All vectors are in $\mathbb{R}^3$ and only $\mathbf{r} = \left[ x; y; z \right]$ is unknown. My question is does the following system define a conic section in the $x-y$ plane and, if so, how can I find it:

$$ \begin{align} \mathbf{r}^\mathrm{T} \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] & = 0 \\ \\ \mathbf{v}_1^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_1} {\Vert \mathbf{r} - \mathbf{r}_1 \Vert } + \mathbf{v}_2^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_2} {\Vert \mathbf{r} - \mathbf{r}_2 \Vert } & = c \end{align} $$

If either $\Vert \mathbf{v}_1 \Vert = 0$ or $\Vert \mathbf{v}_2 \Vert = 0$, then the above is the intersection of a cone and the $z=0$ plane. Likewise if $\mathbf{r} = \mathbf{r}_1$ or $\mathbf{r} = \mathbf{r}_2$. However, I have been unable to figure out what the above represents in the general case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.