The intersection of the sphere $\displaystyle x^2 + y^2 + z^2 = 6$ with the plane $\displaystyle x - 2y + 3z = 7$ is a circle. What is the z-coordinate of the center of the circle?
The intersection of the sphere $\displaystyle x^2 + y^2 + z^2 = 6$ with the plane $\displaystyle x - 2y + 3z = 7$ is a circle. What is the z-coordinate of the center of the circle?
From the plane, $\displaystyle \displaystyle x - 2y + 3z - 1 = 6$, hence
$\displaystyle \displaystyle x^2 + y^2 + z^2 = x - 2y + 3z - 1$
Now, get all the variables on one side, do all that completing the square thing, and then you can find the z-coordinate for the (center of the) circle.
yes, intersection means being at the same point(s), so you set both equations equal to find the common points.
well, there are many points on the circle, and potentially an infinite amount of z-points, meaning, all the coordinates on the circle have a different z-coordinate. the problem makes sense to me if you state a unique point, which makes sense to be the center.
if that's not what they want (did you type the question exactly as it appeared?), then simply solve for z in the last equation i gave you
1. Obviously you have some doubts whether Jhevon's solution is valid or not. It is but ...
2. The equation you got from Jhevon describes a sphere. This sphere is the smallest which fits exactly into the circle which contains the points of intersection between the plane and the original sphere.
3. Therefore the plane cuts the 2nd sphere at it's equator and therefore the center of the sphere is simultaneously the center of the circle you are looking for.
4. In the attached sketch I haven't drawn the lower halfsphere so you have a better "insight".
1. I use Derive 6.0. This program isn't available any more.
2. If I were you I would use Maxima. Have a look here: Maxima, a Computer Algebra System