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: 53



Work Step by Step

Since, we have the parametric equations $x=1-t; y=3t; z=1+t$ Equation of plane $2x-y+3z=6$ Plug all the parametric equations $x=1-t; y=3t; z=1+t$ in the equation of plane, we have $t=\dfrac{-1}{2}$ Thus, we have the Parametric equations as follows: $x=1-(\dfrac{-1}{2})=\dfrac{3}{2}; y=3(\dfrac{-1}{2})=\dfrac{-3}{2}; z=1+(\dfrac{-1}{2})=\dfrac{1}{2}$ Hence, the line will meet at the point: $(\dfrac{3}{2},\dfrac{-3}{2},\dfrac{1}{2})$
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