Tangent plane to a sphere [easy calc 3 question]

[estro]

Homework Statement

I know that x+y+z=0 is a tangent plate to my sphere at (0,0,0).
I want to find the tangent plane for my sphere at (1,-2,3).

Homework Equations

Suppose that his is my sphere $$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
(a,b,c) is the center of my sphere while R is the radius of it.

The Attempt at a Solution

First thing I did is to find the radius: Because (0,0,0) is on my sphere i know that $$(0-a)^2+(0-b)^2+(0-c)^2=a^2+b^2+c^2=R^2$$
So $$R=\sqrt {a^2+b^2+c^2}$$

No I tried to find the center of my sphere:
First thing is to spot that $$((\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}})$$ is orthogonal to the plane x+y+z=0 so it's direction is towards the radius. [I understand that there are 2 orthogonal directions to the plane but the "second" direction is not relevant because it contradicts point (1,-2,3)]
It means that $$(a,b,c)=(0,0,0)+\sqrt{a^2+b^2+c^2} (\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}})$$ from this I concluded that a=b=c.

Now when i plug a=b=c=t into the sphere equation I get: -t+2t-3t+14=0 so t = 7/2 =a=b=c.

It means that the tangent plane to my sphere at (1-2,3) should be z=-5x-11y-14
but when i plug it into my Grapher [program which translate equations into graphics] it showing that i completely missed with my sphere equation and with the tangent plane.

What I did wrong?

 

[HallsofIvy]

So you are saying that (0, 0, 0) and (1, -2, 3) are points on the sphere and that the normal to the sphere at (0, 0, 0) is <1, 1, 1>. That tells us that the center of the sphere is (a, a, a) for some number a. It must also be true that the distance from (a, a, a) to (0, 0, 0) must be the same as the distance from (a, a, a) to (1, -2, 3). That is, we must have
$$a^2+ a^2+ a^2= 3a^2= (a-1)^2+ (a+2)^2+ (a-3)^2= a^2- 2a+ 1+ a^2+ 4a+ 4+ a^2- 6a+ 9= 3a^2-4a+ 14$$
so that -4a+ 14= 0.

 

[estro]

Yes, I came exactly to these calculations and I got that the center of my sphere is (7/2,7/2,7/2).
But when I draw $$(x-7/2)^2+(y-7/2)^2+(z-7/2)^2=3(7/2)^2$$ in Grapher I get that the point (1,-2,3) is inside my sphere, and the plane I found misses the actual point and the sphere by light years.

Is Grapher wrong or it is me?

 

[tiny-tim]

hi estro!

you don't need to find the centre …

you know the normal of one plane, and you want the normal of the other plane

you know they must be coplanar with the chord, and make equal angels with it

 

[estro]

Hello tiny tim!

I'm not quite understand the idea. [maybe it because of my lousy English]
Do you see flaw in thought in what I did at the above posts?

 

[vela]

Yes, I came exactly to these calculations and I got that the center of my sphere is (7/2,7/2,7/2).
But when I draw $$(x-7/2)^2+(y-7/2)^2+(z-7/2)^2=3(7/2)^2$$ in Grapher I get that the point (1,-2,3) is inside my sphere, and the plane I found misses the actual point and the sphere by light years.

Is Grapher wrong or it is me?

Could be both. ;)

Your equations look fine. Are you using a Mac and the Grapher application that comes with it? I just tried plotting the sphere and plane here, and it worked fine. The plane is tangent to the sphere where you'd expect.

 

[estro]

Hi Vela!

Yes this is the software I use.
I found my mistake, I entered incorrect data in Grapher! [Yes this is the cost of stupidity]

But I still would like to hear some better ideas for this problem.

 

[HallsofIvy]

Yes this is the cost of stupidity.

A tax we all pay!

 

[tiny-tim]

hello estro!

(just got up :zzz: …)

I'm not quite understand the idea.

you know the normal is parallel to (1 1 1), and the chord is parallel to (1,-2,3)

so the other normal must be parallel to … ? ​

(it may hep if you first find the normal to the plane they're all in)