Answer
\[\begin{align}
& \left( \mathbf{a} \right)\left( \frac{\sqrt{3}}{2},\frac{\pi }{3},\frac{3\sqrt{3}}{2} \right) \\
& \left( \mathbf{b} \right)\left( \frac{\sqrt{30}}{2},\frac{\pi }{3},\arccos \left( \frac{3\sqrt{10}}{10} \right) \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \left( \frac{\sqrt{3}}{4},\frac{3}{4},\frac{3\sqrt{3}}{2} \right) \\
& \left( \mathbf{a} \right)\text{We have the rectangular coordinates }\left( \frac{\sqrt{3}}{4},\frac{3}{4},\frac{3\sqrt{3}}{2} \right) \\
& \left( x,y,z \right):\text{ }\left( \frac{\sqrt{3}}{4},\frac{3}{4},\frac{3\sqrt{3}}{2} \right) \\
& \text{Rectangular to cylindrical} \\
& {{r}^{2}}={{x}^{2}}+{{y}^{2}}\to r=\sqrt{{{\left( \frac{\sqrt{3}}{4} \right)}^{2}}+{{\left( \frac{3}{4} \right)}^{2}}}=\frac{\sqrt{3}}{2} \\
& \tan \theta =\frac{y}{x}\to \theta ={{\tan }^{-1}}\left( \frac{3/4}{\sqrt{3}/4} \right)=\frac{\pi }{3} \\
& z=z\to z=\frac{3\sqrt{3}}{2} \\
& \text{The cylindrical coordinates are:} \\
& \left( \frac{\sqrt{3}}{2},\frac{\pi }{3},\frac{3\sqrt{3}}{2} \right) \\
& \\
& \left( \mathbf{b} \right)\text{We have the rectangular coordinates }\left( \frac{\sqrt{3}}{4},\frac{3}{4},\frac{3\sqrt{3}}{2} \right) \\
& \left( x,y,z \right):\left( \frac{\sqrt{3}}{4},\frac{3}{4},\frac{3\sqrt{3}}{2} \right) \\
& \text{Rectangular to spherical }\left( \rho ,\theta ,\phi \right) \\
& {{\rho }^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}\to \rho =\sqrt{{{\left( \frac{\sqrt{3}}{4} \right)}^{2}}+{{\left( \frac{3}{4} \right)}^{2}}+{{\left( \frac{3\sqrt{3}}{2} \right)}^{2}}}=\frac{\sqrt{30}}{2} \\
& \tan \theta =\frac{y}{x}\to \theta ={{\tan }^{-1}}\left( \frac{3/4}{\sqrt{3}/4} \right)=\frac{\pi }{3} \\
& \phi =\arccos \left( \frac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \right)\to \phi =\arccos \left( \frac{3\sqrt{3}/2}{\sqrt{30}/2} \right) \\
& \phi =\arccos \left( \frac{3\sqrt{10}}{10} \right) \\
& \text{The spherical }\left( \rho ,\theta ,\phi \right)\text{ coordinates are:} \\
& \left( \frac{\sqrt{30}}{2},\frac{\pi }{3},\arccos \left( \frac{3\sqrt{10}}{10} \right) \right) \\
\end{align}\]
