Orthonormal basis in a cylindrical coordinate system

Question

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

Answer 1

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.

Answer 2

A set of nonzero, pairwise orthogonal vectors is always linearly independent, so you don't have to check that.