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 is 0; the sphere is tangent to the plane.

Step 1: 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.

Step 2: 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.

Step 3: 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, \(r = \sqrt{26}\).

Step 4: 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 \((x_1, y_1, z_1) = (-1, -3, 4)\).

Step 5: 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, \(d = \frac{26}{\sqrt{26}} = \sqrt{26}\).

Step 6: 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.

Key Concepts

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:

• \((h, k, l)\) is the center of the sphere.
• \(r\) represents the radius.

In practical problems, like this one, you're often given the equation in an expanded form, such as \(x^2 + y^2 + z^2 + 2x + 6y - 8z = 0\).
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.

• 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.

Applying this formula helps determine whether the sphere's center is closer, further, or exactly at the plane surface, impacting whether the plane intersects with the sphere.

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:

• 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 \((z-4)^2 - 16\).

This method helps in reformatting the equation to identify the sphere's center and radius, setting a clear path for measuring distances or understanding intersections, like checking if a plane is tangent to a sphere.