/*! 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 9 A line perpendicular to the \(x ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 9

A line perpendicular to the \(x y\) -plane at (2,3,0).

Short Answer

Expert verified
Line equation: \( x=2, y=3, z=t \).

Step by step solution

01

Understanding Perpendicularity to the xy-Plane

A line that is perpendicular to the \(xy\)-plane is a line that runs in the direction of the \(z\)-axis. This means that the line will have no component in the \(x\) or \(y\) directions.
02

Defining the Point of Intersection

The line of interest intersects the \(xy\)-plane at the given point \((2, 3, 0)\). This point serves as a starting point for the line, or the point where it pierces the \(xy\)-plane.
03

Writing the Equation of the Line

Since the line is perpendicular to the \(xy\)-plane and runs parallel to the \(z\)-axis, the equation of the line can be written as: \( (x, y, z) = (2, 3, 0) + t(0, 0, 1) \). Here, \(t\) is a parameter that defines the points on the line as \(t\) varies.
04

Simplifying the Line Equation

The equation above simplifies to: \( x=2, y=3, z=t \). This parameterization shows that as \(t\) changes, \(x\) and \(y\) remain constant at 2 and 3, respectively, while \(z\) can take any real value as \(t\) varies.

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.

xy-plane
The xy-plane is a fundamental element in the three-dimensional coordinate system. It consists of all points where the z-coordinate is zero, which means it extends infinitely along the x and y axes. In simpler terms, imagine the xy-plane as a flat sheet where height (z-coordinate) is irrelevant.

In geometry, understanding the xy-plane is crucial as it serves as a reference for positioning objects in space. It acts as a foundational plane from which heights and depths (z-values) are measured.
  • The xy-plane is perpendicular to the z-axis, creating a right angle with it.
  • It can be thought of as the ground floor or the bottom level in three-dimensional space.
  • All calculations regarding altitude or vertical positioning are based upon this plane.
Recognizing the role of the xy-plane is central to solving problems involving spatial geometry, such as the location of points and lines.
z-axis
The z-axis is an essential counterpart to the xy-plane in the three-dimensional coordinate system. It represents the axis that extends vertically from the xy-plane, providing the third dimension that is responsible for depth or height.

In problems involving perpendicularity, such as the exercise given, the z-axis often comes into play. A line that is perpendicular to the xy-plane is, in fact, aligned parallel to this axis.
  • The z-axis crosses the xy-plane at the origin point \(0, 0, 0\).
  • It provides a directionality for objects above or below the xy-plane.
  • Lines parallel to the z-axis only change in the z direction, keeping x and y constant.
Understanding the z-axis helps in visualizing how objects extend in space, particularly when dealing with heights and depths.
line equation
In the world of coordinate geometry, a line equation is a mathematical expression that shows how a line is aligned in space. It's a way to describe all the points on a line in relation to a starting point and direction.

For a line perpendicular to the xy-plane and parallel to the z-axis, like in our exercise, the line equation becomes straightforward since all changes occur in one dimension along the z-axis.
  • The general form of the line equation here is \( (x, y, z) = (2, 3, 0) + t(0, 0, 1) \).
  • It depicts that the line starts at the point \(2, 3, 0\) and extends infinitely in the direction of the z-axis.
  • With this parameterization, the x and y values remain constant, showing lack of movement in these directions.
Developing a line equation helps in specifying exactly how a line behaves in three-dimensional space, crucial for mapping and visualization.
parameterization
Parameterization is a method to represent a line (or curve) using a parameter, often denoted as \(t\). For lines in geometry, this allows us to express all points along a line based on a starting point and direction.

In the given problem, parameterization illustrates how the line behaves over various values of \(t\). Using the equation \(x=2, y=3, z=t\), the concept becomes real.
  • Here, \(t\) acts as a continuous variable, altering the z-coordinate while leaving x and y constant.
  • This method enables simple calculation of any point along the line by plugging in different \(t\) values.
  • Parameterization simplifies complex geometric concepts, making it easier to visualize changes along lines.
Understanding parameterization enriches your ability to describe and work with lines in spatial problems, emphasizing the dynamic nature of change across coordinates.

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) Let \(\mathbf{u}=\langle 2,1\rangle\) and \(\mathbf{v}=\langle 2,-1\rangle\) be nonzero vectors in \(R^{2} .\) With respect to the standard inner or dot product on \(R^{2},\) $$\mathbf{u} \cdot \mathbf{v}=\langle 2,1\rangle \cdot\langle 2,-1\rangle=2 \cdot 2+1 \cdot(-1)=3.$$ We see that \(\mathbf{u}\) and \(\mathbf{v}\) are not orthogonal with respect to that inner product. But using the inner product in Problem \(37,\) we have $$(\mathbf{u}, \mathbf{v})=2 \cdot 2+4(1) \cdot(-1)=0,$$ and so \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal with respect to that inner product. (b) Consider \(f(x)=\sin x\) and \(g(x)=\cos x\) in \(C[0,2 \pi] .\) Since $$\int_{0}^{2 \pi} \sin x \cos x d x=\frac{1}{2} \int_{0}^{2 \pi} \sin 2 x d x=-\left.\frac{1}{4} \cos 2 x\right|_{0} ^{2 \pi}=-\frac{1}{4}(1-1)=0,$$ these functions are orthogonal in \(C[0,2 \pi]\).

(a) \(\mathrm{With} \omega^{2}=g / l\) and \(K=k / m\) the system of differential equations is \\[ \begin{array}{l} \theta_{1}^{\prime \prime}+\omega^{2} \theta_{1}=-K\left(\theta_{1}-\theta_{2}\right) \\ \theta_{2}^{\prime \prime}+\omega^{2} \theta_{2}=K\left(\theta_{1}-\theta_{2}\right). \end{array} \\] Denoting the Laplace transform of \(\theta(t)\) by \(\Theta(s)\) we have that the Laplace transform of the system is \\[ \begin{array}{l} \left(s^{2}+\omega^{2}\right) \Theta_{1}(s)=-K \Theta_{1}(s)+K \Theta_{2}(s)+s \theta_{0} \\ \left(s^{2}+\omega^{2}\right) \Theta_{2}(s)=K \Theta_{1}(s)-K \Theta_{2}(s)+s \psi_{0}. \end{array} \\] If we add the two equations, we get \\[ \Theta_{1}(s)+\Theta_{2}(s)=\left(\theta_{0}+\psi_{0}\right) \frac{s}{s^{2}+\omega^{2}} \\] which implies \\[ \theta_{1}(t)+\theta_{2}(t)=\left(\theta_{0}+\psi_{0}\right) \cos \omega t. \\] Now solving \\[ \begin{array}{c} \left(s^{2}+\omega^{2}+K\right) \Theta_{1}(s)-K \Theta_{2}(s)=s \theta_{0} \\ -k \Theta_{1}(s)+\left(s^{2}+\omega^{2}+K\right) \Theta_{2}(s)=s \psi_{0} \end{array} \\] gives \\[ \left[\left(s^{2}+\omega^{2}+K\right)^{2}-K^{2}\right] \Theta_{1}(s)=s\left(s^{2}+\omega^{2}+K\right) \theta_{0}+K s \psi_{0}. \\] Factoring the difference of two squares and using partial fractions we get \\[ \Theta_{1}(s)=\frac{s\left(s^{2}+\omega^{2}+K\right) \theta_{0}+K s \psi_{0}}{\left(s^{2}+\omega^{2}\right)\left(s^{2}+\omega^{2}+2 K\right)}=\frac{\theta_{0}+\psi_{0}}{2} \frac{s}{s^{2}+\omega^{2}}+\frac{\theta_{0}-\psi_{0}}{2} \frac{s}{s^{2}+\omega^{2}+2 K}, \\] so \(\theta_{1}(t)=\frac{\theta_{0}+\psi_{0}}{2} \cos \omega t+\frac{\theta_{0}-\psi_{0}}{2} \cos \sqrt{\omega^{2}+2 K} t.\) Then from \(\theta_{2}(t)=-\theta_{1}(t)+\left(\theta_{0}+\psi_{0}\right)\) cos \(\omega t\) we get \\[ \theta_{2}(t)=\frac{\theta_{0}+\psi_{0}}{2} \cos \omega t-\frac{\theta_{0}-\psi_{0}}{2} \cos \sqrt{\omega^{2}+2 K} t. \\] (b) With the initial conditions \(\theta_{1}(0)=\theta_{0}, \theta_{1}^{\prime}(0)=0, \theta_{2}(0)=\theta_{0}, \theta_{2}^{\prime}(0)=0\) we have \\[ \theta_{1}(t)=\theta_{0} \cos \omega t, \quad \theta_{2}(t)=\theta_{0} \cos \omega t. \\] Physically this means that both pendulums swing in the same direction as if they were free since the spring exerts no influence on the motion \(\left(\theta_{1}(t) \text { and } \theta_{2}(t) \text { are free of } K\right)\). With the initial conditions \(\theta_{1}(0)=\theta_{0}, \theta_{1}^{\prime}(0)=0, \theta_{2}(0)=-\theta_{0}, \theta_{2}^{\prime}(0)=0\) we have \\[ \theta_{1}(t)=\theta_{0} \cos \sqrt{\omega^{2}+2 K} t, \quad \theta_{2}(t)=-\theta_{0} \cos \sqrt{\omega^{2}+2 K} t. \\] Physically this means that both pendulums swing in the opposite directions, stretching and compressing the spring. The amplitude of both displacements is \(\left|\theta_{0}\right| .\) Moreover, \(\theta_{1}(t)=\theta_{0}\) and \(\theta_{2}(t)=-\theta_{0}\) at precisely the same times. At these times the spring is stretched to its maximum.

The four points will be coplanar if the three vectors \(\overrightarrow{P_{1} P_{2}}=\langle 3,-1,-1\rangle, \overrightarrow{P_{2} P_{3}}=\langle-3,-5,13\rangle,\) and \(\overrightarrow{P_{3} P_{4}}=\) \langle-8,7,-6\rangle are coplanar. \(\overrightarrow{P_{2} P_{3}} \times \overrightarrow{P_{3} P_{4}}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -5 & 13 \\ -8 & 7 & -6\end{array}\right|=\left|\begin{array}{cc}-5 & 13 \\ 7 & -6\end{array}\right| \mathbf{i}-\left|\begin{array}{cc}-3 & 13 \\ -8 & -6\end{array}\right| \mathbf{j}+\left|\begin{array}{cc}-3 & -5 \\ -8 & 7\end{array}\right| \mathbf{k}=\langle-61,-122,-61\rangle\) \(\overrightarrow{P_{1} P_{2}} \cdot(\overrightarrow{P_{2} P_{3}} \times \overrightarrow{P_{3} P_{4}})=\langle 3,-1,-1\rangle \cdot\langle-61,-122,-61\rangle=-183+122+61=0\) The four points are coplanar.

(a) We have \(\mathbf{u}_{1}=\langle 5,7\rangle\) and \(\mathbf{u}_{2}=\langle 1,-2\rangle .\) Taking \(\mathbf{v}_{1}=\mathbf{u}_{1}=\langle 5,7\rangle,\) and using \(\mathbf{u}_{2} \cdot \mathbf{v}_{1}=-9\) and \(\mathbf{v}_{1} \cdot \mathbf{v}_{1}=74\) we obtain $$\mathbf{v}_{2}=\mathbf{u}_{2}-\frac{\mathbf{u}_{2} \cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}} \mathbf{v}_{1}=\langle 1,-2\rangle-\frac{9}{74}\langle 5,7\rangle=\left\langle\frac{119}{74},-\frac{85}{74}\right\rangle$$ Thus, an orthogonal basis is \(\left\\{\langle 5,7\rangle,\left\langle\frac{119}{74},-\frac{85}{74}\right\rangle\right\\}\) and an orthonormal basis is \(\left\\{\mathbf{w}_{1}^{\prime}, \mathbf{w}_{2}^{\prime}\right\\},\) where $$\mathbf{w}_{1}^{\prime}=\frac{1}{\|\langle 5,7\rangle\|}\langle 5,7\rangle=\frac{1}{\sqrt{74}}\langle 5,7\rangle=\left\langle\frac{5}{\sqrt{74}}, \frac{7}{\sqrt{74}}\right\rangle$$ and $$\mathbf{w}_{2}^{\prime}=\frac{1}{\left\|\left\langle\frac{119}{74},-\frac{85}{74}\right\rangle\right\|}\left\langle\frac{119}{74},-\frac{85}{74}\right\rangle=\frac{1}{17 / \sqrt{74}}\left\langle\frac{119}{74},-\frac{85}{74}\right\rangle=\left\langle\frac{7}{\sqrt{74}},-\frac{5}{\sqrt{74}}\right\rangle$$ (b) We have \(\mathbf{u}_{1}=\langle 5,7\rangle\) and \(\mathbf{u}_{2}=\langle 1,-2\rangle .\) Taking \(\mathbf{v}_{1}=\mathbf{u}_{2}=\langle 1,-2\rangle,\) and using \(\mathbf{u}_{1} \cdot \mathbf{v}_{1}=-9\) and \(\mathbf{v}_{1} \cdot \mathbf{v}_{1}=5\) we obtain $$\mathbf{v}_{2}=\mathbf{u}_{1}-\frac{\mathbf{u}_{1} \cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}} \mathbf{v}_{1}=\langle 5,7\rangle-\frac{9}{5}\langle 1,-2\rangle=\left\langle\frac{34}{5}, \frac{17}{5}\right\rangle$$ Thus, an orthogonal basis is \(\left\\{\langle 1,-2\rangle,\left\langle\frac{34}{5}, \frac{17}{5}\right\rangle\right\\}\) and an orthonormal basis is \(\left\\{\mathbf{w}_{3}^{\prime \prime}, \mathbf{w}_{4}^{\prime \prime}\right\\},\) where $$\mathbf{w}_{3}^{\prime \prime}=\frac{1}{\|\langle 1,-2\rangle\|}\langle 1,-2\rangle=\frac{1}{\sqrt{5}}\langle 1,-2\rangle=\left\langle\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right\rangle$$ and $$\mathbf{w}_{4}^{\prime \prime}=\frac{1}{\left\|\left\langle\frac{34}{5}, \frac{17}{5}\right\rangle\right\|}\left\langle\frac{34}{5}, \frac{17}{5}\right\rangle=\frac{1}{17 / \sqrt{5}}\left\langle\frac{34}{5}, \frac{17}{5}\right\rangle=\left\langle\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right\rangle$$

$$6(x-1 / 2)+8(y-3 / 4)-4(z+1 / 2)=0 ; \quad 6 x+8 y-4 z=11$$.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Work Energy and Power

Read Explanation

Particle Model of Matter

Read Explanation

Fundamentals of Physics

Read Explanation

Classical Mechanics

Read Explanation

Solid State Physics

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.