/*! 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 89 Write an equation whose graph co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 89

Write an equation whose graph consists of the set of points \(P(x, y, z)\) that are twice as far from \(A(0,-1,1)\) as from \(B(1,2,0)\)

Short Answer

Expert verified
The equation of the set of points \(P(x, y, z)\) that are twice as far from \(A(0, -1, 1)\) as they are from \(B(1, 2, 0)\) is \(3x^2 - 8x + 3y^2 + 4y + 3z^2 - 2z = 0\).

Step by step solution

01

Define the Points

Define \(P(x, y, z)\) as the point of interest, \(A(0, -1, 1)\) as point A, and \(B(1, 2, 0)\) as point B.
02

Formulate the Equation

According to the problem statement, the distance from P to A is twice the distance from P to B. Considering the Euclidean distance formula in 3D (i.e., for any two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), their distance d is given by \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}\), we can formulate an equation according to the problem statement: \((x - 0)^2 + (y + 1)^2 + (z - 1)^2 = 2^2((x-1)^2+(y-2)^2+(z-0)^2)\)
03

Simplify the Equation

Simplify further to remove the square root and the brackets, leaving the expression in terms of squared distances: \(x^2 + (y + 1)^2 + (z - 1)^2 = 4 * ((x - 1)^2 + (y - 2)^2 + z^2)\)
04

Expand and Rearrange the Equation

Expand out the terms and consolidate like terms to obtain the final form of the quadratic surface equation: \(3x^2 - 8x + 3y^2 + 4y + 3z^2 - 2z = 0\)

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!

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

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \mathbf{u}=\langle 2,-3,4\rangle, \quad \mathbf{v}=\langle 0,6,5\rangle $$

(a) find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), and (b) find the vector component of u orthogonal to v. $$ \mathbf{u}=\langle 2,1,2\rangle, \quad \mathbf{v}=\langle 0,3,4\rangle $$

Find the component of \(u\) that is orthogonal to \(\mathbf{v},\) given \(\mathbf{w}_{\mathbf{1}}=\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). $$ \mathbf{u}=\langle 8,2,0\rangle, \quad \mathbf{v}=\langle 2,1,-1\rangle, \quad \operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 6,3,-3\rangle $$

Determine whether \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal parallel, or neither. $$ \begin{array}{l} \mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}-\mathbf{k} \\ \mathbf{v}=2 \mathbf{i}+\mathbf{j}-\mathbf{k} \end{array} $$

Determine which of the vectors is (are) parallel to \(\mathrm{z}\). Use a graphing utility to confirm your results. \(\mathbf{z}=\frac{1}{2} \mathbf{i}-\frac{2}{3} \mathbf{j}+\frac{3}{4} \mathbf{k}\) (a) \(6 \mathbf{i}-4 \mathbf{j}+9 \mathbf{k}\) (b) \(-\mathbf{i}+\frac{4}{3} \mathbf{j}-\frac{3}{2} \mathbf{k}\) (c) \(12 \mathbf{i}+9 \mathbf{k}\) (d) \(\frac{3}{4} \mathbf{i}-\mathbf{j}+\frac{9}{8} \mathbf{k}\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.