Projection of a curve on a plane

Question

How to project a curve on an arbitrary plane? I mean, if $\alpha:U\subset\mathbb{R}\rightarrow\mathbb{R}^3$ is a curve defined as $\alpha(t) = (f(t),g(t),h(t))$ where $f,g,h:U\subset\mathbb{R}\rightarrow\mathbb{R}$ and let $\pi$ be the plane defined as $ax+by+cz=d$, how will be the projection of $\alpha$ on $\pi$.

I know that if $\pi$ is $XY$, $YZ$ or $XZ$ the proyection will be of the form $(f(t),g(t),0)$, $(0,g(t),h(t))$ or $(f(t),0,h(t))$, but in general cases how can I determine it?

Answer 1

Let $\hat{n}$ be a unit normal vector to $\pi$ (do you know how to find $\hat{n}$ from the defining equation of $\pi$?). The projection of any vector $\vec{v}$ onto $\pi$ can be calculated as $\vec{v} - (\vec{v} \cdot \hat{n})\hat{n}$. Now, replace $\vec{v}$ with your vector-valued curve.