I understand that if the dot product of two vectors is 0 then they are orthogonal. I'm just having a slight problem conceptualising why we use $\frac{\partial r}{\partial \xi}$ and $\frac{\partial r}{\partial \eta}$
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Usually you consider a smooth curve $\textbf{r}$ in $\mathbb{R}^3$ given in bipolar coordinates. Suppose some point $\textbf{r}(q)$, the tangent of this curve in $q$ can be written as $\textbf{r}'(q) = a\frac{\partial\textbf{r}}{\partial\xi}+b\frac{\partial\textbf{r}}{\partial\eta}$, where $a,b$ are scalars. In fact, you will have a tangent plane in $\textbf{r}(q)$ , which can be viewed as a subspace of $\mathbb{R}^3$ (with dimension 2) with $\left\{\frac{\partial\textbf{r}}{\partial\xi},\frac{\partial\textbf{r}}{\partial\eta}\right\}$ as a basis.
By definition, the coordinate system is orthogonal if this basis is formed by orthogonal vectors.
Of course, this only makes sense when we are talking about curves in regular surfaces.
Integral
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Where does the expression $\textbf{r}'(q) = a\frac{\partial\textbf{r}}{\partial\xi}+b\frac{\partial\textbf{r}}{\partial\eta}$ come from? – May 01 '14 at 10:41
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Any vector in the tangent plane at $\textbf{r}(q)$ is a linear combination of the vectors $\frac{\partial\textbf{r}}{\partial\xi}$ and $\frac{\partial\textbf{r}}{\partial\eta}$ (they are evaluated in $q$). This happens because this tangent plane can be also viewed as a subspace of $\mathbb{R}^3$ with this vectors as basis. – Integral May 01 '14 at 17:34