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Magazine Chapter 11: Problem 77
Find the distance from the sphere \(x^{2}+y^{2}+z^{2}+2 x+\) \(6 y-8 z=0\) to the plane \(3 x+4 y+z=15\).
Short Answer
Expert verified
The distance is 0; the sphere is tangent to the plane.
Step by step solution
01
Identify the center and radius of the sphere
The given equation of the sphere is \(x^2 + y^2 + z^2 + 2x + 6y - 8z = 0\). To find the center and radius, convert the equation to the standard form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). First, complete the square for each variable.
02
Complete the square for each variable
For \(x\), we rewrite \(x^2 + 2x\) as \((x+1)^2 - 1\). For \(y\), \(y^2 + 6y\) becomes \((y+3)^2 - 9\). For \(z\), \(z^2 - 8z\) becomes \((z-4)^2 - 16\). Substitute these back into the equation.
03
Simplify to find the center and radius
Substituting back, we get: \((x+1)^2 - 1 + (y+3)^2 - 9 + (z-4)^2 - 16 = 0\). This simplifies to: \((x+1)^2 + (y+3)^2 + (z-4)^2 = 26\). The center of the sphere is \((-1, -3, 4)\) and the radius is the square root of 26, \(r = \sqrt{26}\).
04
Find distance from center to plane
Use the point-to-plane distance formula. The center of the sphere is \((-1, -3, 4)\), and the plane equation is \(3x + 4y + z = 15\). The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is given by:\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]. Substitute \(A = 3\), \(B = 4\), \(C = 1\), \(D = -15\), and \((x_1, y_1, z_1) = (-1, -3, 4)\).
05
Substitute values into the distance formula
Substitute the values: \[d = \frac{|3(-1) + 4(-3) + 1(4) - 15|}{\sqrt{3^2 + 4^2 + 1^2}}\]\[= \frac{| -3 - 12 + 4 - 15 |}{\sqrt{9 + 16 + 1}}\]\[= \frac{|-26|}{\sqrt{26}}\]. Therefore, \(d = \frac{26}{\sqrt{26}} = \sqrt{26}\).
06
Determine if the sphere and plane intersect
Since the distance from the center to the plane is equal to the radius \(\sqrt{26}\), it means the sphere is tangent to the plane. There is no gap between them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Sphere
The equation of a sphere is fundamental when understanding 3D geometry. A standard form of a sphere's equation is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where:
To convert this into the standard form, we use a method called **completing the square**. This involves rearranging the equation for each variable to isolate squares of the form \((x-a)^2\). Completing the square helps us identify both the center and radius of the sphere, which are critical for further calculations.
- \((h, k, l)\) is the center of the sphere.
- \(r\) represents the radius.
To convert this into the standard form, we use a method called **completing the square**. This involves rearranging the equation for each variable to isolate squares of the form \((x-a)^2\). Completing the square helps us identify both the center and radius of the sphere, which are critical for further calculations.
Point-to-Plane Distance
Understanding how to find the distance from a specific point to a plane is crucial in spatial geometry. The formula to find this distance, \(d\), from a point \((x_1, y_1, z_1)\) to a plane defined by the equation \(Ax + By + Cz + D = 0\) is:
\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]This formula calculates the shortest distance, ensuring a perpendicular measurement from the point to the plane.
\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]This formula calculates the shortest distance, ensuring a perpendicular measurement from the point to the plane.
- The numerator, \(|Ax_1 + By_1 + Cz_1 + D|\), captures the absolute value of the sum of the coordinates multiplied by the plane's coefficients plus the constant.
- The denominator, \(\sqrt{A^2 + B^2 + C^2}\), normalizes the plane's coefficients, making sure the distance is calculated correctly.
Completing the Square
The technique of completing the square is a powerful algebraic method used to simplify quadratic equations and bring them into a form where variables are neatly expressed as squares.
In geometry problems involving spheres, completing the square allows us to convert a given polynomial into the standard sphere equation.
Here's how it works:
In geometry problems involving spheres, completing the square allows us to convert a given polynomial into the standard sphere equation.
Here's how it works:
- For a term like \(x^2 + 2x\), completing the square involves adding and subtracting the square of half the \(x\) coefficient: \((x+1)^2 - 1\).
- Similarly, for \(y^2 + 6y\), you get \((y+3)^2 - 9\).
- For \(z^2 - 8z\), it becomes \((z-4)^2 - 16\).
