Calculus: Early Transcendentals 8th Edition

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

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 831: 18

Answer

$x(t)=-2+13t,y(t)=18-22t,z(t)=31+17t$, for $0\leq t \leq 1$

Work Step by Step

The vector equation of a line segment from $r_0$ to $r_1$ is: $r(t)=(1-t)r_0+tr_1$, for $0\leq t \leq 1$ $r(t)=(1-t)\lt-2,18,31\gt+t\lt11,-4,48\gt$ $r(t)=\lt-2+2t,18-18t,31-31t\gt+\lt11t,-4t,48t\gt$ $r(t)=\lt-2+13t,18-22t,31+17t\gt$ Hence, the parametric equations are: $x(t)=-2+13t,y(t)=18-22t,z(t)=31+17t$, for $0\leq t \leq 1$
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