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: 12

Answer

$x=0,y=0, z=t$

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, $P=(0,0,0)$; the point P lies on z-axis this means that, we have the vector $v=\lt 0,0,1 \gt$ Our parametric equations are: $x=0+0t,y=0+0t, z=0+1t$ or, $x=0,y=0, z=t$
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