Problem 42 - Essential Calculus Early Transcendentals

Question

If \(\mathbf{r}=\langle x, y, z\rangle, \mathbf{a}=\left\langle a_{1}, a_{2}, a_{3}\right\rangle,\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}, b_{3}\right\rangle\) show that the vector equation \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, and find its center and radius.

Step-by-Step Solution

Answer

The equation represents a sphere with center \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\) and radius \(\sqrt{\frac{(a_1-b_1)^2 + (a_2-b_2)^2 + (a_3-b_3)^2}{4}}\)."

Step 1: Define the equation and expand the dot product

The given vector equation is \((\mathbf{r}-\mathbf{a}) \cdot (\mathbf{r}-\mathbf{b}) = 0\). We need to expand this dot product. First, compute the vectors \(\mathbf{r}-\mathbf{a}\) and \(\mathbf{r}-\mathbf{b}\):\[\mathbf{r} - \mathbf{a} = \langle x-a_1, y-a_2, z-a_3 \rangle \] \[\mathbf{r} - \mathbf{b} = \langle x-b_1, y-b_2, z-b_3 \rangle \] Now, calculate the dot product:\[((x-a_1)(x-b_1) + (y-a_2)(y-b_2) + (z-a_3)(z-b_3) = 0)\].

Step 2: Expand the binomials and simplify

Expand the dot product:\[(x-a_1)(x-b_1) = x^2 - (a_1+b_1)x + a_1b_1\]\[(y-a_2)(y-b_2) = y^2 - (a_2+b_2)y + a_2b_2\]\[(z-a_3)(z-b_3) = z^2 - (a_3+b_3)z + a_3b_3\]Substitute back into the dot product expression:\[x^2 - (a_1+b_1)x + a_1b_1 + y^2 - (a_2+b_2)y + a_2b_2 + z^2 - (a_3+b_3)z + a_3b_3 = 0\].

Step 3: Complete the square for each variable

To transform the equation into the standard form of a sphere, complete the square for \(x\), \(y\), and \(z\).For \(x\), complete the square:\[x^2 - (a_1+b_1)x \Rightarrow \left(x - \frac{a_1+b_1}{2}\right)^2 - \left(\frac{a_1+b_1}{2}\right)^2\].For \(y\), complete the square:\[y^2 - (a_2+b_2)y \Rightarrow \left(y - \frac{a_2+b_2}{2}\right)^2 - \left(\frac{a_2+b_2}{2}\right)^2\].For \(z\), complete the square:\[z^2 - (a_3+b_3)z \Rightarrow \left(z - \frac{a_3+b_3}{2}\right)^2 - \left(\frac{a_3+b_3}{2}\right)^2\].

Step 4: Formulate the sphere equation

After completing the square for each variable, the equation takes the form:\[ \left(x - \frac{a_1+b_1}{2}\right)^2 + \left(y - \frac{a_2+b_2}{2}\right)^2 + \left(z - \frac{a_3+b_3}{2}\right)^2 = R^2 \]where \(R^2\) is the sum of the constants and simplifications from completing the square:\[R^2 = \frac{(a_1+b_1)^2}{4} + \frac{(a_2+b_2)^2}{4} + \frac{(a_3+b_3)^2}{4} - (a_1b_1 + a_2b_2 + a_3b_3)\].

Step 5: Identify center and radius of the sphere

From the standard form of the sphere equation: - The center of the sphere is \(\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}, \frac{a_3+b_3}{2}\right)\).- The radius \(R\) is found by taking the square root of \(R^2\). Calculate as:\[R = \sqrt{\frac{(a_1-b_1)^2}{4} + \frac{(a_2-b_2)^2}{4} + \frac{(a_3-b_3)^2}{4}}\].

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental concept in vector calculus, often used to find the angle between two vectors or to project one vector onto another. Given two vectors, \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their dot product is calculated as:\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\]For the exercise at hand, the dot product is being applied to vectors resulted from subtracting given vectors from \( \mathbf{r} \), which leads to the expansion of polynomial expressions. This results in an equation set to zero, a crucial step that ties dot product to geometric interpretations like the sphere.

• The dot product returns a scalar (not a vector).
• It is commutative: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
• If the dot product is zero, the vectors are orthogonal.

In this scenario, the zero result implies orthogonality, paving the way to discovering a geometric entity: a sphere.

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial, facilitating the finding of roots, as well as transforming equations into a recognizable format like the circle or sphere equation. In the context of the exercise, completing the square for each variable \(x\), \(y\), and \(z\) helps factorize and simplify the expanded dot product equation.

To complete the square for the general term \(x^2 + bx\), proceed as follows:\[x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\]Here, by halving the linear coefficient \(b\), squaring it, and adding it to the expression, it results in a perfect square trinomial. This transformation is key in rewriting the equation of a sphere in its standard form.

• This technique reveals the geometric center and radius of spheres in algebraically derived forms.
• Each completed square represents a coordinate shift, aligning the variable terms to the form of a sphere equation.
• It simplifies solving quadratics by making them easier to interpret geometrically.

Sphere Equation

The sphere equation is typically represented in its standard form, signifying a three-dimensional round object where all surface points are equidistant from the center.For a sphere centered at \( (h, k, l) \) with radius \( R \), the equation is:\[(x - h)^2 + (y - k)^2 + (z - l)^2 = R^2\]The solutions from the exercise apply the concept of dot product and completing the square to transition from a polynomial condition to this geometric form.

In our particular context, using the computation from the vectors, the center and radius deduced from the calculation indicate how algebraic manipulations yield geometric insights. Specifically, completing the square provided the adjustments required to center the equation.

• The center is calculated using the midpoint formula from vector components, allowing the sphere's position to be located in space.
• The radius \( R \) is the square root of the result after simplification, indicating the size of the sphere.
• This shows how a vector equation model can depict real spatial properties of objects like spheres.