The vectors are orthonormal Let $(a_{ij})$ be a skew-symmetric $3 \times 3$ matrix (i.e., $a_{ij}=-a_{ji}$  for all $i, j$). 
Let $v_1$, $v_2$ and $v_3$ be smooth functions of a parameter $s$ satisfying the differential equations
$$v'_i= \sum_{j=1}^3 a_{ij} v_j$$
for i = $1, 2$ and $3$, and suppose that for some parameter value $s_0$ the
vectors $v_1(s_0)$, $v_2(s_0)$ and $v_3(s_0)$ are orthonormal. Show that the
vectors $v_1(s), v_2(s)$ and $v_3(s)$ are orthonormal for all values of $s$.
I am facing difficulties. Any hints how to show that  the
vectors $v_1(s), v_2(s)$ and $v_3(s)$ are orthonormal for all values of $s$?
EDIT Nr.1:
$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) = \frac{d}{ds}(v_i(s)^t v_j(s) 
 )=(v_i(s)^t)' v_j(s)+v_i(s)^t (v_j(s))' \\= \sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)+v_i(s)^t \sum_{k=1}^3 a_{jk}v_k(s) $
Any ideas how I can continue? 
$$$$ 
EDIT Nr.2: 
We have $$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) = \frac{d}{ds}(v_i(s)^t v_j(s) 
 )=(v_i(s)^t)' v_j(s)+v_i(s)^t (v_j(s))' \\= \sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)+v_i(s)^t \sum_{k=1}^3 a_{jk}v_k(s) =\sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)-v_i(s)^t \sum_{k=1}^3 a_{kj}v_k(s)$$ 
It is $$v_k(s)^tv_j(s)=v_k(s)v_j(s)^t$$ right?  
Therefore we have $$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) =\sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)-v_i(s)^t \sum_{k=1}^3 a_{kj}v_k(s)=\sum_{k=1}^3 [a_{ik} v_k(s) v_j(s)^t-a_{kj} v_i(s)^t v_k(s)]=\sum_{k=1}^3 v_k(s)[a_{ik} v_j(s)^t-a_{kj} v_i(s)^t]$$ 
What can we do next? 
 A: Let us assume that $v_i \colon \mathbb{R} \rightarrow \mathbb{R}^n$. Then, define $v \colon \mathbb{R} \rightarrow M_{3 \times n}(\mathbb{R})$ by $v(s) = (v_1(s), v_2(s), v_3(s))^t$. The system of equations for $v$ is equivalent to the linear system $v'(s) = Av(s)$ with constant coefficients. Define also $g \colon \mathbb{R} \rightarrow M_3(\mathbb{R})$ by $g(s) = v(s) v(s)^t$. The matrix $g(s)$ is then the Gram matrix for the vectors $v_1(s), v_2(s), v_3(s)$. You want to show that if $g(s_0) = I$, then $g(s) \equiv I$.
If you are familiar with matrix exponential, you can write the solution to $v'(s) = Av(s)$ as $v(s) = e^{A(s-s_0)} v(s_0)$ and then
$$ g(s) = v(s) v(s)^t = e^{A(s - s_0)} v(s_0) v(s_0)^t (e^{A(s - s_0)})^t = e^{A(s - s_0)} g(s_0) e^{A^t(s - s_0)} $$.
If $g(s_0) = I$ and $A$ is anti-symmetric, we see that $g(s) = e^{A(s - s_0)}e^{-A(s - s_0)} = I$ for all $s \in \mathbb{R}$.
If you don't want to use matrix exponentials, you can calculate
$$ g'(s) = v'(s) v(s)^t + v(s) (v'(s))^t = Av(s)v(s)^t + v(s)v(s)^tA^t = Ag(s) - g(s)A  = [A, g(s)]. $$
You can check that $g(s) \equiv I$ satisfies this equation and so, if you know that $g(s_0) = I$ for some $s_0$ then by uniqueness of solutions you must have $g(s) \equiv I$.
