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.

Short 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 its radius is \(\frac{1}{2}||\mathbf{b} - \mathbf{a}||\).

Step-by-step solution

Understanding the Equation

The given equation is \((\mathbf{x}-\mathbf{a}) \cdot (\mathbf{x}-\mathbf{b}) = 0\). In vector terms, \(\mathbf{x} = \left\langle x_1, x_2, x_3 \right\rangle\) is an arbitrary point on the surface of the geometric figure, and this equation represents all points \(\mathbf{x}\) such that the vectors \(\mathbf{x}-\mathbf{a}\) and \(\mathbf{x}-\mathbf{b}\) are orthogonal.

Expanding the Dot Product

The dot product of \((\mathbf{x}-\mathbf{a})\) and \((\mathbf{x}-\mathbf{b})\) expands to \([(x_1 - a_1)(x_1 - b_1) + (x_2 - a_2)(x_2 - b_2) + (x_3 - a_3)(x_3 - b_3)] = 0\). This can be rewritten using \(x_i^2, a_i^2, b_i^2,\) and \(x_i\cdot a_i\) terms, i.e., \( x_1x_1 - x_1b_1 - a_1x_1 + a_1b_1 + x_2x_2 - x_2b_2 - a_2x_2 + a_2b_2 + x_3x_3 - x_3b_3 - a_3x_3 + a_3b_3 = 0 \).

Simplify the Equation

Grouping like terms, the equation becomes \[x_1(x_1) + x_2(x_2) + x_3(x_3) \]-\[x_1(b_1 + a_1) + x_2(b_2 + a_2) + x_3(b_3 + a_3) \]+\[a_1b_1 + a_2b_2 + a_3b_3 = 0\]. This implies that every point \(\mathbf{x}\) satisfying this equation lies on a sphere.

Deriving the Sphere's Center

To rearrange the equation into the standard sphere form, notice that the middle terms \(-2x_i\text{values}(a_i + b_i)\) can be interpreted as a shift to locate the sphere's center. The coefficient \(\mathbf{a} + \mathbf{b}\) can be divided by 2, giving the center: \( \left\langle \frac{a_1 + b_1}{2}, \frac{a_2 + b_2}{2}, \frac{a_3 + b_3}{2} \right\rangle \).

Finding the Sphere's Radius

For a sphere centered at \(\mathbf{c}\), the radius \(r\) satisfies the equation for distance from the origin, derived as \(\sqrt{ \left( \frac{a_1}{2} - \frac{b_1}{2} \right)^2 + \left( \frac{a_2}{2} - \frac{b_2}{2} \right)^2 +\left( \frac{a_3}{2} - \frac{b_3}{2} \right)^2}.\) This radius value explains the points' "equal distance" property about the sphere’s center.

Key concepts

Dot Product

The dot product is an essential operation in vector algebra. It takes two vectors and returns a scalar, showing how much one vector extends in the direction of the other. Given two vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\), the dot product is calculated as:

• \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\)

The dot product connects to many geometric concepts, particularly showing orthogonality when its result is zero. In our exercise, the dot product \((\mathbf{x}-\mathbf{a}) \cdot (\mathbf{x}-\mathbf{b}) = 0\) implies that \((\mathbf{x}-\mathbf{a})\) and \((\mathbf{x}-\mathbf{b})\) are perpendicular. Understanding the dot product is key to visualizing and solving problems involving vectors' angles and lengths.

Orthogonal Vectors

Vectors are orthogonal when their dot product equals zero, indicating they are perpendicular to one another. When vectors \(\mathbf{u}\) and \(\mathbf{v}\) satisfy \(\mathbf{u} \cdot \mathbf{v} = 0\), they form a right angle. In three-dimensional space, this notion helps recognize equilibrium and symmetry among forces or lines.
In the exercise example, the vectors \((\mathbf{x}-\mathbf{a})\) and \((\mathbf{x}-\mathbf{b})\) being orthogonal means they describe directions that maintain balance in forming a geometric shape, such as a sphere. The key takeaway is the critical geometric relation that orthogonality provides for constructing properties like shape and volume.

Vector Space

A vector space is a collection of vectors that are closed under operations of vector addition and scalar multiplication. This forms the basic framework for solving equations involving vectors. Within this space, vectors follow specific rules and properties.

• Closure under addition: If \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in a vector space, then \(\mathbf{u} + \mathbf{v}\) is also in the space.
• Closure under scalar multiplication: If \(c\) is a scalar and \(\mathbf{v}\) is a vector, then \(c\mathbf{v}\) is also in the space.

In analyzing geometric shapes like spheres, the vector space allows transformations and manipulations of vector equations to determine geometric properties. The exercise utilizes the concept of a vector space to efficiently manage the vectors' operations while working toward identifying a sphere’s position and size within this space.

Sphere Center

The center of a sphere is the point from which all points on the surface of the sphere are equidistant. Determining this point is crucial in geometry. In vector terms, for the equation \((\mathbf{x}-\mathbf{a}) \cdot(\mathbf{x}-\mathbf{b})=0\), we find the sphere's center by analyzing shifts in vector positions.
In our exercise, the center \(\mathbf{c}\) is derived by averaging the components of vectors \(\mathbf{a}\) and \(\mathbf{b}\):

• \(\mathbf{c} = \left\langle \frac{a_1 + b_1}{2}, \frac{a_2 + b_2}{2}, \frac{a_3 + b_3}{2} \right\rangle \)

Understanding how these averages form the center helps in visualizing the spatial balance achieved within the sphere, ensuring the center accurately represents this equidistance property. This foundation allows us to progress toward determining the sphere's radius.

Sphere Radius

The radius of a sphere is the constant distance from its center to any point on its surface. It plays a vital role in defining the size of the sphere. Once we have established the center of the sphere, calculating the radius involves measuring this consistent distance.

• Formula: \(r = \sqrt{ \left( \frac{a_1}{2} - \frac{b_1}{2} \right)^2 + \left( \frac{a_2}{2} - \frac{b_2}{2} \right)^2 + \left( \frac{a_3}{2} - \frac{b_3}{2} \right)^2 } \)

With this calculation, the radius confirms the sphere's equidistant property from the center, and it scales proportionally as the vectors \(\mathbf{a}\) and \(\mathbf{b}\) change. Visualizing this radius not only aids in comprehending the measure of a sphere but also how geometric changes affect its dimensional properties. Understanding this concept allows us to see how spheres fit within various vector spaces.