Answer
$\sqrt 6 ; \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$
Work Step by Step
Here, $|v|=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt {6}$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$
Now, $\hat{\textbf{u}}=\dfrac{\lt 1,2,-1 \gt}{\sqrt 6}= \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$
Thus, our final answers are: $\sqrt 6 ; \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$
