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 726: 8

Answer

$x=2+3t,y=4+7t, z=5-5t$

Work Step by Step

As we know the parametric equations of a straight line for a vector $v=v_1i+v_2j+v_3k$ passing through a point $P(a,b,c)$ is given by $x=a+t v_1,y=b+t v_2; z=c+t v_3$ Since, we have the vector $v=\lt 3,7,-5 \gt$ is perpendicular to the plane at point $P(2,4,5)$.Then our parametric equations are: $x=2+3t,y=4+7t, z=5-5t$
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