Answer
$2$
Work Step by Step
The magnitude of a vector $u= \left\langle a,b\right \rangle $ is given by
$$|u|=\sqrt {a^2+b^2}$$ From the given figure, note that the vector $u= \left\langle \sqrt 3,1\right \rangle$.
$u= \left\langle \sqrt 3,1\right \rangle$ implies $a=\sqrt {3}~\text{and}~ b=1$ .
Now use the formula $|u|=\sqrt {a^2+b^2}$ to find magnitude of $u$.
$$|u|=\sqrt {(\sqrt {3})^2+1^2}=\sqrt {3+1}=\sqrt 4=2$$ Therefore, the magnitude of vector $u$ is $2$.
