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: 34


$\sqrt 3(-i+j+k)$

Work Step by Step

Here, $u=\dfrac{1}{2}i -\dfrac{1}{2}j-\dfrac{1}{2}k$ and $|u|=\sqrt{(\dfrac{1}{2})^2+(\dfrac{-1}{2})^2+(\dfrac{-1}{2})^2}=\dfrac{\sqrt {3}}{2}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Now, $\hat{\textbf{u}}=\dfrac{\dfrac{1}{2}i -\dfrac{1}{2}j-\dfrac{1}{2}k}{\dfrac{\sqrt {3}}{2}}=\dfrac{1}{\sqrt 3}(i-j-k)$ Thus, $-3\hat{\textbf{u}}=\dfrac{-3}{\sqrt 3}(i-j-k)=\sqrt 3(-i+j+k)$
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