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: 37



Work Step by Step

Formula to calculate the unit vector $\hat{\textbf{u}}$ is: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ $u=\overrightarrow{P_1P_2}=P_2-P_1$ Here,$P_1$ and $P_2$ are two vectors . or, $u=-i-j-k$ Now $|u|=\sqrt{(1)^2+(-1)^2+(-1)^2}=\sqrt {3}$ Thus, $\hat{\textbf{u}}=\dfrac{-i-j-k}{\sqrt 3}=(\dfrac{-1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k)$ Apply Mid-point formula. Thus, we have $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})=(\dfrac{3+2}{2},\dfrac{4+3}{2},\dfrac{5+ 4}{2})$ or, $=(\dfrac{5}{2},\dfrac{7}{2},\dfrac{9}{2})$
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