## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 40

#### Answer

$(\dfrac{4}{3},\dfrac{-2}{3},\dfrac{-2}{3})$

#### Work Step by Step

As we given are that the parametric equations are: $x=2t; y=-t; z=-t$ The equation of a normal vector to the plane is: $n=\lt 2,-1,-1 \gt$ this implies, $3(2t)-5(-t)+2(-t)=6$ or, $9t =6 \implies t=\dfrac{2}{3}$ and $x=2(\dfrac{2}{3})=\dfrac{4}{3}; y=-\dfrac{2}{3}; z=-\dfrac{2}{3}$ Hence, the line will meet the plane at the point: $(\dfrac{4}{3},\dfrac{-2}{3},\dfrac{-2}{3})$

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