Distance from a point on a sphere to a point on a plane...

[Divergent13]

Hi everyone here is the original question:

Find the length of the shortest line segment which can be drawn from a point on the sphere (x-1)^2 + (y-2)^2 + (z-3)^2 = 9
to a point of the plane x + 2y + 2z = 28.

I am having a lot of difficulty with this:

what I'm trying is to find some way to write the arclength equation for
all the lines between those points lying on the sphere and those in the plane,
then minimize the arclength equation using
d/dt(L(p)) = 0

How could I go about doing this problem?

The answer in the book is 8/3.

[matt grime]

The segment is normal to both the plane and the sphere.

[Divergent13]

ahh well thats understandable now--- but I am still stuck on it.

[matt grime]

You can read off the equation of the nornal of the plane. You can use your vector calc to find the normal to the sphere, those two normal must be parallel. You can thus work out the point(s) where this normal occurs, you've then got to find the distance between two planes (the orginal, and the tangent plane to the sphere), and i'll wager there's a formula for that in your notes...

[mathman]

After you get the normal to the plane, draw a straight line from the center of the sphere along this direction toward the plane. Where this line hits the plane and the sphere will give you the end points of the shortest segment. (I am assuming the plane does not intersect the sphere).

[Divergent13]

My latest (and easiest) idea is:

Find the shortest length between the given plane and the sphere's center, call that D1.
Then the shortest line segment from the plane to a point on the sphere's surface is simply
D2 = D1 - radius.

There's an equation in the Stewart book for finding the shortest distance between a plane and a point, and it uses the scalar projection onto the planes normal vector.

[matt grime]

yep, that seems like a much better way of doing it.