I have trouble understanding the vector interpretation of $\frac{1}{2\pi}$ in the Inverse Fourier Transform Equation: $x(t) = \frac{1}{2\pi}\int_{\Omega=-\infty}^{\infty}{X\left(\Omega\right)e^{j\Omega t}d\Omega}$. My professor said that it is to normalize the vector ${\bf \vec{v}} =\vec{e^{j\Omega t}}$ to make it a unit vector but I'm still not so sure what did he mean.
I understand we can see ${\bf \vec{v}} =\vec{e^{j\Omega t}}$ with different $\Omega$ as an individual vector, and $\vec{e^{j\Omega_1t}}$ is orthogonal to $\vec{e^{j\Omega_2t}}$ if $\Omega_1\neq\Omega_2$.
The component of vector ${\bf \vec{v}} =\vec{e^{j\Omega t}}$ in dimension $t$ is $e^{j\Omega t}$.
I just don't understand why ${\bf \vec{v}}$ divided by $2\pi=\frac{\bf \vec{v}}{2\pi}$ will be a unit vector?
A way to think of the $2\pi$ in your definition of the Fourier transform as a normalization factor of the exponential functions is perhaps to write the identity in the sense of distributions $\widehat{1} = 2\pi\,\delta_0$ which formally can be written $$ \int_{\mathbb R} e^{i\,x\cdot y} \,\mathrm d x = 2\pi\,\delta_0 $$ and so $$ \int_{\mathbb R} \frac{e^{i\,x\cdot y}}{2\pi} \,\mathrm d x = \delta_0 $$ but I do not know if this was the interpretation of your professor.