/*! 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 53 A normal vector to the \(x y\) -... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 53

A normal vector to the \(x y\) -plane is \(\langle 0,0,1\rangle .\) The equation of the parallel plane is \(z-12=0\) or \(z=12\).

Short Answer

Expert verified
The plane equation is \(z=12\) with a normal vector \(\langle 0,0,1 \rangle\), parallel to the \(xy\)-plane.

Step by step solution

01

Identify the Plane Equation

The problem states that the equation of the plane is \(z - 12 = 0\). We can rearrange this equation to \(z = 12\), which describes a plane parallel to the \(xy\)-plane at a height of 12 units along the \(z\)-axis.
02

Determine if Parallel to xy-plane

A plane parallel to the \(xy\)-plane generally has an equation of the form \(z = c\) where \(c\) is a constant. Since our given plane is \(z = 12\), it is indeed parallel to the \(xy\)-plane.
03

Confirm Normal Vector Consistency

A normal vector to a plane \(ax + by + cz = d\) is \(\langle a, b, c \rangle\). For the plane \(z = 12\), the equation can be rewritten as \(0x + 0y + 1z = 12\), so the normal vector is \(\langle 0, 0, 1 \rangle\), confirming it is consistent with being parallel to the \(xy\)-plane.

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.

Parallel Planes
Parallel planes refer to planes that never meet or intersect, no matter how far they extend infinitely. They maintain a constant distance apart. Understanding parallel planes can be easier if you think about sheets of paper stacked neatly, where each sheet is parallel to the one above or below it.
This concept is crucial when dealing with the plane equation
  • If two planes are parallel, their normal vectors are multiples of each other.
  • For example, the normal vector \(\langle 0, 0, 1 \rangle\) means all planes described as \(z = c\) will be parallel to the \(xy\)-plane, since their normal vectors are identical.
A specific example is \(z = 12\). No matter the value of \(c\), all planes with form \(z = c\) maintain the same momentum parallel to the \(xy\)-plane.
xy-plane
The \(xy\)-plane is a fundamental concept used in three-dimensional Cartesian coordinates. It's essentially a flat surface defined by two axes, the \(x\)-axis and \(y\)-axis.
  • The \(xy\)-plane acts as a baseline or reference, located at \(z = 0\).
  • In this plane, there is no height, which means it stretches out like a vast, endless flat sheet.
Crucially, any plane defined by \(z = c\) where \(c eq 0\) is parallel to the \(xy\)-plane. For example, \(z = 12\) implies a plane parallel to this reference at a 12-unit distance along the \(z\)-axis.
Equation of a Plane
An equation of a plane in three-dimensional space is typically given in the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the normal vector components, and \(d\) is a constant. This equation summarizes all the points \((x, y, z)\) that lie on the plane.
  • The coefficients \(a\), \(b\), and \(c\) form the "normal vector," key in determining the plane's orientation.
  • When \(a\) and \(b\) are both zero, the equation simplifies to \(z = c\), defining a horizontal plane parallel to the \(xy\)-plane.
In the given problem, the equation \(z = 12\) directly tells us it's parallel to the \(xy\)-plane and lies at \(z = 12\) from it. The normal vector \(\langle 0, 0, 1 \rangle\) confirms this orientation, pointing directly along the \(z\)-axis.

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

The integral equation is \\[10 i+2 \int_{0}^{t} i(\tau) d \tau=2 t^{2}+2 t\\]. Taking the Laplace transform we obtain \\[\mathscr{L}\\{i\\}=\left(\frac{4}{s^{3}}+\frac{2}{s^{2}}\right) \frac{s}{10 s+2}=\frac{s+2}{s^{2}(5 s+2)}=-\frac{9}{s}+\frac{2}{s^{2}}+\frac{45}{5 s+1}=-\frac{9}{s}+\frac{2}{s^{2}}+\frac{9}{s+1 / 5}\\]. Thus \\[i(t)=-9+2 t+9 e^{-t / 5}\\].

$$\text { parametric: } x=12 t, y=-3-5 t, \quad z=10-6 t ; \text { symmetric: } \frac{x}{12}=\frac{y+3}{-5}=\frac{z-10}{-6}$$

In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both planes, and hence a direction vector for a line parallel to the two planes. Normal vectors are \langle 1,1,-4\rangle and \(\langle 2,-1,1\rangle .\) A direction vector is \\[\langle 1,1,-4\rangle \times\langle 2,-1,1\rangle=\langle-3,-9,-3\rangle=-3\langle 1,3,1\rangle\\] Equations of the line are \(x=5+t, y=6+3 t, z=-12+t\).

Writing the given line in the form \(x / 2=(y-1) /(-3)=(z-5) / 6,\) we see that a direction vector is \langle 2,-3,6\rangle Parametric equations for the line are \(x=6+2 t, y=4-3 t, z=-2+6 t\).

$$\begin{aligned} &\text { (a) Let } A=(1,3,0), B=(2,0,0), C=(0,0,4), \text { and } D=(1,-3,4) . \text { Then } \overrightarrow{A B}=\mathbf{i}-3 \mathbf{j}, \overrightarrow{A C}=-\mathbf{i}-3 \mathbf{j}+4 \mathbf{k}\\\ &\overrightarrow{C D}=\mathbf{i}-3 \mathbf{j}, \text { and } \overrightarrow{B D}=-\mathbf{i}-3 \mathbf{j}+4 \mathbf{k} . \text { since } \overrightarrow{A B}=\overrightarrow{C D} \text { and } \overrightarrow{A C}=\overrightarrow{B D}, \text { the quadrilateral is a parallelogram. } \end{aligned}$$ (b) Computing \\[ \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 0 \\ -1 & -3 & 4 \end{array}\right|=-12 \mathbf{i}-4 \mathbf{j}-6 \mathbf{k} \\] $$\text { we find that the area is }\|-12 \mathbf{i}-4 \mathbf{j}-6 \mathbf{k}\|=\sqrt{144+16+36}=14$$
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Fields in Physics

Read Explanation

Thermodynamics

Read Explanation

Modern Physics

Read Explanation

Energy Physics

Read Explanation

Physics of Motion

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.