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 - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 54



Work Step by Step

Since, we are given that the parametric equations $x=2; y=3+2t; z=-2-2t$ Equation of plane $6x+3y-4z=-12$ Plug all the parametric equations $x=2; y=3+2t; z=-2-2t$ in equation of plane, we have $t=\dfrac{-41}{14}$ Now, the Parametric equations are: $x=2; y=3+2(\dfrac{-41}{14})=\dfrac{-20}{7}; z=-2-2(\dfrac{-41}{14})=\dfrac{27}{7}$ Hence, the line will meet at the point: $(2,\dfrac{-20}{7},\dfrac{27}{7})$
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