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: 26

Answer

$(x)^2 + (y+2)^2 + (z-6)^2 \leq 41$

Work Step by Step

Equation of a sphere with center $(a,b,c)$ and radius $r$. $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ Plugging in the point $(0,-2,6)$ $(x-0)^2 + (y+2)^2 + (z-6)^2 = r^2$ Plugging in the point $(1,4,8)$ $(1)^2 + (4+2)^2 + (8-6)^2 = 1 + 36 + 4 = 41$ Equation of Sphere: $(x)^2 + (y+2)^2 + (z-6)^2 = 41$ Since it's a ball (not a hollow center): $(x)^2 + (y+2)^2 + (z-6)^2 \leq 41$
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