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Chapter 1: Problem 35

In Exercises 35 and \(36,\) find the distance between the parallel lines $$\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]+s\left[\begin{array}{r} -2 \\ 3 \end{array}\right] \text { and }\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 5 \\ 4 \end{array}\right]+t\left[\begin{array}{r} -2 \\ 3 \end{array}\right]$$

Short Answer

Expert verified
The distance between the lines is \( \frac{18}{\sqrt{13}} \).

Step by step solution

01

Identify Direction Vector

Both lines are given in parametric form:\[ \begin{array}{l} x = 1 - 2s \ y = 1 + 3s \end{array} \] and \[ \begin{array}{l} x = 5 - 2t \ y = 4 + 3t \end{array} \]The direction vector for both lines is \((-2, 3)\).
02

Find the Normal Vector

To find the normal vector, we need a vector that is perpendicular to the direction vector. Since our direction vector is \((-2, 3)\), a perpendicular vector to this is \((3, 2)\).
03

Find a Point-Point Vector

Identify a point from each line (e.g., the constant part of each parametric equation). The points are \((1, 1)\) and \((5, 4)\). The vector connecting these points is found by \[ \begin{array}{l} 5 - 1 = 4 \ 4 - 1 = 3 \end{array} \]So, the vector is \((4, 3)\).
04

Compute the Distance Formula

To find the distance \(d\) between the lines, we use the formula for the distance between parallel lines:\[ d = \frac{|\vec{n} \cdot \vec{AB}|}{\| \vec{n} \|} \]Where \(\vec{n}\) is the normal vector \((3, 2)\) and \(\vec{AB}\) is the vector \((4, 3)\).
05

Calculate the Dot Product and Norm

Calculate the dot product \(\vec{n} \cdot \vec{AB}\):\[ 3 \cdot 4 + 2 \cdot 3 = 12 + 6 = 18 \]Calculate the norm of \(\vec{n}\):\[ \| \vec{n} \| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]
06

Determine the Distance

Substitute back into the distance formula:\[ d = \frac{|18|}{\sqrt{13}} = \frac{18}{\sqrt{13}} \]Thus, the distance between the lines is \( \frac{18}{\sqrt{13}} \).

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Key Concepts

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

Parametric Equations of Lines
Parametric equations provide a convenient way to describe a line. They represent the coordinates of points on the line as functions of a parameter. In our exercise, the lines are given as:\[ \begin{array}{l} x = 1 - 2s \ y = 1 + 3s \end{array} \] and \[ \begin{array}{l} x = 5 - 2t \ y = 4 + 3t \end{array} \] where \(s\) and \(t\) are parameters. This means for each value of the parameter, we get a coordinate pair that lies on the line. The part \((1, -2s + 3s)\) and \((5, -2t + 3t)\) helps trace the line starting from a specific point and extending in the direction given by the direction vector. This format is very useful for calculations like finding intersections or distances.
Direction Vector
The direction vector of a line indicates the direction in which the line extends. It's a key part of the parametric equation. In our case, the direction vector is given by \((-2, 3)\). This means the line moves two units left and three units up as the parameter increases.
  • The direction vector remains the same for both lines, confirming they are parallel.
  • It is found within the parametric form as the coefficients of \(s\) and \(t\).
This vector gives orientation to the line, indicating it鈥檚 not vertical or horizontal, but slanted at the angle determined by these components.
Normal Vector
A normal vector is a vector that is perpendicular to another vector, in this case, the direction vector. This is essential for many calculations, such as computing distances between parallel lines. For a direction vector of \((-2, 3)\), the normal vector we used is \((3, 2)\).
  • To find a normal vector, switch the components of the direction vector and change one sign.
  • This is because the dot product between the direction vector and the normal vector should be zero, indicating they are perpendicular.
A normal vector helps project points or vectors orthogonally, making them useful in geometry for finding shortest paths and projections.
Dot Product
The dot product is a way to multiply two vectors, resulting in a scalar. For vectors \(\vec{a} = (a_1, a_2)\) and \(\vec{b} = (b_1, b_2)\), the dot product is calculated as:\[ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 \]In our exercise, the dot product \(\vec{n} \cdot \vec{AB} = 18\) plays a crucial role in determining the distance between lines. The dot product helps in finding projections, determining angles, and checking orthogonality.
  • If the dot product is zero, the vectors are perpendicular.
  • The dot product is used in the distance formula to project vector lengths.
This operation combines vector magnitudes and directions which is useful in many applications like physics and engineering.
Vector Norm
The norm of a vector, often called the magnitude or length, is a measure of its size. For a vector \(\vec{v} = (v_1, v_2)\), the norm is calculated as:\[ \| \vec{v} \| = \sqrt{v_1^2 + v_2^2} \] For our normal vector \((3, 2)\), the norm is \(\sqrt{13}\).
  • The norm is always non-negative.
  • It gives the "length" of the vector in space.
This concept is especially important when you want to normalize vectors, meaning to scale them to unit length. In the context of distance between lines, the norm helps to standardize measurements, allowing us to properly compare distances in the plane.

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Most popular questions from this chapter

A parity check code vector v is given. Determine whether a single error could have occurred in the transmission of v. $$\mathbf{v}=[0,1,0,1,1,1]$$

Find the check digit d in the given Universal Product Code. $$[0,5,9,4,6,4,7,0,0,2,7, d]$$

Prove that \(\mathbf{u} \cdot c \mathbf{v}=c(\mathbf{u} \cdot \mathbf{v})\) for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) in \(\mathbb{R}^{n}\) and all scalars \(c\)

Describe all vectors \(\mathbf{v}=\left[\begin{array}{l}x \\\ y\end{array}\right]\) that are orthogonal to \(\mathbf{u}=\left[\begin{array}{l}3 \\ 1\end{array}\right]\)

Exercises \(47-48\) explore one approach to the problem of finding the projection of a vector onto a plane. As Figure 1.69 shows, if \(\mathscr{P}\) is a plane through the origin in \(\mathrm{R}^{3}\) with normal vector \(\mathbf{n},\) and \(\mathbf{v}\) is a vector in \(\mathbb{R}^{3},\) then \(\mathbf{p}=\operatorname{proj}_{9}(\mathbf{v})\) is \(a\) vector in \(\mathscr{P}\) such that \(\mathbf{v}-c \mathbf{n}=\mathbf{p}\) for some scalar \(c\) Using the fact that \(n\) is orthogonal to every vector in \(\mathscr{P}\) (and hence to \(\mathbf{p}\) ), solve for \(c\) and thereby find an expression for \(\mathbf{p}\) in terms of \(\mathbf{v}\) and \(\mathbf{n}\)
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