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: 35



Work Step by Step

Formula to calculate the unit vector $\hat{\textbf{u}}$ is: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Consider $u=P_2-P_1=\lt 2,5,0 \gt -\lt -1,1,5 \gt =\lt 3,4,-5 \gt$ Here,$P_1$ and $P_2$ are two vectors Also, $u=3 i+4j-5k$; $|u|=\sqrt{(3)^2+(4)^2+(-5)^2}=\sqrt{9+16+25}=\sqrt {50}=5 \sqrt 2$ Thus, $\hat{\textbf{u}}=\dfrac{3 i+4j-5k}{5 \sqrt 2}=\dfrac{3}{5\sqrt 2} i+\dfrac{4}{5\sqrt 2}j -\dfrac{1}{\sqrt 2 }k$ Apply mid-point formula such ass:$(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$ Hence, $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})=(\dfrac{2+(-1)}{2},\dfrac{5+1}{2},\dfrac{0+5}{2})=(\dfrac{1}{2},3,\dfrac{5}{2})$
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