Given the symplectic group $\mathrm{Sp}(2,\mathbb{R})$, represented as real $2\times 2$ matrices, I would like to compute the geodesic from the identity matrix $1\!\!1$ to the group element
\begin{align}
\left(
\begin{array}{cc}
\omega & 0\\
0 & \frac{1}{\omega}
\end{array}\right)\,,
\end{align}
where I define my right-invariant metric on the Lie algebra, such that the following the following basis of the Lie algebra is orthonormal:
\begin{align}
K_1=\left(
\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}\right)\,,\quad K_2=\left(
\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)\,,\quad K_3=\left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)\,.
\end{align}
I believe that the geodesic should be given by just
\begin{align}
\gamma(t)=\exp\left(t\ln{(\omega)}K_1\right)\,,
\end{align}
with $t\in[0,1]$.

Is there a direct way to see from the metric that this is a geodesic? Also, due to the right-invariance of the metric, we should have $3$ Killing vector fields that may help to find the geodesic?