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: 55



Work Step by Step

We have the equation of a plane: $n=\lt 2,-1,2 \gt$ Now, $2x-y+2z=-2$ or, $2(3+2t)-(2-t)+2(1+2t)=-2$ or, $9t =-8 \implies t=\dfrac{-8}{9}$ Thus, our parametric equations become: $x=3+2(\dfrac{-8}{9})=\dfrac{11}{9}; y=2-(-\dfrac{2}{3})=\dfrac{26}{9}; z=1+2(\dfrac{-8}{9})=\dfrac{-7}{9}$ Hence, the line will meet the plane at the point: $(\dfrac{11}{9},\dfrac{26}{9},\dfrac{-7}{9})$
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