If $(e_1, \ldots, e_n)$ and $(f_1, \ldots, f_n)$ are co-oriented, then so are $(e_1, \ldots, e_n, u)$ and $(f_1, \ldots, f_n, u)$

Suppose that $$S$$ is a smooth $$n$$-submanifold of $$M$$ where $$\dim(M) = n+1$$. Suppose also that $$S$$ and $$M$$ are both Riemannian and oriented.

Suppose that $$(e_1, \ldots, e_n)$$ and $$(f_1, \ldots, f_n)$$ are both positively oriented orthonormal bases for $$T_pS$$ and that $$(e_1, \ldots, e_n, u)$$ is positively oriented for $$T_pM$$ where $$u$$ is orthonormal with respect to $$(e_1, \ldots, e_n)$$.

Is it necessarily true that $$(f_1, \ldots, f_n, u)$$ is also positively oriented?

It seems to me possible that $$(f_1, \ldots, f_n, -u)$$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.

Any help is appreciated.

If $$A$$ is the invertible matrix that sends each $$e_i$$ to $$f_i$$, then the the matrix that sends $$(e_1,\dots,e_n,u)$$ to $$(f_1,\dots,f_n,u)$$ has block form $$\begin{pmatrix} A& 0 \\ 0 & 1\end{pmatrix}$$ so its determinant equals the determinant of $$A$$. Hence your assumptions imply that also $$(f_1,\dots,f_n,u)$$ is positively oriented.