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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: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, givingAt $(1,1)$ on the parabola $y=x^2$, choose $T=(1,2{)}^T/\sqrt{5}$ and $N=(-2,1{)}^T/\sqrt{5}$. For $\mathbf{w}=(3,1{)}^T$,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.

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.

For the unit-speed curve, the unit tangent vector, curvature of a space curve, principal normal vector and binormal vector areDifferentiating $t\cdot t=1$ gives $t\cdot {\gamma }^{\prime \prime }=0$, so $t,p,b$ form a positively oriented orthonormal basis. The hypothesis ${\gamma }^{\prime \prime }\ne 0$ makes the principal normal vector well-defined.

To determine the sign of the torsion of a space curve without relying on a pictorial handedness convention, take $R>0$, $h\ne 0$, $L=\sqrt{R^2+h^2}$ and use the unit-speed circular helixWriting $u=s/L$, direct differentiation gives its Frenet frame:ConsequentlyAt $s=0$ the marked point is $(R,0,0)$, with $t=(0,R/L,h/L)$, $p=(-1,0,0)$, $b=(0,-h/L,R/L)$ and $b^{\prime }=(h/L^2,0,0)$. For $h>0$, $b^{\prime }$ points opposite to $p$ and the torsion of a space curve is positive. Replacing $h$ by $-h$ reflects the circular helix across the horizontal plane and makes the torsion of a space curve negative; then $b^{\prime }$ points in the same direction as $p$. The direction of increasing $s$ is fixed in both panels.

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.