Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1162: 65

Answer

The angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is ${{139.8}^{\circ }}$.

Work Step by Step

Consider the vectors, $ v=\text{5}i+2j\text{ and }w=-3i+j $. The angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is, $\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$ To find the value of $\left\| v \right\|\text{ and }\left\| w \right\|$, $\begin{align} & \left\| v \right\|=\sqrt{{{5}^{2}}+{{2}^{2}}} \\ & =\sqrt{25+4} \\ & =\sqrt{29} \end{align}$ And $\begin{align} & \left\| w \right\|=\sqrt{{{\left( -3 \right)}^{2}}+{{1}^{2}}} \\ & =\sqrt{9+1} \\ & =\sqrt{10} \end{align}$ Substitute the values of $\left\| v \right\|\text{ and }\left\| w \right\|$ in $\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$ and solve for $\theta $, $\begin{align} & \theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\ & ={{\cos }^{-1}}\frac{-15+2}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\ & ={{\cos }^{-1}}\frac{-13}{\sqrt{290}}\approx {{139.8}^{\circ }} \end{align}$ Hence, the angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is ${{139.8}^{\circ }}$.
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