The vectors in the TNB-frame form an orthonormal basis, that is, they are linearly independent, mutually orthogonal, and each of unit length. Hence, $\mathbf{N}\prime \=a \mathbf{T}\+b \mathbf{N} \+c \mathbf{B}$, that is, $\mathbf{N}\prime$ can be written as a linear combination of the vectors T, N, and B.

Since N is a unit vector, $\mathbf{N}·\mathbf{N}\=1$ and $\left(\mathbf{N}·\mathbf{N}\right)\prime \=2 \mathbf{N}·\mathbf{N}\prime \=0$, so

$0\= \mathbf{N}·\mathbf{N}\prime \=\mathbf{N}·\left(a \mathbf{T}\+b \mathbf{N} \+c \mathbf{B}\right)\=a \mathbf{N}·\mathbf{T}\+b \mathbf{N}·\mathbf{N}\+c \mathbf{N}·\mathbf{B}\=a\cdot 0\+b\cdot 1\+c\cdot 0$ 

which implies $b\=0$ and $\mathbf{N}\prime \=a \mathbf{T}\+c \mathbf{B}$.

From $\mathbf{T}·\mathbf{N}\=0$ and the following calculation, $a\= -\mathrm{\κ}$.

Hence, $\mathbf{N}\prime \= -\mathrm{\κ} \mathbf{T}\+c \mathbf{B}$. From $B·\mathbf{N}\=0$ and the following calculation, $c\=\mathrm{\τ}$.

Consequently, $\mathbf{N}\prime \= -\mathrm{\κ} \mathbf{T}\+\mathrm{\τ} \mathbf{B}$.