Answer
$2 ; \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
Work Step by Step
As we know that the unit vector $\hat{\textbf{u}}$ is defined as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$
Now, we have, $|v|=\sqrt{(\sqrt 2)^2+(\sqrt 2)^2}=\sqrt {4}=2$
and $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt \sqrt 2, \sqrt 2 \gt}{2}= \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
