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.
Prove that the lines \(OA\) and \(BC\) do not intersect.
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.
Solution:
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.
\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 \(0 < \angle AOB < \tfrac12 \pi\)