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: 36



Work Step by Step

Formula to calculate the unit vector $\hat{\textbf{u}}$ is: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Since, $u=\overrightarrow{P_1P_2}=P_2-P_1$ Here,$P_1$ and $P_2$ are two vectors. or, $u=(4-1)i+(-2-4)j+(7-5)k=3i-6j+2k$; $|u|=\sqrt{(3)^2+(-6)^2+(2)^2}=\sqrt {49}=7$ Thus, $\hat{\textbf{u}}=\dfrac{3i-6j+2k}{7}=(\dfrac{3}{7} i-\dfrac{6}{7}j+\dfrac{2}{7}k)$ Apply Mid-point formula. Thus, we have $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})=(\dfrac{1+4}{2},\dfrac{4+(-2)}{2},\dfrac{5+7}{2})$ or, $=(\dfrac{5}{2},1,6)$
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