Question

A sphere has a center (2,3,5) and has a radius of 3. Find the equation of all planes containing the points (4,0,0) and (0,0,4) that are tangent to this circle.

Hint: Let (a,b,c) be the point on the plane that touches the sphere.

FML

Answer 1

This is a tough problem, but I can give some hints

Answer 2

A point where it's tangent is a solution to both the equation of the plane and the sphere, AND both a solution to the equation to the plane and the sphere's partial derivatives.

Answer 3

Alternatively, you can use the normal vectors of the plane and sphere. The normal of a sphere is simply the point itself (scaled possibly), and the plane is going to be simple to calculate. You just have to do it arbitrarily.

Answer 4

Right now all I have is the equation of the sphere and the normal vector which is <2-a, 3-b, 5-c>. I can not find (a,b,c) I don't know how I can go about solving for each letter to get that point.

Answer 5

I tried dot product or just making the magnitude of the normal vector equal to the radius of the sphere but it becomes a three unknown system of equations and I can't solve that.

Answer 6

So, try this. Let (a,b,c) be the point on the plane that touches the sphere. Then the normal vector of the plane is the line from the center of the sphere to (a,b,c), although you might have to scale this to get absolute value of 1 for your vector. Then you have the normal vector for the plane. You also have some points on the plane. You should be able to find an equation.

Answer 7

I think that's how you do it actually.

Answer 8

That's what I did but it still leaves me with three unknowns. My prof said at the end the equation of the planes should not have any a, b, or c's.

Answer 9

The equation of my plane would be (2-a)x + (3-b) + (5-c)z = 4 - 4a using the point (4,0,0)

Answer 10

That's odd, since it asked for the equation of all planes. How can you have no free variables?

Answer 11

Well you can just not a, b, and c. X, Y, and Z are fine because they are part of the equation of the plane. basically he wants us the fund the point (a,b,c) and I don't know how to find it with the given information.

Answer 12

Ah, here's how: There are only finitely many planes?

Answer 13

Yes. Well at least I believe so.

Answer 14

Because my visualization is telling me there are only two solutions.

Answer 15

I thought so two, to be completely honest I don't know.

Answer 16

Btw thank you so much for your help, I really appreciate all your input.

Answer 17

Ya. This is the most interesting problem I've seen tonight.

Answer 18

Yeah, I don't know why my prof does this to us -.-

The only thing he told me was that the normal vector was correct, <2-a, 3-b, 5-c>, and then he told me to get a second vector from the plane which I got from point (4,0,0) to the point (a,b,c) which is <4-a,-b,-c>.

Answer 19

Ah, so this was where you tried solving for the dot product = 0?

Answer 20

Exactly, they are perpendicular so there for the dot product equals zero but it still leaves me with three unknowns.

Answer 21

I see that ;(

Answer 22

I guess I will just wait until tomorrow, to go to a tutoring session before it's due.

Answer 23

Good luck

Answer 24

Thanks, and thank you for all your input!

Answer 25

Did you try using the other vector from the plane as well?

Answer 26

<2-a, 3-b, 5-c> <4-a,-b,-c> = 0 = <2-a, 3-b, 5-c> <-a,-b,4-c>. That should kill one of the unknowns I'd hope.

Answer 27

anyway, good luck with this.

Answer 28

So, my prof gave us another hint he said that you need 3 equations to solve for all three unknowns. So I have the equation of the sphere at point (a,b,c) which is

(a-2)^2 + (b-3)^2 + (c-5)^2 = 9

Also have the dot product which equals zero which is:

a^2 - 6a + b^2 -3b +c^2 - 5c = 0

Now I just need one more equation, also he told us that both points (4,0,0) and (0,0,4) are a part of both planes that I am solving for.

Answer 29

Okay so I found the other equation for the system of equations which is:

a^2 -2a + b^2 -3b + c^2 -9c = -20

and I found c = 3+a

Answer 30

I am curious to see how you get the answer. I found one generic way which involved using a rotation matrix, and a slightly simpler way for this specific problem. I did not see a nice way to solve for (a,b,c) algebraically... when you find the answer, please post it.