Answer
$(1, -1,1) \Longrightarrow (\sqrt{2}, \frac{7\pi}{4},1) $.
Work Step by Step
We have $$ x=1, \quad y =-1, \quad z=1.$$
Now, we have
$$ r=\sqrt{x^2+y^2}=\sqrt{2},$$
$$\theta =\tan^{-1}y/x=\tan^{-1}\frac{1}{-1}=-\frac{\pi}{4}+2\pi=\frac{7\pi}{4},$$
$$ z=1.$$
So, $(1, -1,1) \Longrightarrow (\sqrt{2}, \frac{7\pi}{4},1) $.
