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Chapter 11: Problem 28

Describe the surface whose equation is given. $$ x^{2}+y^{2}+z^{2}-2 x-6 y-8 z+1=0 $$

Short Answer

Expert verified
The surface is a sphere with center (1, 3, 4) and radius 5.

Step by step solution

01

Recognize the Form of the Equation

The given equation is similar to the general form of a sphere: x^2 + y^2 + z^2 + ax + by + cz + d = 0.
02

Rearrange the Equation for Completing the Square

Rearrange terms to complete the square for each variable:\(x^2 - 2x + y^2 - 6y + z^2 - 8z + 1 = 0\).
03

Complete the Square for x, y, and z

Complete the square for each part:- For \(x\): \((x^2 - 2x)\) becomes \((x-1)^2 - 1\).- For \(y\): \((y^2 - 6y)\) becomes \((y-3)^2 - 9\).- For \(z\): \((z^2 - 8z)\) becomes \((z-4)^2 - 16\).
04

Substitute Completing Squares into the Equation

The equation becomes:\((x-1)^2 - 1 + (y-3)^2 - 9 + (z-4)^2 - 16 + 1 = 0\).
05

Simplify the Equation

Simplify to get:\((x-1)^2 + (y-3)^2 + (z-4)^2 = 25\).
06

Identify the Sphere

The equation \((x-1)^2 + (y-3)^2 + (z-4)^2 = 25\) represents a sphere. It has a center at \((1, 3, 4)\) and a radius of 5, since \(25 = 5^2\).

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Key Concepts

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

Completing the Square
Completing the square is a useful technique to transform a quadratic equation into a more approachable form. This method is particularly handy when dealing with the equation of a sphere in 3D geometry. It allows us to express a quadratic expression in the form of a perfect square. Let's break it down into easily digestible steps.
  • Start by focusing on a single variable. For example, look at an expression like \(x^2 - 2x\).
  • Add and subtract the square of half the coefficient of \(x\), which in this case is \(1\) (since half of \(-2\) is \(-1\)):
  • Thus, we get \((x^2 - 2x + 1) - 1\) or \((x-1)^2 - 1\).
  • Repeat similar steps for the other variables \(y\) and \(z\).
  • This simplifies the equation and reveals more about the geometric nature of the surface.
The entire idea here is to rearrange those quadratic terms into neat, squared forms, which makes it easier to identify and understand the geometry of the sphere.
3D Geometry
3D Geometry is the study of shapes and figures in a three-dimensional space where each point is defined by three coordinates: \(x\), \(y\), and \(z\). This branch of geometry expands the principles of 2D geometry to add depth, enabling a richer analysis of shapes like spheres, cubes, and cylinders.
Understanding 3D geometry is key when dealing with equations of surfaces in space. They describe not just a flat surface but include volume and area in a space that stretches in all directions.
  • Each point in 3D space is determined by three numbers or coordinates \((x, y, z)\) which dictate its position.
  • Spheres, being perfectly symmetrical, are iconic 3D shapes where every point on the surface is equidistant from the center.
  • They provide a clear example of how completing the square can help us visualize these forms.
In exploring the equation \((x-1)^2 + (y-3)^2 + (z-4)^2 = 25\), you are engaging with 3D geometry by identifying the center and radius, which represents a sphere in three-dimensional space.
Center and Radius of a Sphere
Finding the center and radius of a sphere from its equation is a crucial step in identifying the sphere's position and size in 3D space. The equation of a sphere often appears in the form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, \((h, k, l)\) represent the coordinates of the center, while \(r\) is the radius.
In the equation \((x-1)^2 + (y-3)^2 + (z-4)^2 = 25\):
  • The center \((h, k, l)\) can be easily picked from the completed squares: it is at \((1, 3, 4)\).
  • The radius is determined by taking the square root of the number on the right side of the equation. Here, \(r = \sqrt{25} = 5\).
Knowing the center allows us to locate exactly where the sphere lies in the space, and the radius tells us how far it extends from this central point. These insights are invaluable for further calculations and visualizations of the sphere in a 3D environment.

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Most popular questions from this chapter

An equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. $$ r=2 \sec \theta $$

Given two planes, discuss the various possibilities for the set of points they have in common. Then consider the set of points that three planes can have in common.

An equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. $$ x^{2}+y^{2}+z^{2}=9 $$

A vector w is a linear combination of the vectors \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) if \(\mathbf{w}\) can be expressed as \(\mathbf{w}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3},\) where \(c_{1}, c_{2},\) and \(c_{3}\) are scalars. (a) Find scalars \(c_{1}, c_{2},\) and \(c_{3}\) to express \(\langle- 1,1,5\rangle\) as a linear combination of \(\mathbf{v}_{1}=\langle 1,0,1\rangle, \mathbf{v}_{2}=\langle 3,2,0\rangle,\) and \(\mathbf{v}_{3}=\langle 0,1,1\rangle\) (b) Show that the vector \(2 \mathbf{i}+\mathbf{j}-\mathbf{k}\) cannot be expressed as a linear combination of \(\mathbf{v}_{1}=\mathbf{i}-\mathbf{j}, \mathbf{v}_{2}=3 \mathbf{i}+\mathbf{k},\) and \(\mathbf{v}_{3}=4 \mathbf{i}-\mathbf{j}+\mathbf{k} .\)k} .$

Convert from spherical to rectangular coordinates. $$ \begin{array}{ll}{\text { (a) }(5, \pi / 6, \pi / 4)} & {\text { (b) }(7,0, \pi / 2)} \\ {\text { (c) }(1, \pi, 0)} & {\text { (d) }(2,3 \pi / 2, \pi / 2)}\end{array} $$
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