Problem 77 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance from the sphere \(x^{2}+y^{2}+z^{2}+2 x+\) \(6 y-8 z=0\) to the plane \(3 x+4 y+z=15\).

Step-by-Step Solution

Answer

The distance from the sphere to the plane is 0.

Step 1: Identify the center and radius of the sphere

The general equation of a sphere is \( (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 \). We begin by writing the given equation of the sphere in this form. The given sphere's equation is:\[ x^2 + y^2 + z^2 + 2x + 6y - 8z = 0. \]To complete the square, we first organize each variable:- 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 \((z - 4)^2 - 16\).Rewriting the sphere's equation, we get:\[ (x + 1)^2 + (y + 3)^2 + (z - 4)^2 = 26 \].Thus, the center of the sphere is \((-1, -3, 4)\) and its radius is \( \sqrt{26} \).

Step 2: Parameterize the distance as a point-plane distance

The distance from a point, \((x_0, y_0, z_0)\), to a plane, given by \(ax + by + cz + d = 0\), is calculated using:\[ \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}.\]For the plane \(3x + 4y + z = 15\) or \(3x + 4y + z - 15 = 0\), the normal vector is \((3, 4, 1)\). We use \((-1, -3, 4)\) from Step 1 as the point.

Step 3: Calculate the perpendicular distance from the center of the sphere to the plane

Substitute \((-1, -3, 4)\) into the point-plane distance formula:\[\text{Distance} = \frac{|3(-1) + 4(-3) + 1(4) - 15|}{\sqrt{3^2 + 4^2 + 1^2}} = \frac{|-3 - 12 + 4 - 15|}{\sqrt{26}} = \frac{26}{\sqrt{26}} = \sqrt{26}.\]

Step 4: Determine the distance from the sphere to the plane

Since the perpendicular distance from the center of the sphere to the plane is \( \sqrt{26} \), and this is exactly equal to the radius of the sphere, the sphere just touches the plane. Therefore, the distance from the sphere's surface to the plane is 0.

Key Concepts

Sphere Equation

A sphere in a three-dimensional space can be described using its equation, which is typically written in the form \[ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 \]where

• \( (x_0, y_0, z_0) \) is the center of the sphere,
• \( r \) is the radius of the sphere.

For the task at hand, the sphere equation provided is \[ x^2 + y^2 + z^2 + 2x + 6y - 8z = 0 \].To uncover the center and radius, it's essential to reformulate it into the standard sphere equation form by completing the square for each variable. It helps to easily identify key elements like the center and the radius.

Point-Plane Distance

Calculating the distance from a point to a plane involves a handy formula. This formula expresses how far a specific point \( (x_0, y_0, z_0) \)is from a plane defined by \( ax + by + cz + d = 0 \).The point-to-plane distance \( D \) is given by:\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]This ensures we get a positive distance value irrespective of where the point is located in relation to the plane. It's a crucial calculation to determine relationships between lines, planes, and points in geometry, emphasizing spatial understanding and mathematical elegance.

Completing The Square

Completing the square is a technique often used to rewrite quadratic expressions in a way that reveals specific properties, such as the vertex of a parabola or, in this case, the center of a sphere.To effectively apply this technique to the sphere’s equation:

• First, group terms: \((x^2 + 2x)\), \((y^2 + 6y)\), and \((z^2 - 8z)\).
• Then, complete the square for each group:
  • For \(x\): \((x + 1)^2 - 1\),
  • For \(y\): \((y + 3)^2 - 9\),
  • For \(z\): \((z - 4)^2 - 16\).
• Rewrite the equation to find the true center and radius, revealing the equation: \[ (x + 1)^2 + (y + 3)^2 + (z - 4)^2 = 26 \].

This exposes the center as \((-1, -3, 4)\) and the radius as \(\sqrt{26}\), essential for further calculations.

Perpendicular Distance

Perpendicular distance is a concept used to find the shortest possible distance between a point and a geometric entity like a plane. It measures the stretch along a line perpendicular to the plane from the point. In our exercise, calculating this perpendicular distance from the sphere's center \((-1, -3, 4)\) to the plane \(3x + 4y + z - 15 = 0\) involves substituting the center's coordinates into the point-plane distance formula. This step provides a direct understanding of whether or not the sphere and plane are intersecting or just touching. In this scenario, after performing the calculation, it is found that the distance is exactly the same as the sphere's radius, which implies the sphere just grazes the plane surface, touching it at a single point.