University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Section 11.2 - Vectors - Exercises - Page 608: 10

Answer

$\lt -1,1 \gt$

Work Step by Step

Let us consider $O$ as origin and $P$ as the mid-point of segment $RS$. The mid-point can be found as: $(\dfrac{x_1+y_1}{2},\dfrac{x_2+y_2}{2})$ $P(\dfrac{x_1+y_1}{2},\dfrac{x_2+y_2}{2})=P(\dfrac{2+(-4)}{2},\dfrac{-1+3}{2})= \lt -1,1 \gt$ $OP=P-O=\lt -1,1 \gt -\lt 0,0 \gt =\lt -1,1 \gt$ Hence, our final answer is: $\lt -1,1 \gt$
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