Compute the normal of a 2D spline using animation nodes

Question

In this question I asked how to deform an object with an spline using animation nodes. The technique explained there the deformation considers all 3 dimensions tilting the output object, but what if I want to deform this object locking the tilt as if it was done with a 2D spline but still deform in a 3d space?, exactly as it is done here.

EDIT

Using @Omar Ahmad answer I managed to get the results I needed. but I'm still a little bit confused. The following image shows the resulting graph:

I'm not quite sure if the new computed normal is correctly linked as in order to get the desired results I was forced to increment its Z value up to 1000.

Answer 1

The normal to a point on a spline is equal to the tangent at this point rotated by 90 degrees around the z axis. Let the tangent be $\vec{t} = (x, y)$, then the normal is equal to $R\vec{t}$, where $R$ is a 2D rotation matrix with an angle of $\frac{\pi}{2}$. So the normal is equal to:

$$ \begin{bmatrix} \cos \frac{\pi}{2} & -\sin \frac{\pi}{2} \\ \sin \frac{\pi}{2} & \cos \frac{\pi}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -y \\ x \end{bmatrix} $$

So, given the evaluated tangents, you can compute the normals as follows:

To deform an object as described in the answer you linked, you will need a third vector, the one we computed using the cross product. In this case, it will always be the vector $(0, 0, 1)$.