Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Problems Plus - Problems - Page 884: 1


$r=(\sqrt 3-\dfrac{3}{2})m$

Work Step by Step

Suppose $1$ ball has center at $(r,r,r)$ $(x-r)^2+(y-r)^2+(z-r)^2=r^2$ $ \lt \frac{m}{2},\frac{m}{2},\frac{m}{2} \gt =(1+\frac{2}{\sqrt 3}) \cdot \lt r,r,r \gt$ $\frac{m}{2}=r(\frac{\sqrt 3+2}{\sqrt3})$ $r=\dfrac{\sqrt 3}{4+2\sqrt3}m$ Hence, $r=(\sqrt 3-\dfrac{3}{2})m$
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