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Chapter 9: Problem 16

Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}+4 x-6 y+2 z=10$$

Short Answer

Expert verified
The equation represents a sphere centered at \((-2, 3, -1)\) with radius \(2\sqrt{6}\).

Step by step solution

01

Rearrange the Equation

Write the equation in the form that separates the quadratic terms from the linear and constant terms: \ \[ x^2 + 4x + y^2 - 6y + z^2 + 2z = 10 \]\. This will help us apply the technique of completing the square.
02

Complete the Square for the x-Terms

For the terms involving \(x\), \(x^2 + 4x\), complete the square by adding and subtracting \((\frac{4}{2})^2 = 4\). This transforms the expression into \((x+2)^2 - 4\).
03

Complete the Square for the y-Terms

For the terms involving \(y\), \(y^2 - 6y\), complete the square by adding and subtracting \((\frac{-6}{2})^2 = 9\). This transforms the expression into \((y-3)^2 - 9\).
04

Complete the Square for the z-Terms

For the terms involving \(z\), \(z^2 + 2z\), complete the square by adding and subtracting \((\frac{2}{2})^2 = 1\). This transforms the expression into \((z+1)^2 - 1\).
05

Insert and Rearrange Completed Squares

Substitute the completed squares from Steps 2-4 back into the equation: \ \[(x+2)^2 - 4 + (y-3)^2 - 9 + (z+1)^2 - 1 = 10\] \ Combine the constants on the right side of the equation: \ \[(x+2)^2 + (y-3)^2 + (z+1)^2 = 24\].
06

Identify Center and Radius of the Sphere

The equation \((x+2)^2 + (y-3)^2 + (z+1)^2 = 24\) is in the standard form of a sphere \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\) where \((h, k, l)\) is the center and \(r\) is the radius.\Thus, the center of the sphere is \((-2, 3, -1)\), and the radius is \(\sqrt{24}\) or \(2\sqrt{6}\).

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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 technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful in geometry, including when identifying the equation of a sphere.
Here's a simple breakdown:
  • First, identify the quadratic term (e.g., \(x^2+4x\)).
  • Take the linear term, divide it by 2, and square the result \((\left(\frac{b}{2}\right)^2)\).
  • Add this square inside the bracket and subtract it outside to balance the equation.
Let鈥檚 consider the term involving \(x\):
\[x^2 + 4x\]
By completing the square, this expression becomes:
\[(x+2)^2 - 4\]
Repeat this process for other variables to transform the entire quadratic equation. This simplification is crucial because it allows you to express the equation in a form that reveals more about the geometry of a shape, such as a sphere.
Center of a Sphere
A sphere's equation looks like \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\) where \((h, k, l)\) is the center. After transforming an equation using completing the square, the resulting form helps identify the center.
In the problem exercise, through completing the square, the equation is rewritten as:
\[(x+2)^2 + (y-3)^2 + (z+1)^2 = 24\]
To find the center of the sphere:
  • The expression \((x+2)^2\) tells us that the \(x\)-coordinate of the center is \(-2\).
  • Similarly, \((y-3)^2\) gives the \(y\)-coordinate as \(3\), and \((z+1)^2\) gives the \(z\)-coordinate as \(-1\).
Therefore, the center of the sphere is \((-2, 3, -1)\). This point represents the symmetry center from which every surface point of the sphere is equidistant.
Radius of a Sphere
The radius of a sphere is the distance from its center to any point on its surface. It's crucial for understanding the size of the sphere in three-dimensional space.
In a standard sphere equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r^2\) is the squared radius.
Once the equation is transformed into its completed square form:
\[(x+2)^2 + (y-3)^2 + (z+1)^2 = 24\]
The expression \(24\) represents \(r^2\).
To find the radius \(r\), take the square root of 24:
  • \(r = \sqrt{24}\)
  • Further simplified, \(r = 2\sqrt{6}\)
This gives us the length of the radius, confirming that every point on the sphere is \(2\sqrt{6}\) units from its center. This helps understand not only the position but the scale of the sphere.

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Most popular questions from this chapter

Three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are given. \(\mathbf{(a)}\) Find their scalar triple product \(\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) .\) \(\mathbf{(b)}\) Are the vectors coplanar? If not, find the volume of the parallelepiped that they determine. $$\mathbf{a}=\langle 1,2,3\rangle, \quad \mathbf{b}=\langle- 3,2,1\rangle, \quad \mathbf{c}=\langle 0,8,10\rangle$$

Three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are given. \(\mathbf{(a)}\) Find their scalar triple product \(\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) .\) \(\mathbf{(b)}\) Are the vectors coplanar? If not, find the volume of the parallelepiped that they determine. $$\mathbf{a}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \quad \mathbf{b}=-\mathbf{j}+\mathbf{k}, \quad \mathbf{c}=\mathbf{i}+\mathbf{j}+\mathbf{k}$$

Rubik's Cube, a puzzle craze of the 1980 s that remains popular to this day, inspired many similar puzzles. The one illustrated in the figure is called Rubik's Tetrahedron; it is in the shape of a regular tetrahedron, with each edge \(\sqrt{2}\) inches long. The volume of a regular tetrahedron is one-sixth the volume of the parallelepiped determined by any three edges that meet at a corner. (a) Use the triple product to find the volume of Rubik's Tetrahedron. (b) Construct six identical regular tetrahedra using modeling clay. Experiment to see how they can be put together to create a parallelepiped that is determined by three edges of one of the tetrahedra (thus confirming the above statement about the volume of a regular tetrahedron).

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\mathbf{i}+\mathbf{j}$$

Three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are given. \(\mathbf{(a)}\) Find their scalar triple product \(\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) .\) \(\mathbf{(b)}\) Are the vectors coplanar? If not, find the volume of the parallelepiped that they determine. $$\mathbf{a}=\langle 1,-1,0\rangle, \quad \mathbf{b}=\langle- 1,0,1\rangle, \quad \mathbf{c}=\langle 0,-1,1\rangle$$
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