Chapter 9.2, Problem 53E

a.

To determine

Show that the points $Q\left(0,2\right)$ and $R\left(2,1\right)$ lies on the line.

Answer to Problem 53E

The points $Q\left(0,2\right)$ and $R\left(2,1\right)$ lies on the line.

Explanation of Solution

Given information:

Let $L$ be the line $2x+4y-8$ and the point $P\left(3,4\right)$ .

Given figure,

  

Calculation:

We have to verify that $Q\left(0,2\right)$ and $R\left(2,1\right)$ lie on $L$ .

Consider the firuge,

  

Now substitute $Q\left(0,2\right)$ in equation $2x+4y-8$ we get,

  $\begin{array}{l}2x+4y=8\\ 2\left(0\right)+4\left(2\right)=8\\ 8=8\\ 0\end{array}$

  $Q\left(0,2\right)$ Satisfy the equation of lin. Hence it lies on line $\text{L}$ .

Now substitute $R\left(2,1\right)$ in equation $2x+4y-8$ we get,

  $\begin{array}{l}2x+4y=8\\ 2\left(2\right)+4\left(1\right)=8\\ 8=8\end{array}$

  $R\left(2,1\right)$ Satisfy the equation of lin. Hence it lies on line $\text{L}$ .

Hence, the points $Q\left(0,2\right)$ and $R\left(2,1\right)$ lie on line $L$ .

b.

To determine

Find $w=pro{j}_{v}u$ .

Answer to Problem 53E

  $w={\text{proj}}_{v}u=\langle 1.6,-0.8\rangle$

Explanation of Solution

Given information:

Let $L$ be the line $2x+4y-8$ and let $P$ be the point $\left(3,4\right)$ .

  

  $u=\overrightarrow{QP}$ and $v=\overrightarrow{QR}$

Calculation:

We have to given that $u=\overrightarrow{QP}$ and $v=\overrightarrow{QR}$

Consider the firuge,

  

Projection of vector $u$ on the $\text{v}$ is given as,

  ${\text{proj}}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$

Vector $\text{v}$ represented in plane with initial and final points, $P\left(x_1,y_1\right),Q\left(x_2,y_2\right)$ then,

  $\begin{array}{l}v=\langle x_2-x_1,y_2-y_1\rangle \\ we,have,u,as\\ u=\langle 3-0,4-2\rangle \\ =\langle 3,2\rangle \\ v=\langle 2-0,1-2\rangle \\ =\langle 2,-1\rangle \end{array}$

Now dot product of vector $u$ and $\text{v}$ will be,

  $\begin{array}{l}u\cdot v=a_1{a}_2+b_1{b}_2\\ \text{put the calculated values,}\\ u\cdot v=3\cdot 2+2\cdot \left(-1\right)\\ =4\\ now,magnitude,\left|u\right|=\langle a_1,b_1\rangle is,\\ \left|u\right|=\sqrt{a_1{}^2+b_1{}^2}\\ similarly.\\ \left|v\right|=\sqrt{4+1}\\ =\sqrt{5}\end{array}$

Now substitute the calculated values in equation, ${\text{proj}}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$

  $\begin{array}{l}{\text{proj}}_{v}u=\left(\frac{4}{{\left(\sqrt{5}\right)}^2}\right)\left(2,-1\right)\\ =\left(\frac{4}{5}\right)\left(2,-1\right)\\ =0.8\left(2,-1\right)\\ {\text{proj}}_{v}u=1.6,-0.8\\ w={\text{proj}}_{v}u=\langle 1.6,-0.8\rangle \end{array}$

Hence, $w={\text{proj}}_{v}u=\langle 1.6,-0.8\rangle$

c.

To determine

Find the distance.

Answer to Problem 53E

  $3.13units$

Explanation of Solution

Given information:

Let $L$ be the line $2x+4y-8$ and let $P$ be the point $\left(3,4\right)$ .

  

Sketch a graph that explains why $\left|u-w\right|$ is the distance from $P$ to $L$ .

Calculation:

Consider projection vector $u$ on the $\text{v}$ as $u_1$

  $u_1={\text{proj}}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$

  $u$ can be resolve into two components as $u_1,u_2$

  $u_2=u-{\text{proj}}_{v}u$ , $u_2$ is perpendicular to $\text{v}$ and is component of $u$ .

We have calculated $w={\text{proj}}_{v}u=\langle 1.6,-0.8\rangle$

  

Vector $u_2=u-w$ is orthogonal to vector $v$

The distance $u-w$ gives the distance the point $P$ and line $L$ .

  $\begin{array}{l}u=\langle 3,2\rangle ,w=\langle 1.6,-0.8\rangle \\ \left|u-w\right|=\langle 3,2\rangle -\langle 1.6,-0.8\rangle \\ =\langle 1.4,2.8\rangle \\ \left|u-w\right|=\sqrt{{\left(1.4\right)}^2+{\left(2.8\right)}^2}\\ =3.13\end{array}$

The distance the point $P$ and line $L$ is $3.13units$ .

d.

To determine

Write a short paragraph.

Answer to Problem 53E

The magnitude of this component represents the distance between the point and the line.

Explanation of Solution

Given information:

Let $L$ be the line $2x+4y-8$ and let $P$ be the point $\left(3,4\right)$ .

  

Write a short paragraph describing the steps to measure the distance from a given point to a given line.

Calculation:

The distance from a given point to a given line.

  

The distance between a line and point is the perpendicular distance between them. A vector $v$ along the line $L$ and vector $u$ from the line to point $P_1$ . Then resolve vector $u$ in to two components $u_{1}and{u}_2$ such that $u_1$ is parallel to vector $v$ and $u_2$ is orthogonal to $v$ .

The component $u_1$ is same as the projection of vector $u$ on vector $v$ is given as,

  $u_1={\text{proj}}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)$

The orthogonal component $v$ is given as

  $u_2=u-u_1$ .

This component is perpendicular to $v$ and is a component of $u$ .

Hence, the magnitude of this component represent the distance between the point and the line.