How do I find the intersection of a sphere and an arbitrary plane?
asked 2014-06-30 15:46:39 -0600
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 for the answer.
