/*! 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 86 Let \(\mathbf{r}=\langle x, y, z... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(\mathbf{r}=\langle x, y, z\rangle\) and \(\mathbf{r}_{0}=\langle 1,1,1\rangle .\) Describe the set of all points \((x, y, z)\) such that \(\left\|\mathbf{r}-\mathbf{r}_{0}\right\|=2\)

Short Answer

Expert verified
The set of all points \((x, y, z)\) such that \(\|\mathbf{r}-\mathbf{r}_{0}\|=2\) is described by the equation \((x-1)^2 + (y-1)^2 + (z-1)^2 = 4\), which is a sphere of radius 2 centered at the point \((1,1,1)\).

Step by step solution

01

Find the vector difference

Find the vector \(\mathbf{r}-\mathbf{r}_{0}\) by subtracting the coordinates of \(\mathbf{r}_{0}\) from \(\mathbf{r}\). This results in the vector \(\mathbf{r}-\mathbf{r}_{0} = \langle x-1, y-1, z-1 \rangle\).
02

Calculate the magnitude

Calculate the magnitude (length) of the vector from step 1 using the formula \(\sqrt{(x-1)^2 + (y-1)^2 + (z-1)^2}\). This is equivalent to finding the distance from the point \((x,y,z)\) to the point \((1,1,1)\).
03

Set the magnitude equal to the given value

Set the magnitude calculated in step 2 equal to 2 (the given length of the vector from \((1,1,1)\) to \((x,y,z)\)) to obtain the equation \((x-1)^2 + (y-1)^2 + (z-1)^2 = 2^2\).
04

Simplify the equation

Simplify the equation found in step 3 to its final form. Thus, we have the equation of a sphere of radius 2 centered at the point \((1,1,1)\) as \((x-1)^2 + (y-1)^2 + (z-1)^2 = 4\).

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 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.