Suppose a great circle passes through a plane with normal $[a,2a,3a]$, so by converting it to a unit vector, the normal of the plane is $\frac{1}{\sqrt{14a^2}}[a,2a,3a]$.
My teacher explained that the distance between the great circle (that passes through the above mentioned plane) and the North Pole $(0,0,1)$ is the latitutde of the unitary normal vector of the plane. (We are only considering the unit sphere)
Hence, the distance is $arcssin(\frac{3a}{\sqrt{14a^2}})$. Why is this so? Did she write something wrong?
Your teacher is correct.
Let $\mathbf{n}$ is your normal. If $\mathbf{n}\cdot\mathbf{N}=1$ then the circle is the equator and $\mathbf{n}=\mathbf{N}$ has latitude $\pi/2$.
So assume $\mathbf{n}\cdot\mathbf{N}\in[0,1)$. There is a unique closet point $\mathbf{P}$ to $\mathbf{N}$ which is the rotation of $\mathbf{n}$ by an angle of $\pi/2$ in the plane containing $\mathbf{N}$ and $\mathbf{n}$ (orientation chosen so that the course of rotation by $\theta\in[0,\frac\pi2]$ passes through $\mathbf{N}$). Then $$ d(\mathbf{N},\mathbf{P})+d(\mathbf{N},\mathbf{n})=\frac\pi2 $$ and you know $d(\mathbf{N},\mathbf{n})=\arccos(\mathbf{N}\cdot\mathbf{n})=\arccos n_z$. Hence $d(\mathbf{N},\mathbf{P})=\arcsin n_z$.