# Solving system of non-linear equations in two variables

I am trying to solve the following system of equations

$$\begin{eqnarray} \|p + \alpha n\| & = &\|q + \beta n\| \\ (p + \alpha n) \cdot (q + \beta n) & = & 0 \end{eqnarray}$$

for the unknowns $\alpha, \beta \in \mathbb{R}$, with $n$, $p$ and $q$ known vectors in $\mathbb{R}^n$ (the second line contains a dot product), and $\|n\|=1$.

Using some algebra, these equations can be written as

$$\begin{eqnarray} \alpha^2 - \beta^2 + A\alpha + B\beta + C & = & 0 \\ \alpha\beta + D\alpha + E\beta + F & = & 0 \end{eqnarray}$$

for suitably defined scalar constants $A$,..., $F$. Further manipulation leads to a quartic equation.

My question: is there some "elegant" closed-form solution, or, otherwise, by what method could this be solved numerically (it seems there are two pairs of solutions)?

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It is not "really" a quartic equation! Let me explain.

The first equation expands to $$\alpha^2+(2p\cdot n) \alpha +p\cdot p=\beta^2+(2q\cdot n )\beta +q\cdot q=0.$$

The second equation expands to $$\alpha\beta+ (q\cdot n)\alpha +(p\cdot n)\beta +p \cdot q=0.$$

Complete the squares in the first equation. We get something of shape $$(\alpha +p\cdot n)^2-(\beta +q\cdot n)^2=A.$$

The second equation can be written as $$(\alpha +p\cdot n)(\beta +q\cdot n)=B.$$

Let $x=\alpha+p\cdot n$ and $y=\beta +q\cdot n$.

Our system of equations then becomes $$x^2-y^2=A, \qquad xy=B.$$

So $y=B/x$. Substitute in the first equation, and multiply through by $x^2$. We get a quadratic equation in $x^2$. Solve this equation for $x^2$ in the usual way, then take the square roots.

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That's brilliant... I wonder how you came up with this. Thanks! –  koletenbert Oct 21 '11 at 20:49
There is very clear symmetry, so for general algebraic reasons, it is inevitable that there will be collapse. Then it was just a question of describing the collapse without fancy tools. –  André Nicolas Oct 21 '11 at 20:57