Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 735: 55



Work Step by Step

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