Orthonormal basis in a cylindrical coordinate system So I am supposed to show if these vectors make an orthonormal basis in a cylindrical coordinate system.
$\vec e_p=\bigl(\begin{smallmatrix}
cos(\theta )\\ 
sin(\theta )\\0 
\end{smallmatrix}\bigr); \vec e_\phi=\bigl(\begin{smallmatrix}
-sin(\theta )\\ 
cos(\theta )\\0 
\end{smallmatrix}\bigr); \vec e_z=\bigl(\begin{smallmatrix}
0\\ 
0\\1 
\end{smallmatrix}\bigr);$
In order for a set of vectors to be an orthonormal basis they need to
1) have length one
2) be orthogonal to each other (dot product=0)
3) (be linearly independent. Not sure about this one)
To check the first point I used the dot product. I know that:
$\sqrt{<\vec u,\vec u>}=|\vec u|$ 
$\sqrt{<\vec e_p,\vec e_p>}=\sqrt{cos^2(\theta)+sin^2(\theta)+0}=1$
$\sqrt{<\vec e_\phi,\vec e_\phi>}=\sqrt{sin^2(\theta)+cos^2(\theta)+0}=1$
$\sqrt{<\vec e_z,\vec e_z>}=\sqrt{0^2+0^2+1^2}=1$
Now i'll check the second point:
$<\vec e_p,\vec e_\phi>=-sin(\theta)cos(\theta)+sin(\theta)cos(\theta)+0\cdot1=0$
$<\vec e_p,\vec e_z>=cos(\theta)\cdot0+sin(\theta)\cdot0+0\cdot1=0$
$<\vec e_\phi,\vec e_z>=-sin(\theta)\cdot0+cos(\theta)\cdot0+0\cdot1=0$
Is my reasoning okay? Do I still have to show they are linearly independant?
Thanks

The vectors are linearly independent if determinant of the matrix, composed by taking the vectors as columns in the matrix, is non-zero.
In other words:
$$\left|\begin{matrix}\cos \theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{matrix}\right| = 1 \neq 0,$$
so the vectors are linearly independent. 
