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: 37


$\dfrac{-1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$ and $(\dfrac{5}{2},\dfrac{7}{2},\dfrac{9}{2})$

Work Step by Step

Let us consider $P_1$ and $P_2$ as vectors and, a vector $u=\overrightarrow{P_1P_2}=P_2-P_1=-i-j-k$ and $|u|=\sqrt{(1)^2+(-1)^2+(-1)^2}=\sqrt {3}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Now, $\hat{\textbf{u}}=\dfrac{-i-j-k}{\sqrt 3}=\dfrac{-1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$ The mid-point can be found as: $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$ Thus, $(\dfrac{3+2}{2},\dfrac{4+3}{2},\dfrac{5+ 4}{2})=(\dfrac{5}{2},\dfrac{7}{2},\dfrac{9}{2})$
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