Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 11 - Three-Dimensional Space; Vectors - 11.1 Rectangular Coordinates In 3-Space; Spheres; Cylindrical Surfaces - Exercises Set 11.1 - Page 772: 50

Answer

Sphere centered at $C\left(-\frac{1}{3}, 2, \frac{4}{3}\right)$ of radius $r=\frac{2 \sqrt{11}}{3}$

Work Step by Step

It's given that \[ 2 d(P, B)=d(P, A) \] where $A, B, P$ have coordinates \[ P(x, y, z), \quad B(0,1,1) \text { and } A(1,-2,0) \] and then \[ \begin{aligned} 4 d(P, B) = d^{2}(P, A) \\ (2+y)^{2}+(-1+x)^{2}+(z-0)^{2}=4\left[(-1+z)^{2}+(x-0)^{2}+(-1+y)^{2}\right] \\ \Rightarrow y^{2}+4 y+4+z^{2}+x^{2}-2 x+1=& 4 y^{2}-8 y+4+4 z^{2}-8 z+4+4 x^{2} \\ \Rightarrow 3 z^{2}+2 x-12 y+3 x^{2}+3 y^{2}-8 z=& \\ \Rightarrow y^{2}-4 y+x^{2}+\frac{2}{3} x+z^{2}-\frac{8}{3} z=-1 & \\ \Rightarrow\left(\frac{2}{3} x+x^{2}+\frac{1}{9}\right)+\left(y^{2}-4 y+4\right)+\left(z^{2}-\frac{8}{3} z+\frac{16}{9}\right)=\frac{1}{9}+4+\frac{16}{9}-1 & \\ \Rightarrow\left(\frac{1}{3}+x\right)^{2}+(-2+y)^{2}+(-4 / 3+z)^{2}=\frac{44}{9}=\left(\frac{2 \sqrt{11}}{3}\right)^{2} \end{aligned} \] This equation corresponds to a sphere centered at $C\left(-\frac{1}{3}, 2, \frac{4}{3}\right)$ with the radius \[ \frac{2 \sqrt{11}}{3}=r \]
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