## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.2 - Vectors - Exercises 12.2 - Page 705: 38

#### Answer

$(1,-1,-1)$

#### 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 and $u=2i-2j-2k$; $|u|=\sqrt{(2)^2+(-2)^2+(-2)^2}=\sqrt{4+4+4}= 2\sqrt {3}$ Thus, $\hat{\textbf{u}}=\dfrac{2i-2j-2k}{2 \sqrt 3}=(\dfrac{1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k)$ Apply Mid-point formula. Thus, we have $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})=(\dfrac{0+2}{2},\dfrac{0+(-2)}{2},\dfrac{0+(-2)}{2})$ or, $=(1,-1,-1)$

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