Answer
$\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
0 & 0 & 0 & 1
\end{array}\right]$
Work Step by Step
\[
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
0 & \frac{\sqrt{3}}{2} & \frac{1}{2}
\end{array}\right]
\]
First get the $3 \times 3$ matrix for the rotation.
The vector $\mathbf{e}_{1}$ on $\mathrm{x}$ axis does not move, vector $\mathbf{e}_{2}$ stop at $\left(0, \cos 60^{\circ},-\sin 60^{\circ}\right)$ and vector $\mathbf{e}_{3}$ stops at
$\left(0, \sin 60^{\circ}, \cos 60^{\circ}\right) .$ so the standard matrix
for this rotation is here.
\[
\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\]
The required rotation matrix for homogeneous coordinates.