# Tangent plane to a sphere [easy calc 3 question]

• estro
In summary: The normal of one plane, and you want the normal of the other planeyou know they must be coplanar with the chord, and make equal angels with it
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?

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.

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?

Last edited:
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

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?

estro said:
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.

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.

estro said:
Yes this is the cost of stupidity.
A tax we all pay!

hello estro!

(just got up :zzz: …)
estro said:
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)

## 1. What is a tangent plane to a sphere?

A tangent plane to a sphere is a flat surface that touches the sphere at only one point. It is perpendicular to the radius of the sphere at that point, and therefore has no curvature at that specific point.

## 2. How is the tangent plane to a sphere calculated?

The tangent plane to a sphere can be calculated by finding the normal vector to the surface of the sphere at the point of tangency. This normal vector will be perpendicular to the tangent plane. Then, using this normal vector and the coordinates of the point of tangency, the equation of the tangent plane can be found using the formula Ax + By + Cz = D.

## 3. What is the purpose of finding the tangent plane to a sphere?

The tangent plane to a sphere is important in the study of calculus and geometry, as it helps to understand the behavior of surfaces and curves. It is also used in applications such as computer graphics and engineering.

## 4. Can there be more than one tangent plane to a sphere?

No, there can only be one tangent plane to a sphere at a given point. This is because the tangent plane is defined as being perpendicular to the radius of the sphere at that specific point, and a sphere has a unique radius at each point.

## 5. How does the size of the sphere affect the tangent plane?

The size of the sphere does not affect the orientation or location of the tangent plane, as long as the point of tangency remains the same. However, a larger sphere will have a larger tangent plane compared to a smaller sphere, since the tangent plane expands as the sphere grows in size.

• Calculus and Beyond Homework Help
Replies
1
Views
430
• Calculus and Beyond Homework Help
Replies
13
Views
497
• Calculus and Beyond Homework Help
Replies
14
Views
868
• Calculus and Beyond Homework Help
Replies
1
Views
378
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
909
• Calculus and Beyond Homework Help
Replies
2
Views
687
• Calculus and Beyond Homework Help
Replies
6
Views
849
• Calculus and Beyond Homework Help
Replies
21
Views
3K
• Calculus and Beyond Homework Help
Replies
5
Views
1K