Answer
$\sqrt 2 ; \lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$
Work Step by Step
Here, $|v|=\sqrt{(-1)^2+(-1)^2}=\sqrt {2}$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$
Now, $\hat{\textbf{u}}=\dfrac{\lt -1,-1 \gt}{\sqrt 2}= \lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$
Thus, our final answers are: $\sqrt 2 ; \lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$
