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 704: 7


$\lt \dfrac{1}{5},\dfrac{14}{5}\gt$ and $\dfrac{\sqrt{197}}{5}$

Work Step by Step

The magnitude of a vector is:$|n|=\sqrt{n_1^2+n_2^2}$ Here, $u=\lt 3,-2 \gt; v= \lt -2,5 \gt$ Now, $\dfrac{3}{5}u +\dfrac{4}{5}v=\dfrac{3}{5}\lt 3,-2 \gt +\dfrac{4}{5} \lt -2,5 \gt$ or, $ =\lt \dfrac{1}{5},\dfrac{14}{5} \gt$ and $|\lt \dfrac{1}{5},\dfrac{14}{5}\gt|=\sqrt{(\dfrac{1}{5})^2+(\dfrac{14}{5})^2}=\sqrt{\dfrac{1}{25}+\dfrac{196}{25}}=\dfrac{\sqrt{197}}{5}$
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