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Chapter 12: Problem 28

Find an equation of the sphere passing through \(P(-4,2,3)\) and \(Q(0,2,7)\) with its center at the midpoint of \(P Q\)

Short Answer

Expert verified
Answer: The equation of the sphere is \((x+2)^2 + (y-2)^2 + (z-5)^2 = 8\).

Step by step solution

01

Find the midpoint/center of the sphere

To find the center of the sphere, which is the midpoint M of points P and Q, we can use the midpoint formula: \(M(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})\). Plugging in the coordinates of points P and Q, we get: \(M(\frac{-4+0}{2}, \frac{2+2}{2}, \frac{3+7}{2}) = M(-2, 2, 5)\) So, the center of the sphere is at point M(-2, 2, 5).
02

Find the radius of the sphere

Now we need to find the radius of the sphere. We can do this by finding the distance between points M(-2, 2, 5) and P(-4, 2, 3) using the distance formula: \(r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\). Plugging in the coordinates, we get: \(r = \sqrt{(-4-(-2))^2 + (2-2)^2 + (3-5)^2}\) \(r = \sqrt{(-2)^2 + (0)^2 + (-2)^2}\) \(r = \sqrt{4 + 0 + 4}\) \(r = \sqrt{8}\) So, the radius of the sphere is \(\sqrt{8}\).
03

Write the equation of the sphere

Finally, we can write down the equation of the sphere in its standard form: \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), where M(a, b, c) is the center and r is the radius. Plugging in the values M(-2, 2, 5) and radius \(\sqrt{8}\), we get: \((x-(-2))^2 + (y-2)^2 + (z-5)^2 = (\sqrt{8})^2\) \((x+2)^2 + (y-2)^2 + (z-5)^2 = 8\) So, the equation of the sphere passing through points P(-4, 2, 3) and Q(0, 2, 7) and having its center at the midpoint of PQ is \((x+2)^2 + (y-2)^2 + (z-5)^2 = 8\).

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Most popular questions from this chapter

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).

Consider the lines $$\begin{aligned} \mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \text { and } \\ \mathbf{R}(s) &=\langle 6+s, 10-2 s, 16-s\rangle. \end{aligned}$$ a. Determine whether the lines intersect (have a common point) and if so, find the coordinates of that point. b. If \(\mathbf{r}\) and \(\mathbf{R}\) describe the paths of two particles, do the particles collide? Assume \(t \geq 0\) and \(s \approx 0\) measure time in seconds, and that motion starts at \(s=t=0\).

Prove the following identities. Assume that \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) and \(\mathbf{x}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$(\mathbf{u} \times \mathbf{v}) \cdot(\mathbf{w} \times \mathbf{x})=(\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{x})-(\mathbf{u} \cdot \mathbf{x})(\mathbf{v} \cdot \mathbf{w})$$

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). If an object decelerates in the sense that \(d^{2} s / d t^{2}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N}\)

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3,0\rangle, \mathbf{v}=\langle 1,1,1\rangle\)
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