Answer
$2 ; \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
Work Step by Step
Here, $|v|=\sqrt{(\sqrt 2)^2+(\sqrt 2)^2}=\sqrt {4}=2$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$
Now, $\hat{\textbf{u}}=\dfrac{\lt \sqrt 2, \sqrt 2 \gt}{2}= \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
Thus, our final answers are: $2 ; \lt \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
