Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 894: 18

Answer

$r(t) =\lt -1-2t,2+3t,-2+3t \gt$ and $ \\0 \leq t \leq 1$ $x=-1-2t$ and $ y= 2+3t $and $z=-2+3t$ ; \\0 \leq t \leq 1$

Work Step by Step

The General vector line equation for the given two points is defined as: $r(t)=(1-t) r_0+t \times r_1$ Now, we have $r(t) =\lt -1, 2,-2 \gt +t \lt -3, 5,1 \gt$ $\implies \lt -1+t,2-2 \times t, -2+2 \times t \gt + \lt -(3) \times t,5 \times t,t \gt$ $\implies \lt -1-2t,2+3t,-2+3t \gt$ Hence, $r(t) =\lt -1-2t,2+3t,-2+3t \gt$ and $ \\0 \leq t \leq 1$ Also, the parametric equations are as follows: $x=-1-2t$ and $ y= 2+3t $and $z=-2+3t$ ;$0 \leq t \leq 1$
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