Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 1 - Vectors - 1.1 The Geometry and Algebra of Vectors - Exercises 1.1: 13

Answer

$u = [\frac{1}{2}, \frac{\sqrt{3}}{2}]$ $v = [-\frac{\sqrt{3}}{2}, -\frac{{1}}{2}]$ $u+v = [\frac{1-\sqrt{3}}{2}, \frac{\sqrt{3}-1}{2}]$ $u-v = [\frac{1+\sqrt{3}}{2}, 1+\frac{\sqrt{3}}{2}]$

Work Step by Step

Knowing that $\mid u \mid = 1$ and $u$ is $60$ degrees counterclockwise away from the x-axis and into the first quadrant, $u = 1 [\cos(60), \sin(60)] = [\frac{1}{2}, \frac{\sqrt{3}}{2}]$. Since that $\mid v \mid = 1$ and $v$ is $30$ degrees counterclockwise away from the x-axis and into the third quadrant, $v = 1 [-\cos(30), - \sin(30)] = [-\frac{\sqrt{3}}{2}, -\frac{{1}}{2}]$. Given the above components of $u$ and $v$, we can compute $u+v = [\frac{1}{2}, \frac{\sqrt{3}}{2}] + [-\frac{\sqrt{3}}{2}, -\frac{{1}}{2}]$, which equals $ [\frac{1-\sqrt{3}}{2}, \frac{\sqrt{3}-1}{2}]$. Similarly, we compute $u - v = [\frac{1}{2}, \frac{\sqrt{3}}{2}] - [-\frac{\sqrt{3}}{2}, -\frac{{1}}{2}]$, which equals $ [\frac{1+\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}]$.
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