Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.1 Vectors in the Plane - 11.1 Exercises - Page 768: 26

Answer

$\overrightarrow{PQ}=\overrightarrow{RS}$, $\overrightarrow{TU}\not=\overrightarrow{PQ}$, $\overrightarrow{TU}\not=\overrightarrow{RS}$

Work Step by Step

We are given the points: P(−3,−1) Q(−1,2) R(1,2) S(3,5) T(4,2) U(6,4) Determine $\overrightarrow{PQ},\overrightarrow{RS},\overrightarrow{TU}$: $\overrightarrow{PQ}(-1-(-3),2-(-1))=(2,3)$ $\overrightarrow{RS}(3-1,5-2)=(2,3)$ $\overrightarrow{TU}(6-4,4-2)=(2,2)$ This means that $\overrightarrow{PQ}=\overrightarrow{RS}$, but $\overrightarrow{TU}\not=\overrightarrow{PQ}$, $\overrightarrow{TU}\not=\overrightarrow{RS}$
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