http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 30 Jun 2014 15:46:39 -0500http://calc3.askbot.com/question/29/how-do-i-find-the-intersection-of-a-sphere-and-an-arbitrary-plane/It is very easy to find the intersection of a sphere and a plane like the $xy$ plane, but harder to find the intersection of a plane in general. The example given in class today was: > Describe the intersection between the unit sphere and the plane $x + 2y + z = 2$. Here is the work I have done on this problem so far. The general parametrization of a circle in space is: $$ \vec p(t) = \vec c + r\cos(t) \vec u + r\sin(t) \vec v$$ Where $\vec c$ is the center, $r$ is the radius, $\vec u \perp \vec v$, and $||\vec u|| = ||\vec v||$. The normal vector of the plane is: $$ \vec n = \langle 1, 2, 1 \rangle $$ I now let $\vec u$ and $\vec v$ equal a random vector on the plane. $$ \vec u = \langle 1, 0, -1 \rangle $$ $$ \vec v = \vec u \times \vec n = \langle 2, -2, 2 \rangle $$ Normalizing these vectors, I obtain: $$ \vec u = \langle 1, 0, -1 \rangle / \sqrt{2} $$ $$ \vec v = \langle 2, -2, 2 \rangle / \sqrt{12} $$ Then to find the center of the circle, I create a line that has the normal vector as its direction. $$ \vec l(t) = \langle 1, 2, 1 \rangle t $$ Next, I find the $t$-value intersection of this line and the plane, by plugging in the components into the equation of the plane. $$ t + 2(2t) + t = 2 $$ $$ t = \frac{1}{3} $$ The center is therefore equal to $\frac{1}{3} \langle 1, 2, 1 \rangle = \langle \frac{1}{3}, \frac{2}{3}, \frac{1}{3} \rangle $. Are these steps correct? From here, where do I go? *Comment*: Good question and looks great so far. As far as where to go next, take a look at your general parametrization of a circle and address what you're missing. Update ====== See [this question](http://calc3.askbot.com/question/36/sphereswill-this-workin-class-3/) for the answer.JustinMon, 30 Jun 2014 15:46:39 -0500http://calc3.askbot.com/question/29/