Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

To d\({\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)escribe all set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\).

All set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\) is described and it forms a sphere with radius \(1\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of Subtraction of Vector

Subtraction of vectors:

Consider the two three-dimensional vectors such as \(a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \) and .

The vector subtraction of two vectors \((a - b)\) is,

\(\begin{aligned}{l}(a - b) &= \left\langle {{a_1},{a_2},{a_3}} \right\rangle - \left\langle {{b_1},{b_2},{b_3}} \right\rangle \\ &= \left\langle {{a_1} - {b_1},{a_2} - {b_2},{a_3} - {b_3}} \right\rangle \end{aligned}\)

Given:

Two three-dimensional vectors \({\bf{r}} = \langle x,y,z\rangle \) and \({{\bf{r}}_0} = \left\langle {{x_0},{y_0},{z_0}} \right\rangle \).

Step 2: Calculate Subtraction of Vector

Consider the expression for magnitude of vector \(a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle (|{\rm{a}}|)\).

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} .......(1)\)

Write the expression for sphere with center \(C(h,k,l)\) and radius \(r\).

\({(x - h)^2} + {(y - k)^2} + {(z - l)^2} = {r^2}.......(2)\)

Write the expression for relation between vectors \(r\) and \({{\bf{r}}_0}\).

\(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\)

From definition, substitute \(\langle x,y,z\rangle \) for \(r\) and \(\left\langle {{x_0},{y_0},{z_0}} \right\rangle \) for \({{\bf{r}}_0}\),\(\begin{aligned}{l}\left| {\langle x,y,z\rangle - \left\langle {{x_0},{y_0},{z_0}} \right\rangle } \right| = 1\\\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right| = 1.......(3)\end{aligned}\)

Find the value of \(\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right|\) by using equation (1).

\(\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right| = \sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} \)

Substitute \(\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} \) for \(\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right|\) in equation (3), \(\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} = 1\)

Take square on both sides of equation.

\(\begin{aligned}{l}{\left( {\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} } \right)^2} &= {1^2}\\{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} + {\left( {z - {z_0}} \right)^2} &= {1^2}.......(4)\end{aligned}\)

Compare the equations (2) and (4), the equation (4) represents a sphere with radius \(1\).

Thus, all set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\) is described and it forms a sphere with radius \(1\).

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Essential Calculus: Early Transcendentals

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Short Answer

To d\({\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)escribe all set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\).

Step by Step Solution

Step 1: Concept of Subtraction of Vector

Step 2: Calculate Subtraction of Vector

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

To d\({\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)escribe all set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\).

Step by Step Solution

Step 1: Concept of Subtraction of Vector

Step 2: Calculate Subtraction of Vector

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