Orthonormal Basis Proofs Suppose $u, v, w$ are three differentiable functions from $R \to R^3$ such that for every $t ∈ R$ the vectors $u(t), v(t), w(t)$ form an orthonormal basis in $R^3$.
i) Prove that $u′(t) ⊥ u(t)$ and $u′(t)·v(t) = −u(t)·v′(t)$ for all $t$.
I did
$\theta=cos^{-1}((u_1'(t),u_2'(t),u_3'(t))\bullet(u_1(t),u_2(t),u_3(t)))$
$=cos^{-1}(u_1'(t)u_1(t)+u_2'(t)u_2(t)+u_3'(t)u_3(t))\ne0$.
Aren't they meant to be parallel because the tangent vector of an orthonormal basis takes the same direction as the basis itself?
For the second part, since $\frac{d}{dt} (u(t) \bullet v(t))=0 =u′(t)·v(t) + u(t)·v′(t) $
Therefore $u′(t)·v(t) = −u(t)·v′(t)$ 

so we will show that if $u$ and $v$ constant inner product then $u \cdot v' = -v \cdot u'.$ suppose we have $$u \cdot v = constant.  $$ then differentiating this using product rule we get $$u' \cdot v + u \cdot v' = 0\implies u'\cdot v = -u \cdot v' $$
if we have orthogonal vectors $u, v$ then $u \cdot v = 0$ and we can apply the previous result. if we have a unit vector $u$, then we take $v = u$ and the constant is $1.$ the result $u \cdot u' = 0$ follows.
