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 609: 39



Work Step by Step

Since, we have $\overrightarrow{AB}=i+4j-2k$ and suppose $A=(p,q,r)$ $\overrightarrow{AB}=(5-p)i+(1-q)j+(3-r)k=i+4j-2k$ Now, $5-p=1, 1-q=4, 3-r=-2$ Thus, $p=4,q=-3,r=5$ Hence, $A=(4,-3,5)$
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