Problem 51 - Algebra and Trigonometry

Question

Vector Equation of a Sphere 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 \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) rep- resents 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. \([\text { Hint: Draw a diagram showing the }\) endpoints of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{r},\) noting that the end- points of \(\mathbf{a}\) and \(\mathbf{b}\) are the endpoints of a diameter and the endpoint of \(\mathbf{r}\) 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

The equation represents a sphere with center \( (0, 0, 1) \) and radius \( 3 \).

Step 1: Expand the dot product

Let \( \mathbf{a} = \langle 2,2,2 \rangle \), \( \mathbf{b} = \langle -2,-2,0 \rangle \), \( \mathbf{r} = \langle x,y,z \rangle \).
\( (\mathbf{r}-\mathbf{a}) \cdot (\mathbf{r}-\mathbf{b}) = 0 \)
\( (x-2)(x+2) + (y-2)(y+2) + (z-2)(z) = 0 \).

Step 2: Expand each term

\( x^2-4 + y^2-4 + z^2-2z = 0 \)
\( x^2 + y^2 + z^2 - 2z - 8 = 0 \).

Step 3: Complete the square

\( x^2 + y^2 + (z-1)^2 - 1 - 8 = 0 \)
\( x^2 + y^2 + (z-1)^2 = 9 \).

Step 4: Identify the sphere

Wait, let me recheck. \( (z-1)^2 = z^2 - 2z + 1 \), so \( x^2 + y^2 + z^2 - 2z = x^2 + y^2 + (z-1)^2 - 1 \).
So \( x^2 + y^2 + (z-1)^2 = 9 \). Hmm, but answer says radius 2. Let me recheck the expansion.
Actually \( (x-2)(x+2) = x^2 - 4 \), \( (y-2)(y+2) = y^2 - 4 \), \( (z-2)(z-0) = z^2 - 2z \).
Sum: \( x^2 + y^2 + z^2 - 2z - 8 = 0 \), so \( x^2 + y^2 + (z-1)^2 = 9 \).
Center \( (0, 0, 1) \), radius \( 3 \).

Key Concepts

Vector Algebra

Vector algebra is a fascinating branch of mathematics that deals with quantities having both magnitude and direction. When working with vectors, you'll encounter operations such as addition, subtraction, and various forms of multiplication.

• Add and Subtract Vectors: To add vectors, simply add their corresponding components. Subtraction works similarly, but involves subtracting the components.
• Multiplication: Unlike scalar quantities, vectors have two different ways to be multiplied: the dot product and the cross product, each serving unique purposes.

Vectors can be represented in component form, such as \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\), allowing for straightforward manipulation using algebraic methods. Understanding these fundamental operations is essential for solving problems involving vector equations, such as the one in our exercise.

Dot Product

The dot product is an essential operation in vector algebra. It yields a scalar value from two vectors, providing insights into their geometric relationship.

• The formula for the dot product is given by \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\).
• This operation is commutative, meaning \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\).

The dot product is particularly useful because it can tell us if two vectors are perpendicular. If the dot product is zero, like in our exercise, the vectors are orthogonal. This property is crucial in verifying geometric interpretations, such as determining whether points in a space can form a sphere.

Geometric Interpretation of Vectors

Vectors are not just abstract quantities—they have valuable geometric interpretations. In our context, when vectors \(\mathbf{r}-\mathbf{a}\) and \(\mathbf{r}-\mathbf{b}\) have a dot product of zero, it tells us that these vectors are perpendicular.

• Perpendicular vectors correspond to intersecting at 90 degrees, forming a right angle.
• In three-dimensional space, this orthogonal relationship helps locate points, like the endpoints of a diameter, critical in establishing spherical geometry.

Additionally, visualizing vectors in diagrams can greatly aid in understanding. By drawing the vectors and their resultant relationships, as suggested in our original exercise, you clarify how they provide geometric insights into the structure of spheres and other shapes.

Sphere Center and Radius

In the context of a sphere's vector equation, understanding how to find its center and radius is fundamental. A sphere in three-dimensional space can be visualized as a set of points equidistant from a central point.

• Center: The midpoint of the diameter is the center of the sphere. For vectors \(\mathbf{a}\) and \(\mathbf{b}\), calculate their midpoint as \(\mathbf{c} = \frac{\mathbf{a} + \mathbf{b}}{2}\).
• Radius: Half the length of the diameter gives the radius. It can be calculated using the distance formula for vectors, \(r = \frac{1}{2} \|\mathbf{a} - \mathbf{b}\|\).

These calculations lead to a clear picture of the sphere's properties—its size and position in space. Mastering these concepts allows you to represent spheres algebraically and graphically, simplifying complex geometric problems.