Let's assume $\textbf{g}(t)$ is a position vector of point $A$, where $t$ is a variable. In addition, $\textbf{g}'(t)$ is the position vector of point $B$. I have to prove that if $\textbf{g}(t)\parallel \textbf{g}''(t)$, the area of the triangle $OAB$ does not depend on $t$ assuming that $\textbf{g}(t)$ is differentiable $2$ times.
This is what I tried to do:
Let's say
$$\textbf{g}(t)=\begin{bmatrix} a_{1} \\a_{2} \\ a_{3} \end{bmatrix}$$
$$\textbf{g}'(t)=\begin{bmatrix} a'_{1} \\a'_{2} \\ a'_{3} \end{bmatrix}$$ $$\textbf{g}''(t)=\begin{bmatrix} a''_{1} \\a''_{2} \\ a''_{3} \end{bmatrix}$$
When $\textbf{g}(t)\parallel \textbf{g}''(t)$ the cross product of the two vectors becomes
$$\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ a_{1} & a_{2} & a_{3} \\ a''_{1}& a''_{2} &a''_{3} \\ \end{vmatrix}=\textbf{i}(a_{2}a''_{3}-a''_{2} a_{3})+\textbf{j}(a_{3} a''_{1}-a_{1}a''_{3})+\textbf{k}(a_{1}a''_{2}- a''_{1}a_{2})=\textbf{0}$$
The area of $OAB$ is
$$S_{OAB}=\frac{1}{2}\left|\textbf{g}(t)\times \textbf{g}'(t)\right|$$
The derivative of the inner product of $\textbf{g}(t)$ and $\textbf{g}'(t)$ is
$$(\textbf{g}(t)\cdot \textbf{g}'(t))'=\textbf{g}'(t)\cdot \textbf{g}'(t)+\textbf{g}(t)\cdot \textbf{g}''(t)=\left\| \textbf{g}'(t)\right\|^2+\textbf{g}(t)\cdot \textbf{g}''(t)$$
That's all of the information I could extract and now I don't know what to do. Any suggestions?
You have to differentiate the cross product rather than the inner product since $S_{OAB}(t)$ is defined in terms of the former as you noted. Differentiation of cross products dependent on a single variable satisfy a analogous product rule: \begin{align*} \frac{d}{dt}(\mathbf{g}(t) \times \mathbf{g}'(t)) &= \Big(\frac{d}{dt}\mathbf{g}(t) \times \mathbf{g}'(t)\Big) + \Big(\mathbf{g}(t) \times \frac{d}{dt}\mathbf{g}'(t)\Big) \\ &= (\underbrace{\mathbf{g}'(t) \times \mathbf{g}'(t)}_{\mathbf{0} \text{ as } \mathbf{g}'(t) \ || \ \mathbf{g}'(t) \text{ trivially}}) + (\underbrace{\mathbf{g}(t) \times \mathbf{g}''(t)}_{\mathbf{0} \text{ as } \mathbf{g}(t) \ || \ \mathbf{g}''(t) \text{ is given}})) = \mathbf{0} \end{align*} In other words $\frac{d}{dt}S_{OAB}(t) = 0$ i.e. the area does not change with $t$.