Answer
$\lt \dfrac{\sqrt {10}}{5},\dfrac{3 \sqrt {10}}{5} \gt$
Work Step by Step
We are given that $ i+3j=\lt 1, 3 \gt$ and $||\overrightarrow{v}|| =2$
So, $\sqrt {1^2+(3)^2}=\sqrt {10}$
and $ \cos \theta =\dfrac{1}{\sqrt {10}}; \sin \theta =\dfrac{3}{\sqrt {10}}$
We know that $v=||v|| \lt \cos \theta , \sin \theta \gt $
Thus, $v=2 \lt \dfrac{1}{\sqrt {10}},\dfrac{3}{\sqrt {10}} \gt \\= \lt \dfrac{2 \sqrt {10}}{10},\dfrac{6 \sqrt {10}}{10} \gt \\= \lt \dfrac{\sqrt {10}}{5},\dfrac{3 \sqrt {10}}{5} \gt$
