how to handle a gradient expression

Question

How to prove in a rigorous way that: $$|u|=1 \implies \nabla|u|^2 = 0 \implies (\nabla u)^Tu=0$$ and then $\forall v$ $$\nabla u : \nabla((u.v)\cdot u)= |\nabla u|^2 (u \cdot v)$$

Answer 1

Using the Einstein summation convention:

$$\nabla u : \nabla((u.v)\cdot u) = \frac{\partial u_i}{\partial x_j} \frac{\partial}{\partial x_j}((u_k v_k) u_i) = \frac{\partial u_i}{\partial x_j}\frac{\partial u_i}{\partial x_j} (u_k v_k) + \frac{\partial u_i}{\partial x_j} u_i \frac{\partial}{\partial x_j}(u_k v_k) = |\nabla u|^2 (u\cdot v) $$ because $$ \frac{\partial u_i}{\partial x_j} u_i = \frac12 \frac{\partial}{\partial x_j}|u_i|^2 = 0 .$$