# How to show $\alpha(t)=P\cos t+X\sin t$ is geodesic of $S^3$?

$S^3=\{(x_1,x_2,x_3,x_4)\in R^4 ~|~~ x_1^2+x_2^2+x_3^2+x_4^2=4\}$ with the induced metric.$\forall p\in S^3 , X\in T_pS^3 , ||X||=2$, how to show $\alpha(t)=P\cos t+X\sin t$ is geodesic of $S^3$ ?

I find a local coordinate $$u:(A,B,C)\rightarrow(2\cos A \cos B, 2\cos A\sin B, 2\sin A\cos C , 2\sin A\sin C)$$ then compute the $\frac{\partial u}{\partial A},\frac{\partial u}{\partial B},\frac{\partial u}{\partial C}$, (it is a little complex to write)

then compute the $g_{ij}$ , only $g_{11}=4 , g_{2}=4\cos^2 A , g_{33}=4\sin^2 A$ and others are zero.

then connection $$\Gamma^1_{ij}=0 ~(i\ne j ~or~ i=j=1 )~,~\Gamma_{22}^1=\Gamma^1_{33}=\sin A\cos A \\ \Gamma^2_{12}=\Gamma^2_{21}=-\tan A ~,~ other ~\Gamma^2_{ij}=0 \\ \Gamma^3_{13}=\Gamma_{31}^3=\cot A ~,~ other ~\Gamma^3_{ij}=0.$$

In fact , I have compute out the $\alpha(t) ~\text{and}~ \alpha'(t)$ in local coordinate, but $\alpha '(t)$ can't be represented by $\frac{\partial u}{\partial A},\frac{\partial u}{\partial B},\frac{\partial u}{\partial C}$ , I must have done some wrong in somewhere .

I think this is a good excise , but I fail to work out.

Note that $|\alpha'(t)|$ is constant So we have a claim that $\alpha$ is in image of a geodesic.
Note that image of $\alpha$ is in great circle of sphere For simplicity consider $S:=S^2$ Note that great circle of $S$ is intersection of cylinder $C:=S^1\times \mathbb{R}$ and $S$
Note : $c(t)$ is curve which is an intersection of two surfaces $S_i$ in $\mathbb{R}^3$ In further assume that $$T_{c(t)}S_1=T_{c(t)} S_2$$ Then $$\nabla_{c'(t)}^{S_1}c'(t)= \nabla_{c'(t)}^{S_2} c'(t)$$
From this note, since great circle is image of geodesic in $C$, so is in $S$ So we complete the proof