Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.2 Vectors in Three Dimensions - 11.2 Exercises: 27

Answer

$(x-\frac{3}{2})^2 + (y-\frac{3}{2})^2 + (z-7)^2 = \frac{13}{2}$

Work Step by Step

Midpoint: $(\frac{1+2}{2},\frac{0+3}{2},\frac{5+9}{2}) = (\frac{3}{2},\frac{3}{2},7)$ Distance between $P$ and $Q$: $\sqrt {(2-1)^2+(3-0)^2+(9-5)^2} = \sqrt {1 + 9 + 16} = \sqrt {26}$ Center = Midpoint Radius = $\frac{(distance\ between\ P\ and\ Q)}{2}$ Equation of Sphere with Center $(a,b,c)$ and radius $r$ $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ Plug in the center and radius: $(x-\frac{3}{2})^2 + (y-\frac{3}{2})^2 + (z-7)^2 = (\frac{\sqrt {26}}{2})^2$ Simplify: $(x-\frac{3}{2})^2 + (y-\frac{3}{2})^2 + (z-7)^2 = \frac{13}{2}$
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