/*! 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 42 If \(\mathbf{r}=\langle x, y, z\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 42

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

Short Answer

Expert verified
The equation represents a sphere with center \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\) and radius \(\sqrt{\frac{(a_1-b_1)^2 + (a_2-b_2)^2 + (a_3-b_3)^2}{4}}\)."

Step by step solution

01

Define the equation and expand the dot product

The given vector equation is \((\mathbf{r}-\mathbf{a}) \cdot (\mathbf{r}-\mathbf{b}) = 0\). We need to expand this dot product. First, compute the vectors \(\mathbf{r}-\mathbf{a}\) and \(\mathbf{r}-\mathbf{b}\):\[\mathbf{r} - \mathbf{a} = \langle x-a_1, y-a_2, z-a_3 \rangle \] \[\mathbf{r} - \mathbf{b} = \langle x-b_1, y-b_2, z-b_3 \rangle \] Now, calculate the dot product:\[((x-a_1)(x-b_1) + (y-a_2)(y-b_2) + (z-a_3)(z-b_3) = 0)\].
02

Expand the binomials and simplify

Expand the dot product:\[(x-a_1)(x-b_1) = x^2 - (a_1+b_1)x + a_1b_1\]\[(y-a_2)(y-b_2) = y^2 - (a_2+b_2)y + a_2b_2\]\[(z-a_3)(z-b_3) = z^2 - (a_3+b_3)z + a_3b_3\]Substitute back into the dot product expression:\[x^2 - (a_1+b_1)x + a_1b_1 + y^2 - (a_2+b_2)y + a_2b_2 + z^2 - (a_3+b_3)z + a_3b_3 = 0\].
03

Complete the square for each variable

To transform the equation into the standard form of a sphere, complete the square for \(x\), \(y\), and \(z\).For \(x\), complete the square:\[x^2 - (a_1+b_1)x \Rightarrow \left(x - \frac{a_1+b_1}{2}\right)^2 - \left(\frac{a_1+b_1}{2}\right)^2\].For \(y\), complete the square:\[y^2 - (a_2+b_2)y \Rightarrow \left(y - \frac{a_2+b_2}{2}\right)^2 - \left(\frac{a_2+b_2}{2}\right)^2\].For \(z\), complete the square:\[z^2 - (a_3+b_3)z \Rightarrow \left(z - \frac{a_3+b_3}{2}\right)^2 - \left(\frac{a_3+b_3}{2}\right)^2\].
04

Formulate the sphere equation

After completing the square for each variable, the equation takes the form:\[ \left(x - \frac{a_1+b_1}{2}\right)^2 + \left(y - \frac{a_2+b_2}{2}\right)^2 + \left(z - \frac{a_3+b_3}{2}\right)^2 = R^2 \]where \(R^2\) is the sum of the constants and simplifications from completing the square:\[R^2 = \frac{(a_1+b_1)^2}{4} + \frac{(a_2+b_2)^2}{4} + \frac{(a_3+b_3)^2}{4} - (a_1b_1 + a_2b_2 + a_3b_3)\].
05

Identify center and radius of the sphere

From the standard form of the sphere equation: - The center of the sphere is \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\).- The radius \(R\) is found by taking the square root of \(R^2\). Calculate as:\[R = \sqrt{\frac{(a_1-b_1)^2}{4} + \frac{(a_2-b_2)^2}{4} + \frac{(a_3-b_3)^2}{4}}\].

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product, also known as the scalar product, is a fundamental concept in vector calculus, often used to find the angle between two vectors or to project one vector onto another. Given two vectors, \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their dot product is calculated as:\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\]For the exercise at hand, the dot product is being applied to vectors resulted from subtracting given vectors from \( \mathbf{r} \), which leads to the expansion of polynomial expressions. This results in an equation set to zero, a crucial step that ties dot product to geometric interpretations like the sphere.
  • The dot product returns a scalar (not a vector).
  • It is commutative: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
  • If the dot product is zero, the vectors are orthogonal.
In this scenario, the zero result implies orthogonality, paving the way to discovering a geometric entity: a sphere.
Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial, facilitating the finding of roots, as well as transforming equations into a recognizable format like the circle or sphere equation. In the context of the exercise, completing the square for each variable \(x\), \(y\), and \(z\) helps factorize and simplify the expanded dot product equation.

To complete the square for the general term \(x^2 + bx\), proceed as follows:\[x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\]Here, by halving the linear coefficient \(b\), squaring it, and adding it to the expression, it results in a perfect square trinomial. This transformation is key in rewriting the equation of a sphere in its standard form.

  • This technique reveals the geometric center and radius of spheres in algebraically derived forms.
  • Each completed square represents a coordinate shift, aligning the variable terms to the form of a sphere equation.
  • It simplifies solving quadratics by making them easier to interpret geometrically.
Sphere Equation
The sphere equation is typically represented in its standard form, signifying a three-dimensional round object where all surface points are equidistant from the center.For a sphere centered at \( (h, k, l) \) with radius \( R \), the equation is:\[(x - h)^2 + (y - k)^2 + (z - l)^2 = R^2\]The solutions from the exercise apply the concept of dot product and completing the square to transition from a polynomial condition to this geometric form.

In our particular context, using the computation from the vectors, the center and radius deduced from the calculation indicate how algebraic manipulations yield geometric insights. Specifically, completing the square provided the adjustments required to center the equation.
  • The center is calculated using the midpoint formula from vector components, allowing the sphere's position to be located in space.
  • The radius \( R \) is the square root of the result after simplification, indicating the size of the sphere.
  • This shows how a vector equation model can depict real spatial properties of objects like spheres.

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

Let \(L_{1}\) be the line through the origin and the point \((2,0,-1)\) Let \(L_{2}\) be the line through the points \((1,-1,1)\) and \((4,1,3)\) ) Find the distance between \(L_{1}\) and \(L_{2} .\)

Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=(1+t) \mathbf{i}+\left(t^{2}-2 t\right) \mathbf{j}$$

A rocket burning its onboard fuel while moving through space has velocity \(\mathbf{v}(t)\) and mass \(m(t)\) at time \(t .\) If the exhaust gases escape with velocity \(\mathbf{v}_{e}\) relative to the rocket, it can be deduced from Newton's Second Law of Motion that $$m \frac{d \mathbf{v}}{d t}=\frac{d m}{d t} \mathbf{v}_{e}$$ (a) Show that \(\mathbf{v}(t)=\mathbf{v}(0)-\ln \frac{m(0)}{m(t)} \mathbf{v}_{e}\) (b) For the rocket to accelerate in a straight line from rest to twice the speed of its own exhaust gases, what fraction of its initial mass would the rocket have to burn as fuel?

Let's consider the problem of designing a railroad track to make a smooth transition between sections of straight track. Existing track along the negative \(x\) -axis is to be joined smoothly to a track along the line \(y=1\) for \(x \geqslant 1\) (a) Find a polynomial \(P=P(x)\) of degree 5 such that the function \(F\) defined by $$F(x)=\left\\{\begin{array}{ll}{0} & {\text { if } x \leqslant 0} \\ {P(x)} & {\text { if } 0

\(33-38=\) (a) Sketch the plane curve with the given vector equation. (b) Find \(\mathbf{r}^{\prime}(t) .\) (c) Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}^{\prime}(t)\) for the given value of \(t\) . $$ \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j}, \quad t=0 $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

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.