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: 28

Answer

$(x-2)^2 + (y+2)^2 + (z-5)^2 = 8$

Work Step by Step

Midpoint: $(\frac{-4+0}{2},\frac{2+2}{2},\frac{3+7}{2}) = (-2,2,5)$ Distance between $P$ and $Q$: $\sqrt {(0+4)^2+(2-2)^2+(7-3)^2} = \sqrt {16 + 0 + 16} = \sqrt {32}$ 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: $(x-2)^2 + (y+2)^2 + (z-5)^2 = (\frac{\sqrt {32}}{2})^2$ Simplify: $(x-2)^2 + (y+2)^2 + (z-5)^2 = 8$
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