Suppose $V$ is a vector fields on a geodesic $\alpha$. Show that $V$ is parallel if, and only if $\| V\| $ is constant and the angle between $V$ and $\alpha'$ is constant.
I have done the following: Suppose $V$ is parallel. Then $\alpha'[\langle V, V\rangle]=2\left\langle \frac{DV}{dt}, V\right\rangle=0$. Therefore, $\| V\| ^2$ is constant. Also, $\alpha'[\langle V, \alpha'\rangle]=\left\langle\frac{DV}{dt}, \alpha'\right\rangle+\left\langle \frac{D\alpha'}{dt}, V\right\rangle=0$ since $\alpha$ is a geodesic, therefore this internal product is constant. Since $\alpha$ is a geodesic, $\| \alpha'\| $ is constant. Since $\| \alpha'\| $, $V$ and $\langle V, \alpha'\rangle$ are all constant, it follows that the angle between $\alpha'$ and $V$ is constant.
I'm having trouble with the converse.
Let's do the converse.
First case: Suppose that $V$ is not a multiple of $\alpha'$. Let $\varphi$ be the constant angle between $V$ and $\alpha'$. Suppose that $\alpha$ is parametrized by arc-length, that is, $\|\alpha '\| = 1$. Then $$\cos \varphi = \frac{\langle V, \alpha' \rangle}{\|V\|}$$ is constant too. Since $\|V\|$ is assumed constant, we have that $\langle V, \alpha' \rangle$ is constant, hence: $$\alpha'[\langle V, \alpha '\rangle] = \left\langle\frac{DV}{dt}, \alpha'\right\rangle + \left\langle V, \frac{D \alpha'}{dt}\right\rangle = 0$$ Since $\alpha$ is a geodesic, we get that $\left\langle\frac{DV}{dt}, \alpha'\right\rangle = 0$. On the other hand, $\|V\|$ is constant, so: $$\alpha'[\langle V,V\rangle] = 2\left\langle \frac{DV}{dt},V\right\rangle = 0,$$ and so $\left\langle \frac{DV}{dt},V\right\rangle = 0$. Since $V$ and $\alpha'$ are not parallel, they form a basis for the tangent space, hence: $$\frac{DV}{dt} = A\alpha' + B V.$$ Taking the inner products with $V$ and $\alpha'$, we can solve for $A$ and $B$: both zero. Remember that $\langle V, \alpha'\rangle = \|V\|\cos\varphi$. Then:
$$\begin{cases} A + B\|V\|\cos \varphi = 0 \\ A\|V\|\cos \varphi + B\|V\|^2 = 0\end{cases}$$ We have that $A = B = 0$ solves the system, and $$\begin{vmatrix} 1 & \|V\|\cos \varphi \\ \|V\|\cos\varphi & \|V\|^2\end{vmatrix} = \|V\|^2\sin^2\varphi \neq 0,$$ because we're supposing that $V$ and $\alpha'$ are not parallel. Then, the solution $A = B = 0$ is the only one. Hence, $\frac{DV}{dt} = 0$, and $V$ is parallel.
Second case: Suppose that $V$ and $\alpha'$ are parallel, and that $\alpha$ is parametrized by arc-length. Then, $V = \lambda\alpha '$, for some function $\lambda$. It happens that $\lambda$ is constant, and it follows from the angle $\varphi$ between $V$ and $\alpha'$ being constant. We do have that: $$\cos \varphi = \frac{\langle V, \alpha'\rangle}{\|V\|} = \frac{\lambda}{\|V\|}\langle \alpha' , \alpha' \rangle = \frac{\lambda}{\|V\|}$$ is constant. Since $\|V\|$ is assumed constant, $\lambda$ is too.
This way: $$V = \lambda \alpha' \implies \nabla_{\alpha'}V = \nabla_{\alpha'}\lambda\alpha' = \lambda \nabla_{\alpha'}\alpha' = 0,$$using that $\alpha$ is a geodesic, and so we conclude that $V$ is parallel.