Problem 51 - Precalculus Mathematics for Calculus

Question

Let a \(=\langle 2,2,2\rangle\) \(\mathbf{b}=\langle- 2,-2,0\rangle,\) and \(\mathbf{r}=\langle x, y, z\rangle\) (a) Show that the vector equation \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, by expanding the dot product and simplifying the resulting algebraic equation. (b) Find the center and radius of the sphere. (c) Interpret the result of part (a) geometrically, using the fact that the dot product of two vectors is 0 only if the vectors are perpendicular. endpoints of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{r},\) noting that the end. points of a and b are the endpoints of a diameter and the endpoint of \(\mathbf{r} \text { is an arbitrary point on the sphere. }]\) (d) Using your observations from part (a), find a vector equation for the sphere in which the points \((0,1,3)\) and \((2,-1,4)\) form the endpoints of a diameter. Simplify the vector equation to obtain an algebraic equation for the sphere. What are its center and radius?

Step-by-Step Solution

Answer

(a) Sphere: \((x-0)^2 + (y-0)^2 + (z-1)^2 = 9\); Center: (0,0,1); Radius: 3. (b) Different sphere: Center: (1,0,3.5); Radius: \(\sqrt{3}\)

Step 1: Expand the Dot Product

To show that the vector equation \((\mathbf{r}-\mathbf{a})\cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, start by expanding the dot product. Let \(\mathbf{r}-\mathbf{a} = \langle x-2, y-2, z-2 \rangle\) and \(\mathbf{r}-\mathbf{b} = \langle x+2, y+2, z \rangle\).The dot product is \[ \langle x-2, y-2, z-2 \rangle \cdot \langle x+2, y+2, z \rangle = (x-2)(x+2) + (y-2)(y+2) + (z-2)z. \]

Step 2: Simplify the Expanded Expression

Simplify the expression from Step 1:\[ (x-2)(x+2) + (y-2)(y+2) + (z-2)z = x^2-4 + y^2-4 + z^2 - 2z. \]Combine like terms, which gives:\[ x^2 + y^2 + z^2 - 2z - 8 = 0. \]

Step 3: Reorganize into Equation of a Sphere

Reorganize the equation from Step 2 as:\[ x^2 + y^2 + (z-1)^2 = 9. \]This is the standard form of the sphere equation: \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2, \]where \((h, k, l)\) is the center and \(r\) is the radius.

Step 4: Identify the Center and Radius

From the equation \[ (x-0)^2 + (y-0)^2 + (z-1)^2 = 9, \]we identify the center as \((0, 0, 1)\) and the radius as \(3\).

Step 5: Interpret Geometric Result

Geometrically, the result shows that the vectors are perpendicular, indicating diametrically opposite points on the sphere, with the center at their midpoint. This confirms the spherical nature of the locus of point \(\mathbf{r}\).

Step 6: Using Part (a) for New Sphere

Given points \((0, 1, 3)\) and \((2, -1, 4)\) as endpoints of a diameter, determine the center \((h, k, l)\) by midpoint calculation: \\[ \left(\frac{0+2}{2}, \frac{1+(-1)}{2}, \frac{3+4}{2}\right) = (1, 0, 3.5). \]

Step 7: Find and Simplify the New Sphere's Equation

The vector equation of the sphere is given by midpoint and diameter length. Calculate radius: \\[ r = \frac{\sqrt{(0-2)^2 + (1-(-1))^2 + (3-4)^2}}{2} = \sqrt{3}. \]The sphere's equation is \[ (x-1)^2 + y^2 + \left(z-3.5\right)^2 = \left(\sqrt{3}\right)^2. \]Simplifying, we get \[ (x-1)^2 + y^2 + (z-3.5)^2 = 3. \]

Step 8: Identify the Center and Radius of the New Sphere

The sphere's center is \((1, 0, 3.5)\) and the radius is \(\sqrt{3}\).

Key Concepts

Sphere Equation

The equation of a sphere is a fascinating and versatile formula that can be understood in both an algebraic and geometric context. The general form of a sphere's equation is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \(h, k,\) and \(l\) represent the coordinates of the center of the sphere, and \(r\) signifies its radius. Expanding on this form reveals how any point \(x, y, z\) can be assessed to determine if it lies on the sphere's surface.

Initially, you might see it written as \(x^2 + y^2 + z^2 - 2hx - 2ky - 2lz + (h^2 + k^2 + l^2 - r^2) = 0\), which can sometimes appear in certain problems and requires some manipulation to turn back into the familiar sphere equation form. Each part of this equation ties directly back to the sphere's geometry, providing a clear understanding of its structure in space.

Dot Product

In vector mathematics, the dot product is a crucial concept that connects vectors to geometric properties. It measures the extent to which two vectors align with one another. The dot product of two vectors \(f{a}\) and \(f{b}\) is defined as \(f{a} \cdot \bf{b} = a_1b_1 + a_2b_2 + a_3b_3\), where each component of the vectors is multiplied and the results are summed. A zero result from a dot product reveals orthogonality, meaning the vectors are perpendicular.

This property is fundamental in the study of spheres when using vector equations to determine the relationships between points. In problems like the one provided, identifying when vectors are perpendicular can simplify determining the geometric properties of the sphere, such as finding the center and radius derived from these spatial relationships.

Geometric Interpretation

Understanding the geometrical interpretation of mathematical concepts, such as the dot product, is vital in visualizing vectors in space. When dealing with the sphere equation, we often use vectors to understand the spatial layout that gives rise to this equation.

In the exercise, when \( \bf{r-a}\) and \(\bf{r-b}\) are used in a dot product that equates to zero, it implies that these vectors are perpendicular, thus defining a special relationship where the points \(\bf{a}\) and \(\bf{b}\) form a diameter of the sphere. This perpendicularity establishes \(\bf{r}\), the arbitrary point on the sphere as being equidistant from \(\bf{a}\) and \(\bf{b}\), grounding the position of any 'r' within the sphere's boundary.

Center of a Sphere

The center of a sphere, expressed in the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), is given by the coordinates \(h, k, l\). The center is crucial because it serves as the reference point from which all radial distances are measured. In this problem, by reorganizing the terms from the expanded form of the dot product equation, you can identify the center.

For instance, solving the given exercise leads to finding the center at \( (0, 0, 1) \), indicating the complete symmetry around this point in 3D space. Knowing the center allows for the construction or graphical representation of the sphere, especially when midpoints between diametrically positioned vectors are utilized.

Radius of a Sphere

The radius is a fundamental characteristic of a sphere, consistently measuring the distance from the center to any point on its surface. In the standard equation form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r\) represents the sphere's radius. This exercise uncovered the radius via the simplification of the derived equation from the vector equation, with \(r=3\) for the initial sphere.

When working with endpoints that form a diameter, the radius is simply half of the calculated distance between these endpoints. Using the provided points, for example, provided a more unconventional sphere centered at \( (1, 0, 3.5) \) with a radius of \(\sqrt{3}\). Being able to find the radius builds confidence in solving more complex geometrical shapes and strengthens spatial reasoning skills.