Problem source

The distinct points $O\,(0,0,0),$ $A\,(a^{3},a^{2},a),$ $B\,(b^{3},b^{2},b)$ and $C\,(c^{3},c^{2},c)$ lie in 3-dimensional space. 
\begin{questionparts}
\item Prove that the lines $OA$ and $BC$ do not intersect. 
\item Given that $a$ and $b$ can vary with $ab=1,$ show that $\angle AOB$ can take any value in the interval $0<\angle AOB<\frac{1}{2}\pi$, but no others. 
\end{questionparts}

Solution source

\begin{questionparts}
\item The line $OA$ is $\lambda \begin{pmatrix} a^3 \\ a^2 \\ a \end{pmatrix}$. The line $BC$ is $\begin{pmatrix} b^3 \\ b^2 \\ b \end{pmatrix} + \mu \begin{pmatrix} c^3-b^3 \\ c^2-b^2 \\ c-b \end{pmatrix}$. If these lies are concurrent then there would be $\mu$ and $\lambda$ such that they are equal, and in particular,

\begin{align*}
&& \frac{b^2 + \mu(c^2-b^2)}{b + \mu (c-b)} &= \frac{b^3 + \mu(c^3-b^3)}{b^2 + \mu (c^2-b^2)} \\
\Leftrightarrow && (b^2 + \mu(c^2-b^2))^2 &= (b+\mu(c-b))(b^3+\mu(c^3-b^3)) \\
&& b^4 +2\mu b^2 (c^2-b^2) + \mu^2 (c^2-b^2) &= b^4 + \mu(c-b)b^3 + \mu b(c^3-b^3) + \mu^2 (c-b)(c^3-b^3) \\
\Leftrightarrow && 2\mu b^2 (c+b) + \mu^2(c-b)(c+b)^2 &= \mu (b^3 + b(c^2+bc+b^2)) + \mu^2 (c^3-b^3) \\
&& \mu = 0 & \Rightarrow a = b \\
\Leftrightarrow && b^2c - bc^2 &= \mu (c^3-b^3-(c-b)(c+b)^2) \\
\Leftrightarrow && bc(b-c) &= \mu (c-b)(c^2+bc+b^2-c^2-2bc-b^2) \\
\Leftrightarrow && bc &= \mu (bc) \\
\Leftrightarrow && \mu &= 1 \\
&& \mu = -1 & \Rightarrow a = c
\end{align*}

Therefore there are no solutions.

\item 
\begin{align*}
\cos \angle AOB &= \frac{ab+a^2b^2+a^3b^3}{\sqrt{a^2+a^4+a^6}\sqrt{b^2+b^4+b^6}} \\
&= \frac{3}{\sqrt{1 + a^2 + a^4} \sqrt{1 + b^2 + b^4}} \\
&> 0
\end{align*}

Therefore the angle satisfies $\angle AOB < \tfrac12 \pi$. We cannot achieve $0$, since that would require $a = b = 1$, therefore it falls in the range $0 < \angle AOB < \tfrac12 \pi$
\end{questionparts}