/*! 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 30 Write an equation of the indicat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 30

Write an equation of the indicated plane. Through \(P(5,1,4)\) and parallel to the plane with equation \(x+y-2 z=0\)

Short Answer

Expert verified
The equation is \(x + y - 2z + 2 = 0\).

Step by step solution

01

Identify the given plane equation

The given plane equation is \(x + y - 2z = 0\). This plane has a normal vector \(\mathbf{n} = \langle 1, 1, -2 \rangle\). This normal vector will also be the normal vector for the plane we want to find, since the planes are parallel.
02

Use the point-normal form of a plane equation

The equation of a plane can be expressed as \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\) where \((x_0, y_0, z_0)\) is a point on the plane and \(\langle a, b, c \rangle\) is the normal vector. In this problem, the point \((x_0, y_0, z_0)\) is \(P(5, 1, 4)\) and the normal vector is \(\langle 1, 1, -2 \rangle\).
03

Substitute the point and normal vector into the equation

By substituting the given point and normal vector into the point-normal form, we get:\[1(x - 5) + 1(y - 1) - 2(z - 4) = 0\] which simplifies to:\[1x - 5 + 1y - 1 - 2z + 8 = 0\] Combine the constants to get:\[x + y - 2z + 2 = 0\].

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.

Normal Vector
A normal vector is a key component when discussing planes in 3D space. Simply put, it is a vector that is perpendicular to the plane.
When you have a plane given by the equation, for example, \(x + y - 2z = 0\), the coefficients of \(x\), \(y\), and \(z\) define its normal vector: \(\mathbf{n} = \langle 1, 1, -2 \rangle\).
This means that this vector points directly away from the surface of the plane, representing its orientation.
  • Understanding the normal vector helps in identifying the orientation of the plane.
  • The length of the normal vector does not affect the plane's geometry. It purely dictates the direction.
  • Knowing it is crucial for seeing whether planes are parallel or perpendicular.
Point-Normal Form
The point-normal form of a plane provides a handy formula to find a plane's equation.
It uses a specific point ewline \( (x_0, y_0, z_0) \) on the plane and the plane's normal vector \( \langle a, b, c \rangle \).
The general form of this equation is:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]To apply it:
  • Identify your point on the plane.
  • Use the corresponding normal vector.
  • Blend these into the point-normal formula.
This method turns these components into a neat equation that describes the plane.
For students, mastering this form makes it easier to derive planes' equations from the minimal information provided, like a point and a direction.
Parallel Planes
Parallel planes share a fascinating relationship: they're identical in orientation but positioned differently in 3D space.
A key aspect of this relationship is that parallel planes have identical normal vectors.
If a plane is represented by \(x + y - 2z = 0\), then a parallel plane will also have a normal vector \(\langle 1, 1, -2 \rangle\).
This signifies that while they do not intersect, the direction in which they face remains unchanged.
  • Parallelism in planes is dictated by identical normal vectors.
  • The distance between parallel planes can vary, defined by shifting along the direction of the normal vector.
  • This information is vital when finding a new plane parallel to an existing one but passing through a different point.
Grasping parallel planes enables an effortless transition between known and derived plane scenarios.
Plane Equation Derivation
Deriving the equation of a plane through a known point, with direction given by a normal vector, is a straightforward process.
Consider a scenario: finding a plane parallel to \(x + y - 2z = 0\) passing through the point \((5, 1, 4)\).
You begin by confirming the normal vector: \(\langle 1, 1, -2 \rangle\).Next:
  • Insert the point and the normal vector into the point-normal form equation:
\[1(x - 5) + 1(y - 1) - 2(z - 4) = 0\]This simplifies after arithmetic operations:
Combine terms to finalize the plane's equation:
\[x + y - 2z + 2 = 0\]This solution hinges on understanding each piece's role, from the normal vector's orientation to incorporating the point's position.

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

Determine whether the graph of the given equation is a paraboloid or a hyperboloid. Check your answer graphically if you have access to a computer algebra system with a "contour plotting" facility. \(3 x^{2}+2 y^{2}+5 z^{2}-2 x y-4 x z-2 y z=20\)

Describe and sketch the surface or solid described by the given equations and/or inequalities. \(0 \leqq \rho \leqq 2,0 \leqq \phi \leqq \pi / 2\)

Describe the graph of the given equation. (It is understood that equations including \(r\) are in cylindrical coordinates and those including \(\rho\) or \(\phi\) are in spherical coordinates.) \(\rho=5\)

Give the equation of a curve in one of the coordinate planes. Write an equation for the surface generated by revolving this curve around the indicated axis. Then sketch the surface. The line \(z=2 x ;\) the \(x\) -axis

The hyperbola \(y^{2}-z^{2}=1, x=0\) is rotated around the \(z\) -axis. Write a cylindrical equation for the surface thereby generated.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

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.