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 $u′(t)·v(t) = −u(t)·v′(t)$ 

$\{u(t),v(t),w(t)\}$ orthonormal basis implies $u(t)\cdot u(t)=1.$ Then
$$0=\frac{d}{dt}1=\frac{d}{dt}(u(t)\cdot u(t))=u'(t)\cdot u(t)+u(t)\cdot u'(t)=2u'(t)\cdot u(t)\implies u'(t)\cdot u(t)=0.$$
$\{u(t),v(t),w(t)\}$ orthonormal basis implies $u(t)\cdot v(t)=0.$ Then
$$0=\frac{d}{dt}0=\frac{d}{dt}(u(t)\cdot v(t))=u'(t)\cdot v(t)+u(t)\cdot v'(t)\implies u'(t)\cdot v(t)=-u(t)\cdot v'(t).$$
