## Algebra and Trigonometry 10th Edition

$||v||=\sqrt {65}$ $θ=119.74°$
$v=ai+bj=-4i+7j~~$ ($a\lt0$ and $b\gt0~→~Quadrant~II$) $||v||=\sqrt {v·v}=\sqrt {(-4i+7j)·(-4i+7j)}=\sqrt {(-4)^2+7^2}=\sqrt {16+49}=\sqrt {65}$ $tan~θ=\frac{b}{a}=\frac{7}{-4}$ $θ=arctan(\frac{7}{-4})+180°=-60.26°+180°=119.74°$ Notice that $-60.26°$ lies in Quadrant IV.