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 832: 45

Answer

$(-2,6,3)$

Work Step by Step

Substitute $x=2-2t$, $y=3t$, $z=1+t$ in $x+2y-z=7$ $(2-2t)+2(3t)-(1+t)=7$ $2-2t+6t-1-t=7$ $t=2$ Substitute $t=2$ in the equation of the line to get the point. $x=2-2.2$, $y=3.2$, $z=1+2$ $x=-2$, $y=6$, $z=3$ Hence, $(-2,6,3)$
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