Answer
$(1, -1,1) \Longrightarrow (\sqrt{8}, \frac{\pi}{4},1) $.
Work Step by Step
We have $$ x=2, \quad y =2, \quad z=1.$$
Now, we have
$$ r=\sqrt{x^2+y^2}=\sqrt{8},$$
$$\theta =\tan^{-1}y/x=\tan^{-1}\frac{1}{1}=\frac{\pi}{4}=\frac{\pi}{4},$$
$$ z=1.$$
So, $(1, -1,1) \Longrightarrow (\sqrt{8}, \frac{\pi}{4},1) $.
