Derivative of unit normal vector is tangential

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Frankel in his The geometry of physics writes on page 203:

$\mathrm{N}=x_u \times x_v/||x_u\times x_v||$ be the unit normal to $M^2$, where M is a two dimensional manifold in $\mathbb{R}^3$.

Moreover, he says, $$d\mathrm{N}/dt=(\partial N/\partial u^{\alpha})(du^{\alpha}/dt)$$ and this vector is a tangent vector to $M^2$ since $\mathrm{N}$ is a unit vector.

My question is, any linear combination of tangent vectors will remain tangent, so why does he mention since $\mathrm{N}$ is a unit vector; i.e. where/how does he use the unity of $\mathrm{N}$?

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In the expression

$$ \frac{\partial \mathrm N}{\partial u_\alpha} \frac{du_{\alpha}}{dt},$$

$\frac{du_\alpha}{dt}$ is a scalar and $\frac{\partial \mathrm N}{\partial u_\alpha}$ is a vector in $\mathbb R^3$. It is in general not in the tangent plane $T_pM \subset \mathbb R^3$. However, if $\|\mathrm N\|=1$, then

$$ 0 = \partial_t \|\mathrm N\|^2 = 2 \left\langle \mathrm N , \frac{d\mathrm N}{dt}\right\rangle.$$

Thus $\frac{d\mathrm N}{dt}$ is orthogonal to $\mathrm N$ and thus tangent to $M$.