Calculus 8th Edition

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

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises - Page 883: 22


$\frac{3}{\sqrt 2}$ or, $\frac{3\sqrt 2}{2} $ or $\approx 2.12$

Work Step by Step

$1(1+t)-1(2-t)+2(-1+2t)=0$ $0=1+t-2+t-2+4t$ $6t-3=0$ $t= \frac{1}{2}$ Point on the line closet to the origin is: $(1+1/2,2-1/2,-1+2(1/2))=(3/2,3/2,0)$ Distance$=\sqrt {(3/2-0)^2+(3/2-0)^2+(0-0)^2}$ $=\sqrt \frac{9}{2}$ $=\frac{3}{\sqrt 2}$ $=\frac{3\sqrt 2}{2} \approx 2.12$ Answers are: $\frac{3}{\sqrt 2}$ or, $\frac{3\sqrt 2}{2} $ or $\approx 2.12$
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