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Expert-verified Found in: Page 550 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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 $$1$$.

See the step by step solution

## 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 $$1$$. ### Want to see more solutions like these? 