Problem 83 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=\left\langle a_{1}, a_{2}, a_{3}\right\rangle\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}, b_{3}\right\rangle\) be fixed vectors. Show that \((\mathbf{x}-\mathbf{a}) \cdot(\mathbf{x}-\mathbf{b})=0\) is the equation of a sphere, and find its center and radius.

Step-by-Step Solution

Answer

The sphere's center is \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\) and the radius is \(\sqrt{\frac{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}{4}}\)."

Step 1: Expand the Dot Product Equation

Start from the equation \((\mathbf{x}-\mathbf{a}) \cdot(\mathbf{x}-\mathbf{b})=0\). Expand the dot product:\[(\mathbf{x}-\mathbf{a}) \cdot (\mathbf{x}-\mathbf{b}) = (x_1-a_1)(x_1-b_1) + (x_2-a_2)(x_2-b_2) + (x_3-a_3)(x_3-b_3)\].Now expand each term: \[(x_1-a_1)(x_1-b_1) = x_1^2 - (a_1+b_1)x_1 + a_1b_1,\] and similarly for the other terms.

Step 2: Rearrange the Equation

Combine the expanded terms from Step 1 to get the complete expanded form:\[x_1^2 - (a_1+b_1)x_1 + a_1b_1 + x_2^2 - (a_2+b_2)x_2 + a_2b_2 + x_3^2 - (a_3+b_3)x_3 + a_3b_3 = 0\].Combine like terms:\[x_1^2 + x_2^2 + x_3^2 - (a_1+b_1)x_1 - (a_2+b_2)x_2 - (a_3+b_3)x_3 + a_1b_1 + a_2b_2 + a_3b_3 = 0\].

Step 3: Complete the Square for Each Variable

For each variable, complete the square to convert the equation into a standard sphere form.Example for \(x_1\):\[x_1^2 - (a_1+b_1)x_1 \rightarrow \left(x_1 - \frac{a_1+b_1}{2}\right)^2 - \left(\frac{a_1+b_1}{2}\right)^2\].Apply similar transformations for \(x_2\) and \(x_3\).

Step 4: Gather the Transformed Terms

Gather the squared terms:\[\left(x_1 - \frac{a_1+b_1}{2}\right)^2 + \left(x_2 - \frac{a_2+b_2}{2}\right)^2 + \left(x_3 - \frac{a_3+b_3}{2}\right)^2\].Calculate the constant terms for each completed square and adjust the equation accordingly.

Step 5: Identify the Center and Radius

Use the gathered and squared terms to identify the center and radius of the sphere:- The center is \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\).- Calculate the radius from the adjusted equation:\[\text{Radius} = \sqrt{\left(\frac{a_1-b_1}{2}\right)^2 + \left(\frac{a_2-b_2}{2}\right)^2 + \left(\frac{a_3-b_3}{2}\right)^2}\].

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra that results in a scalar. Given two vectors, it helps determine the angle between them or project one vector onto another. The mathematical definition for 3D vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\) is given by the equation:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

The dot product can reveal interesting geometric properties:
if it's zero, the vectors are orthogonal (at 90 degrees to each other). In our equation of a sphere,

• \( (\mathbf{x}-\mathbf{a}) \cdot (\mathbf{x}-\mathbf{b}) = 0 \)

we utilize the dot product setting it to zero to characterize orthogonality, or perfect perpendicularity, which is a crucial step in determining the center and radius of the sphere from intersection of geometric spaces.

Vector Algebra

Vector algebra is a cornerstone of geometry and physics, dealing with quantities that have both magnitude and direction. Basic operations include addition, subtraction, and multiplication (dot and cross products). In solving our sphere problem, we're primarily concerned with linear combinations and transformations of the vectors.

When we examine the expression \((\mathbf{x} - \mathbf{a})\) and \((\mathbf{x} - \mathbf{b})\), we're subtracting vectors,
leading to a geometric interpretation regarding the relative positions. This helps break down the given problem into manageable parts,
allowing us to rewrite the expression in terms of squares and products of terms stemming from these vectors.

Vector algebra provides a powerful toolkit: both in establishing foundational relationships within our equations like lengths and angles, and in facilitating algebraic manipulations, such as in transforming the given expression into standard identifiable forms like that of a sphere.

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic polynomial into a perfect square trinomial, making it easier to solve equations or analyze geometrical properties. This is particularly useful in the context of spheres,
where the equation can be simplified into a standard form.

In each part of our equation

• \( x_1^2 - (a_1+b_1)x_1 \)
• \( x_2^2 - (a_2+b_2)x_2 \)
• \( x_3^2 - (a_3+b_3)x_3 \)

we complete the square to find relationships between the variables and can visually comprehend the center coordinates. Completing the square transforms these terms into:

• \( \left(x_1 - \frac{a_1+b_1}{2}\right)^2 \)
• \( \left(x_2 - \frac{a_2+b_2}{2}\right)^2 \)
• \( \left(x_3 - \frac{a_3+b_3}{2}\right)^2 \)

By aggregating these squares alongside appropriate constant adjustments, the equation assumes a neat form that reveals the sphere's center and radius directly. This technique, often taught in the context of solving quadratic equations, proves immensely versatile even in multi-variable settings.