Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 646: 72

Answer

$ \ Radius, r= \sqrt 1=1$ and Center at $(-1,0,1)$

Work Step by Step

Let $(a,b,c)$ be the center of the sphere, $(x,y,z)$ be an arbitrary point of the sphere, and $r$ be the distance between $(a,b,c)$ and $(x,y,z)$. Then, using the Pythagorean Theorem, the standard form of the equation of the sphere can be expressed as: $r=\sqrt {(x-a)^2+(y-b)^2+(z-c)^2}$ or, $(x-a)^2+(y-b)^2+(z-c)^2=r^2 ...(1)$ We are given: $x^2+y^2+z^2+2x-2z=-1$ We complete the square as: follows: $(x^2+2x) +(y^2) + (z^2-2z) =-1 \\ (x^2+2x+1) +y^2 +(z^2-2z+1) =-1+1+1 \\ (x+1)^2+(y-0)^2+(z-1)^2 =1$ On comparing the above equation with equation (1), we have: $ \ Radius, r= \sqrt 1=1$ and Center at $(-1,0,1)$
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