Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.7 - Maximum and Minimum Values - 14.7 Exercise - Page 969: 41

Answer

$\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$

Work Step by Step

Formula to calculate the shortest distance is: $D=|\dfrac{pa+qb+rc-s}{\sqrt{p^2+q^2+r^2}}|$ From the given question, we have $D=|\dfrac{(1)(2)+(1)(0)+(1)(-3)-1}{\sqrt{1^2+1^2+1^2}}|$ $D=\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$
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