/*! 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 43 If \(\mathbf{r}=[x, y, z], \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 43

If \(\mathbf{r}=[x, y, z], \mathbf{a}=\left[a_{1}, a_{2}, a_{3}\right],\) and \(\mathbf{b}=\left[b_{1}, b_{2}, b_{3}\right],\) show that the vector equation \((\mathbf{r}-\) a) \(\cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, and find its center and radius.

Short Answer

Expert verified
The center is \(\left(\frac{a_1 + b_1}{2}, \frac{a_2 + b_2}{2}, \frac{a_3 + b_3}{2}\right)\) and the radius is \(\frac{1}{2}\sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2 + (b_3 - a_3)^2}\)."

Step by step solution

01

Expand the Dot Product

Start by writing down the dot product \((\mathbf{r} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{b})\). This can be expanded as:\[(x - a_1)(x - b_1) + (y - a_2)(y - b_2) + (z - a_3)(z - b_3)\]
02

Distribute Each Term

Distribute each of the terms to simplify the equation:For the x-coordinates: \[(x - a_1)(x - b_1) = x^2 - (a_1 + b_1)x + a_1 b_1\]For the y-coordinates:\[(y - a_2)(y - b_2) = y^2 - (a_2 + b_2)y + a_2 b_2\]For the z-coordinates: \[(z - a_3)(z - b_3) = z^2 - (a_3 + b_3)z + a_3 b_3\]
03

Combine and Simplify the Equation

Combine all terms obtained:\[x^2 - (a_1 + b_1)x + a_1 b_1 + y^2 - (a_2 + b_2)y + a_2 b_2 + z^2 - (a_3 + b_3)z + a_3 b_3 = 0\]Re-arrange this equation:\[x^2 + y^2 + z^2 -(a_1 + b_1)x -(a_2 + b_2)y -(a_3 + b_3)z + (a_1 b_1 + a_2 b_2 + a_3 b_3) = 0\]
04

Recognize the Equation of a Sphere

The equation can be rewritten into the standard form of a sphere by completing the square for each variable term. The simplified equivalent sphere equation is:\[(x - \frac{a_1 + b_1}{2})^2 + (y - \frac{a_2 + b_2}{2})^2 + (z - \frac{a_3 + b_3}{2})^2 = R^2\]
05

Determine Sphere Center and Radius

The center of the sphere is given by the point:\[\left(\frac{a_1 + b_1}{2}, \frac{a_2 + b_2}{2}, \frac{a_3 + b_3}{2}\right)\]To find the radius \(R\), calculate the square root of the previously rearranged equation.Since constant terms canceled, substitute known values to solve for \(R^2\). The general form reveals:\[R = \frac{1}{2}\sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2 + (b_3 - a_3)^2}\]

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.

Vector Equations
Vector equations are a concise way to describe geometric entities and their relationships using vectors. They involve expressions where each term is a vector. In the exercise, we have the equation
  • (\(\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\), which is a vector equation describing a geometric condition involving a sphere.
In such expressions, vectors like \(\mathbf{r}\), \(\mathbf{a}\), and \(\mathbf{b}\) encapsulate information about coordinates in a three-dimensional space:
  • \(\mathbf{r} = [x, y, z]\) represents a point in space.
  • \(\mathbf{a}\) and \(\mathbf{b}\) are fixed position vectors.
The equation can be related to a sphere, as shown by expanding and simplifying it, revealing its center and radius. This showcases how vector equations offer a powerful means to describe spheres, lines, planes, and more in vector calculus.
Dot Product
The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (e.g., vectors) and returns a single number. It is calculated as:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]In this equation, the components of two vectors are multiplied pairwise, and their sums are taken:
  • The dot product is useful for finding angles between vectors and projecting one vector onto another.
  • If \(\mathbf{a} \cdot \mathbf{b} = 0\), the vectors are orthogonal.
In our specific exercise, the dot product is used to express a geometric condition. When the dot product \((\mathbf{r}-\mathbf{a}) \cdot (\mathbf{r}-\mathbf{b}) = 0\), it indicates vectors \((\mathbf{r}-\mathbf{a})\) and \((\mathbf{r}-\mathbf{b})\) are perpendicular. This helps us connect the vector equation to the geometry of a sphere.
Equation of a Sphere
An equation of a sphere in its standard form is given by:\[(x - h)^2 + (y - k)^2 + (z - l)^2 = R^2\]Here, \((h, k, l)\) is the center of the sphere and \(R\) is its radius.To derive this form from our vector equation, we:
  • Expand and combine the terms from the dot product \((\mathbf{r}-\mathbf{a}) \cdot (\mathbf{r}-\mathbf{b}) = 0\).
  • Rearrange to recognize the terms \((x - \frac{a_1 + b_1}{2})^2 + (y - \frac{a_2 + b_2}{2})^2 + (z - \frac{a_3 + b_3}{2})^2\).
  • Identify the center as \((\frac{a_1 + b_1}{2}, \frac{a_2 + b_2}{2}, \frac{a_3 + b_3}{2})\) and derive the radius \(R = \frac{1}{2}\sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2 + (b_3 - a_3)^2}\).
These steps transform a general quadratic equation into the familiar form of a sphere, allowing us to visualize the geometric shape described by our initial vector equation.
Vector Expansions
Vector expansions involve breaking down vector operations into basic components to simplify calculations and reveal underlying geometric interpretations.In our problem, vector expansion is employed by:
  • Breaking down \((\mathbf{r} - \mathbf{a})\) and \((\mathbf{r} - \mathbf{b})\), then finding their dot product.
  • Expanding \((x - a_1)(x - b_1), (y - a_2)(y - b_2), (z - a_3)(z - b_3)\) into expressions like \(x^2 - (a_1 + b_1)x + a_1 b_1\).
  • Recognizing these forms contribute to understanding the geometry described by the vector equation.
By expanding vectors this way, we can transition from abstract vector expressions to concrete geometric forms, aiding clarity in studying vector calculus and its applications.

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

Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Find an eigenvector associated with the given eigenvalue of \(A .\) $$\begin{array}{l}{\text { (a) } A=\left[ \begin{array}{ll}{9} & {0} \\ {2} & {3}\end{array}\right] \quad \lambda=9} \\ {\text { (b) } A=\left[ \begin{array}{ll}{1} & {5} \\ {2} & {7}\end{array}\right] \quad \lambda=4+\sqrt{19}}\end{array}$$ $$(\mathrm{c}) A=\left[ \begin{array}{lll}{1} & {0} & {3} \\ {2} & {0} & {0} \\\ {0} & {2} & {1}\end{array}\right] \quad \lambda=3$$ $$(\mathrm{d})A=\left[ \begin{array}{ll}{0} & {1} \\ {3} & {2}\end{array}\right] \quad \lambda=-1$$ $$\begin{array}{l}{\text { (e) } A=\left[ \begin{array}{ll}{0} & {1} \\ {1} & {1}\end{array}\right] \quad \lambda=\frac{1+\sqrt{5}}{2}} \\ {\text { (f) } A=\left[ \begin{array}{lll}{1} & {2} & {3} \\ {0} & {1} & {7} \\ {0} & {2} & {1}\end{array}\right] \quad \lambda=1+\sqrt{14}}\end{array}$$

Suppose that \(A^{2}=0\) for some matrix \(A .\) Prove that the only possible eigenvalues of \(A\) are then \(0 .\)

\(\begin{array}{c}{\text { Mutation Suppose that, as a result of mutation, the }} \\ {\text { number of individuals in a population carrying allele } A \text { and }} \\ {\text { allele } B \text { changes in a way described by the equation }} \\ {\mathbf{y}_{t+1}=\left[ \begin{array}{cc}{0.95} & {0} \\\ {0.05} & {1}\end{array}\right] \mathbf{y}_{t}}\end{array}\) \(\begin{array}{c}{\text { where } \mathbf{y}_{t} \text { is the vector whose components are the numbers }} \\ {\text { of individuals carrying each allele at time } t . \text { An equilibrium }} \\ {\text { is a value of the vector for which no change occurs (that is, }} \\ {\mathbf{y}_{t+1}=\mathbf{y}_{t} ) . \text { Denoting such values by } \hat{\mathbf{y}}, \text { they must therefore }} \\ {\text { satisfy the equation }} \\ {\hat{\mathbf{y}}=\left[ \begin{array}{cc}{0.95} & {0} \\ {0.05} & {1}\end{array}\right] \hat{\mathbf{y}}}\end{array}\) Find all possible equilibrium values.

$$\begin{array}{l}{\text { (a) Draw the vectors } \mathbf{a}=[3,2], \mathbf{b}=[2,-1], \text { and }} \\ {\mathbf{c}=[7,1] .} \\ {\text { (b) Show, by means of a sketch, that there are scalars } s \text { and } t} \\\ {\text { such that } \mathbf{c}=s \mathbf{a}+t \mathbf{b} .} \\ {\text { (c) Use the sketch to estimate the values of } s \text { and } t} \\ {\text { (d) Find the exact values of } s \text { and } t .}\end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Probability and 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.