## University Calculus: Early Transcendentals (3rd Edition)

$\lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$
Use formula $v=\lt |v| \cos \theta, |v| \sin \theta \gt$ Here, $|v|=1$ and $\theta=\dfrac{-3\pi}{4}$ Thus, $v=\lt (1) \cos (\dfrac{-3\pi}{4}), (1) \cos (\dfrac{-3\pi}{4}) \gt=\lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$