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

$\lt \dfrac{-\sqrt 3}{2},\dfrac{-1}{2} \gt$
The vector $\lt 0,1 \gt$ is equivalent to rotating the unit vector $<1,0>$ $90^{\circ}$ counterclockwise about the origin. So, the unit vector makes an angle with the positive x-axis of $120^{\circ} + 90^{\circ}=210^{\circ}$ Therefore, $\lt 1*\cos (210^{\circ}), 1*\sin (210^{\circ}) \gt=\lt \dfrac{-\sqrt 3}{2},\dfrac{-1}{2} \gt$