University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 638: 32

Answer

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

Work Step by Step

The parametric equations of a straight line can be found by knowing the value of a vector such as $v=v_1i+v_2j+v_3k$ passing through a point $P(x_0,y_0,z_0)$ as follows: $x=x_0+t v_1,y=y_0+t v_2; z=z_0+t v_3$ Here, we have the vector $v=\lt 0,1,-1 \gt$ and $P=(1,2,0)$ . Thus, we get the parametric equations: $x=1+(0)t,y=2+t(1), z=0-t$ Hence, $x=1,y=2+t, z=-t$
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