By definition, the torsion is $\mathrm{\τ}\= -∥\mathbf{B}\prime \left(s\right)∥$. Since B is a unit vector, $\mathbf{B}·\mathbf{B}\=1$, and $\left(\mathbf{B}·\mathbf{B}\right)\prime \=2 \mathbf{B}·\mathbf{B}\prime \=0$, so $\mathbf{B}\prime$ is orthogonal to B. But B is orthogonal to the plane containing T and N, so $\mathbf{B}\prime$ must lie in that plane.

Now B is orthogonal to T, so $\mathbf{B}·\mathbf{T}\=0$. Differentiating gives $\mathbf{B}\prime ·\mathbf{T}\+\mathbf{B}·\mathbf{T}\prime \=0$. Rearranging gives

where $\mathbf{T}\prime \=\mathrm{\κ} \mathbf{N}$ is the basis for the definition of the curvature $\mathrm{\κ}$.

Thus, $\mathbf{B}\prime$, already in the osculating plane, is now orthogonal to T. Hence, it must be proportional to N, with the constant of proportionality taken as $-\mathrm{\τ}$, that is, as $\parallel \mathbf{B}\prime \parallel$