$(X_1,X_2,X_3)$ is a global frame on $S^3$

Question

Id like you to check this:

Given

$X_1(w,x,y,z)=(-x,w,-z,y)$

$X_2(w,x,y,z)=(-y,z,w,-x)$

$X_3(w,x,y,z)=(-z,-y,x,w)$

prove that:

1. they are $\mathcal{C}^{\infty}$ vector fields on $S^3$;

2. $(X_1,X_2,X_3)$ is a global frame on $S^3$.

3. What can we say about $TS^3$?

My idea of the solution is:

1. $X_1,X_2,X_3$ are $\mathcal{C}^{\infty}$ vector fields on $S^3$ since I can compute every partial derivatives, for all $p\in S^3$. Moreover if $p=(w,x,y,z)\in S^3$ then \begin{eqnarray} & &(w,x,y,z)\cdot(-x,w,-z,y)=-wx+xw-zy+zy=0\\ & &(w,x,y,z)\cdot(-y,z,w,-x)=-wy+xz+yw-zx=0\\ & &(w,x,y,z)\cdot(-z,-y,x,w)=-wz-xy+xy+zw=0\\ \end{eqnarray} so $X_1,X_2,X_3$ are tangent to $S^3$ for all $p\in S^3$.
2. $(X_1,X_2,X_3)$ is a global frame because they are linearly independent since $aX_1+bX_2+cX_3=0$ iff $a=b=c=0$ for all $p\in S^3$.
3. Since there exists a global frame on $S^3$, that is a global frame on $TS^3$, we can say that $TS^3$ is trivial.

Is everything all right?

Thank you so much