University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Practice Exercises - Page 639: 54



Work Step by Step

We have the equation of a plane: $n=\lt 2,-1,-1 \gt$ Now, $3(2t)-5(-t)+2(-t)=6$ or, $9t =6 \implies t=\dfrac{2}{3}$ Thus, our parametric equations become: $x=2(\dfrac{2}{3})=\dfrac{4}{3}; y=-\dfrac{2}{3}; z=-\dfrac{2}{3}$ Hence, the line will meet the plane at the point: $(\dfrac{4}{3},\dfrac{-2}{3},\dfrac{-2}{3})$
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