## Thomas' Calculus 13th Edition

$\lt \dfrac{-1}{2},\dfrac{-\sqrt 3}{2} \gt$
Since, we have $v=\lt |v| \cos \theta, |v| \sin \theta \gt$ to compute the components of vector $v$ Given: $|v|=1$ and $\theta=\dfrac{2\pi}{3}$ Thus, $v=\lt (1) \cos (\dfrac{2\pi}{3}), (1) \cos (\dfrac{2\pi}{3}) \gt$ or, $v=\lt \dfrac{-1}{2},\dfrac{-\sqrt 3}{2} \gt$