= Normal and tangential components
At a point on a <plane curve>, or on a <hypersurface> with <unit normal> $\mathbf n$, a <vector> $\mathbf w$ splits into a <normal component> and a <tangential component>:
$$
\boxed{\mathbf w_N=(\mathbf w\cdot\mathbf n)\mathbf n,\qquad
\mathbf w_T=\mathbf w-\mathbf w_N,\qquad
\mathbf w=\mathbf w_T+\mathbf w_N.}
$$
The terms are <orthogonal projections>, and $\mathbf w_T\cdot\mathbf w_N=0$. Changing $\mathbf n$ to $-\mathbf n$ reverses the scalar coefficient $\mathbf w\cdot\mathbf n$, but leaves the <normal component> <vector> unchanged.
For a <plane curve>, a <unit tangent vector> $T$ and a <unit normal> $N$ form an <orthonormal basis>, giving
$$
\mathbf w_T=(\mathbf w\cdot T)T,\qquad
\mathbf w_N=(\mathbf w\cdot N)N.
$$
At $(1,1)$ on the <parabola> $y=x^2$, choose $T=(1,2)^T/\sqrt5$ and $N=(-2,1)^T/\sqrt5$. For $\mathbf w=(3,1)^T$,
$$
\mathbf w\cdot T=\sqrt5,\qquad
\mathbf w\cdot N=-\sqrt5,\qquad
\boxed{\mathbf w_T=(1,2)^T,\quad \mathbf w_N=(2,-1)^T.}
$$
Their sum is $(3,1)^T$ and their <inner product> is zero. A negative normal coefficient means that the <normal component> points opposite the chosen $N$, rather than giving it a negative <Euclidean norm>.
\Image[/tangent-normal-vector-decomposition.png]
{title=An arbitrary vector at a point of a parabola splits into tangent and normal projections}
{description=At $(1,1)$, the <vector> $(3,1)^T$ decomposes as $(1,2)^T+(2,-1)^T$. Arrows are based at the same point; dashed connectors show their <vector addition>.}
These are local directions: being tangent at a point does not mean that a straight displacement stays on the <curve>. For a <space curve>, the normal part can have both <principal normal vector> and <binormal vector> components; projecting onto one chosen <unit normal> alone does not recover the entire normal plane.
Back to article page
