determining the distance between parallel lines.
Perpendicular Distance
The perpendicular distance between two parallel lines in three-dimensional space is found by creating a connecting vector between specific points on each line. In this exercise, a vector \( \mathbf{AB} \) is constructed by choosing initial points from both given lines:
  • Point from Line 1: \( \begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} \)
  • Point from Line 2: \( \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \)
  • Connecting vector \( \mathbf{AB} = \begin{bmatrix} -1 \ 1 \ 2 \end{bmatrix} \)
To find the perpendicular distance, one must calculate the component of this connecting vector that is perpendicular to the line's direction vector. This perpendicular distance is specifically the length of the vector that is orthogonal to the direction vector, emphasizing the shortest path connecting the lines.
Projection
Projection is a fundamental concept in vector calculus and is used to measure how much of one vector lies in the direction of another vector. When determining the perpendicular distance between the parallel lines, projection helps in isolating the component of a vector along another vector.In this exercise:
  • The direction vector \( \mathbf{d} \) of the lines is \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \).
  • Find the projection of \( \mathbf{AB} \) onto \( \mathbf{d} \) using: \[\text{proj}_{\mathbf{d}} \mathbf{AB} = \frac{\mathbf{AB} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}\]
  • The calculation turns out to be \( \begin{bmatrix} \frac{2}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \).
Essentially, projection helps in breaking down the connecting vector \( \mathbf{AB} \) into parts that are parallel and perpendicular to the direction vector, pinpointing the perpendicular component required for distance calculation.
Magnitude of a Vector
The magnitude of a vector is a measure of its length. For a vector \( \mathbf{v} = \begin{bmatrix} a \ b \ c \end{bmatrix} \), the magnitude is calculated using the formula:\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]This represents how long the vector is and is essential in determining distances.In the context of parallel lines, after isolating the perpendicular component of the connecting vector \( \mathbf{AB} \) using the concept of projection, the next step is to compute its magnitude to find the distance:
  • Perpendicular vector: \( \mathbf{AB}_\perp = \begin{bmatrix} -\frac{5}{3} \ \frac{1}{3} \ \frac{4}{3} \end{bmatrix} \).
  • Magnitude computation: \[|\mathbf{AB}_\perp| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{42}{9}} = \frac{\sqrt{42}}{3}\]
Finally, the magnitude provides the physical distance between the two sets of parallel lines, effectively describing the shortest path directly linking them.

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Most popular questions from this chapter

Give the vector equation of the plane passing through \(P, Q,\) and \(R\) $$P=(1,1,1), Q=(4,0,2), R=(0,1,-1)$$

Solve the given equation or indicate that there is no solution. $$x+3=2 \text { in } \mathbb{Z}_{5}$$

Write the equation of the plane passing through \(P\) with direction vectors u and v in (a) vector form and (b) parametric form. $$P=(6,-4,-3), \mathbf{u}=\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right], \mathbf{v}=\left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right]$$

Perform the indicated calculations. $$[2,1,2] \cdot[2,2,1] \text { in } \mathbb{Z}_{3}^{3}$$

(a) For which values of \(a\) does \(a x=1\) have a solution in \(\mathbb{Z}_{5}\) ? (b) For which values of \(a\) does \(a x=1\) have a solution in \(\mathbb{Z}_{6}\) ? (c) For which values of \(a\) and \(m\) does \(a x=1\) have a solution in \(\mathbb{Z}_{m} ?\)
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