Algebra and Trigonometry 10th Edition

$\lt \dfrac{9}{5},\dfrac{12}{5} \gt$
We are given that $||\overrightarrow{v}|| = 3$. So, $\cos \theta =\dfrac{3}{5}; \sin \theta =\dfrac{4}{5}$ We know that $v=||v|| \lt \cos \theta , \sin \theta \gt$ Thus, $v=3 \lt \dfrac{3}{5},\dfrac{4}{5} \gt \\= \lt (3) (\dfrac{3}{5}),(3) (\dfrac{4}{5} ) \gt \\= \lt \dfrac{9}{5},\dfrac{12}{5} \gt$