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 - Practice Exercises - Page 734: 32


$x=1,y=2+t, z=-t$

Work Step by Step

The parametric equation of a straight line is defined as $v=v_1i+v_2j+v_3k$, and when it passes through a point $P(x_0,y_0,z_0)$, it is given as: $x=x_0+t v_1,y=y_0+t v_2; z=z_0+t v_3$ Here, $v=\lt 0,1,-1 \gt$ and $P=(1,2,0)$ . Now, the parametric equations are: $x=1+(0)t,y=2+t(1), z=0-t$ Hence, $x=1,y=2+t, z=-t$
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