= Solution
From part (a),
$$
\alpha=-\frac1\kappa n+\frac{\kappa_g}{\kappa}b.
$$
As $\alpha$ is orthogonal to $t$,
$$
\begin{aligned}
\alpha\cdot\beta
&=\left(-\frac1\kappa n+\frac{\kappa_g}{\kappa}b\right)
\cdot\left(-\varphi t-\frac{\dot\varphi}{\kappa}n
-\frac{\dot\varphi}{\kappa\kappa_g}b\right)\\
&=\frac{\dot\varphi}{\kappa^2}
-\frac{\dot\varphi}{\kappa^2}=0.
\end{aligned}
$$
To compute $\dot\beta\cdot\dot\alpha=\dot\beta\cdot t$, only its $t$ component matters. Differentiating $-\varphi t$ contributes $-\dot\varphi t$, while differentiating $-\kappa^{-1}\dot\varphi n$ contributes $+\dot\varphi t$ through $\dot n=-\kappa t-\tau b$. Every other term is normal to $t$. Hence
$$
\boxed{\beta\perp\alpha,\qquad\dot\beta\perp\dot\alpha}.
$$
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