/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 Use projections to find a genera... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 90

Use projections to find a general formula for the (smallest) distance between the point \(\left.P\left(x_{0}, y_{0}\right) \text { and the line } a x+b y=c . \text { (See Exercises } 62-65 .\right)\)

Short Answer

Expert verified
Answer: The general formula for the smallest distance between point P and line ax + by = c is given by: $ d(P,Q) = \frac{|ax_0+by_0-c|}{\sqrt{a^{2}+b^{2}}} $

Step by step solution

01

Find the normal vector of the line

The normal vector of a line ax + by = c is given by the coefficients of x and y. So the normal vector is N = (a, b).
02

Find the unit normal vector of the line

To find the unit normal vector, we need to divide the normal vector by its magnitude. The magnitude of N is given by \(\sqrt{a^{2}+b^{2}}\). Therefore, the unit normal vector \(\hat{N}\) is given by: \(\hat{N}=\frac{1}{\sqrt{a^{2}+b^{2}}}(a, b)\)
03

Project the point P onto the line

To project the point P onto the line, we subtract its coordinates from the coordinates of the point (x0, y0) and dot the result with the unit normal vector. We will call the projection point Q: $ Q = \text{proj}_{\hat{N}}(P) = \left( x_0 - a\frac{\left\cdot\left}{a^2+b^2}, y_0 - b\frac{\left\cdot\left}{a^2+b^2}\right) $
04

Calculate the distance between point P and its projection Q

To find the distance between point P and its projection Q, we need to find the difference between their coordinates and take the magnitude of the resulting vector: $ d(P,Q) = \sqrt{\left(x_0 - a\frac{\left\cdot\left}{a^2+b^2}\right)^{2}+\left(y_0 - b\frac{\left\cdot\left}{a^2+b^2}\right)^{2}} $
05

Simplify the expression and obtain the general formula for the distance

We know that the required distance between point P and the line is given by d(P, Q). Now, let's simplify the expression: $ d(P,Q) = \frac{1}{\sqrt{a^{2}+b^{2}}} \sqrt{(ax_0+by_0-c)^{2}} $ Since we are calculating the smallest distance between the point and the line, we can drop the square root and apply the absolute value for the given expression. So the general formula to find the smallest distance between point P and line ax + by = c is: $ d(P,Q) = \frac{|ax_0+by_0-c|}{\sqrt{a^{2}+b^{2}}} $

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Projection
Projection in the context of geometry, particularly when discussing lines and points, essentially measures how one vector relates to a line as it "projects" or "drops down" onto it.
The projection of a point onto a line helps us determine the perpendicular distance to the line from that point.
  • To project a point onto a line means finding the closest point on that line.
  • This involves utilizing vectors and dot products for accuracy.
  • It describes how far along the line a particular point is when viewed directly from a perpendicular perspective.
Imagine shining a light directly above point P that casts a shadow straight onto the line鈥攖his shadow is the projection. Through calculations, notably by employing the normal vector, you'll locate this shadow point mathematically.
Normal Vector
A normal vector in terms of a line in 2D is a vector that is perpendicular to the line.
In the equation of the line given by \[ ax + by = c \], the normal vector is represented by the coefficients of \( a \) and \( b \).
  • For a line \( ax + by = c \), the normal vector is \( N = (a,b) \).
  • This vector points perpendicularly from the line, determining direction and slope characteristics of the line.
  • The normal vector is crucial as it becomes a baseline for more complex vector calculations like finding distances and projections.
Understanding how the coefficients shape this vector helps visualize how lines are oriented in plane geometry. Grasping the role of the normal vector is essential when using it to carry out projections.
Distance Formula
The Distance Formula is a mathematical equation used to compute the distance between two points in some coordinate space.
For our specific task of finding the distance between a point and a line, this formula is slightly adapted.
  • For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the Euclidean distance is given by \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].
  • In determining the distance from a point to a line, we utilize the formula: \[ \frac{|ax_0 + by_0 - c|}{\sqrt{a^2 + b^2}} \].
  • This results from projecting the point onto the line and calculating the magnitude of that point's vector perpendicular to the line.
By using this streamlined approach, the formula aids in efficiently calculating the shortest distance, crucial in optimizing many geometric problems.
Vector Magnitude
Vector magnitude is akin to the "length" or "size" of a vector.
It represents how long or large a vector is in a given space, irrespective of its direction.
  • The magnitude of a vector \( (a, b) \) is computed via \[ \sqrt{a^2 + b^2} \].
  • This forms the basis for computing a unit normal vector, which scales the normal vector to just a unit length.
  • When finding distances, this magnitude factors into normalizing vectors, leading to simpler calculations.
Comprehending vector magnitude is fundamental to grasping other vector operations. It not only quantifies vectors but also normalizes them鈥攎aking calculations neat and practical in different mathematical and physics applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A golfer launches a tee shot down a horizontal fairway and it follows a path given by \(\mathbf{r}(t)=\left\langle a t,(75-0.1 a) t,-5 t^{2}+80 t\right\rangle,\) where \(t \geq 0\) measures time in seconds and \(\mathbf{r}\) has units of feet. The \(y\) -axis points straight down the fairway and the z-axis points vertically upward. The parameter \(a\) is the slice factor that determines how much the shot deviates from a straight path down the fairway. a. With no slice \((a=0),\) sketch and describe the shot. How far does the ball travel horizontally (the distance between the point the ball leaves the ground and the point where it first strikes the ground)? b. With a slice \((a=0.2),\) sketch and describe the shot. How far does the ball travel horizontally? c. How far does the ball travel horizontally with \(a=2.5 ?\)

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

Consider the lines $$\begin{aligned} \mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \text { and } \\ \mathbf{R}(s) &=\langle 6+s, 10-2 s, 16-s\rangle. \end{aligned}$$ a. Determine whether the lines intersect (have a common point) and if so, find the coordinates of that point. b. If \(\mathbf{r}\) and \(\mathbf{R}\) describe the paths of two particles, do the particles collide? Assume \(t \geq 0\) and \(s \approx 0\) measure time in seconds, and that motion starts at \(s=t=0\).

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\). c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Show that \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.