Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.2 - Vectors - Exercises 12.2 - Page 705: 39



Work Step by Step

Since, we are given that $\overrightarrow{AB}=i+4j-2k$ Now, let us suppose $A=(p,q,r)$ This implies that $\overrightarrow{AB}=(5-p)i+(1-q)j+(3-r)k$ or, $(5-p)i+(1-q)j+(3-r)k=i+4j-2k$ Equate the terms. $5-p=1, 1-q=4, 3-r=-2$ So, $p=4,q=-3,r=5$ Hence, we have out point: $A=(4,-3,5)$
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