## Calculus 8th Edition

Published by Cengage

# Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 894: 19

#### Answer

$r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$ $x=\dfrac{t}{2}$ and $y= -1+\dfrac{4t}{3}$ and $z=1-\dfrac{3t}{4}$ ; $0 \leq t \leq 1$

#### 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 $r(t)=(1-t) \times \lt 0,-1, 1 \gt +t \times \lt \dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4} \gt$ $\implies \lt 0,-1+t,1-t \gt + \lt (\dfrac{1}{2})t,(\dfrac{1}{3}) t,(\dfrac{1}{4})t \gt$ $\implies \lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$ Answer: $r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$ Hence, the parametric equations are: $x=\dfrac{t}{2}$ and $y= -1+\dfrac{4t}{3}$ and $z=1-\dfrac{3t}{4}$ ; $0 \leq t \leq 1$

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