University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 60

Answer

$x=1+10t; y=-3+25t; z=1+20t$

Work Step by Step

Let us take $z=1$ We have the equation of a plane: $5x-2y=11$ and $4y-5=-17$ $\implies y=-3$ and $x=\dfrac{11+2y}{5}=\dfrac{11+2(-3)}{5}=1$ Thus, $r_0=\lt 1,-3,1 \gt$ We know that $r=r_0+tv$ and $v=\lt 10,25,20 \gt$ Thus, our parametric equations become: $x=1+10t; y=-3+25t; z=1+20t$ Hence, our parametric equations are: $x=1+10t; y=-3+25t; z=1+20t$
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