This one stumps me: A circle in 3D space given by its center = $(0.15, 0.5, 1.0)$, its radius $=64$ and an orientation vector that points away from the circle's plane $(0.251, -0.796, 0.551)$

How would I go about finding the $2$ point coordinates where it touches the ground plane?

As a I am math beginner (and always will be), I have given some example numbers because I hoped I'll understand better when I see how they're transformed into the solution.

I'll reply to myself here in case I find the solution.


Is the ground plane the plane $z=0$?

The plane of the circle is given by $r. \left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right) = \left( \begin{array} \\ 0.15\\ 0.5\\ 1.0 \end{array} \right) . \left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right) =0.19065 $

The problem is asking for a point $(x,y,0)$ that is on the plane i.e.

$\left( \begin{array} \\ x\\ y\\ 0 \end{array} \right).\left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right)=0.19065$

and of distance $64$ from the centre, i.e.

$\left| \begin{array} \\ x-0.15\\ y-0.5\\ -1.0 \end{array} \right|=64$





$y=\frac{-3.817621 \pm \sqrt{3.817621^2+4\times 11.057237 \times 4095.3501}}{2 \times 11.057237}=-19.4186,19.0733$

The points are $(-60.82,-19.42,0)$ and $(61.25,19.07,0)$

  • $\begingroup$ Thank you! This is the first internet page I've printed out in years :) -Dizzy $\endgroup$
    – Dizzy
    May 23 '13 at 0:59

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.