## Linear Algebra: A Modern Introduction

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