# Unit normal to a sphere

1. Jul 10, 2009

### jeff1evesque

1. The problem statement, all variables and given/known data
The unit normal to a sphere is defined as, $$\hat{a}_{n} = x\hat{x} + y\hat{y} + z\hat{z}$$

But I thought it would be defined as, $$\hat{a}_{n} = 1\hat{x} + 1\hat{y} + 1\hat{z}$$

Could someone explain to me why am I thinking incorrectly?

Thanks,

JL

2. Jul 10, 2009

### diazona

If you compute the norm a.k.a. magnitude a.k.a. length of $1\hat{x} + 1\hat{y} + 1\hat{z}$, it's $\sqrt{3}$. So that's not a unit vector.

3. Jul 10, 2009

### jeff1evesque

Yes sir.

4. Jul 10, 2009

### jeff1evesque

Yes sir. But I thought the magnitude of the unit vector is a scalar, in which case $\sqrt{3}$ is fine?

5. Jul 10, 2009

### flatmaster

I believe the top equation is the normal for a general sphere of radius r = Sqrt(x^2+y^2+z^2). The second one is specifically for a unit sphere.

Also, the top equation is not necessarily of unit length. Imagine you had a sphere of radius of 5. The sphere would include the pointing (5,0,0), giving the above vector a magnitude of 5.

6. Jul 10, 2009

### diazona

The magnitude of a vector is a scalar. BUT: the magnitude of a unit vector is 1. That's the definition of a unit vector, a vector that has magnitude 1. A vector with any other magnitude is not a unit vector.

As flatmaster said, the magnitude of $x\hat{x} + y\hat{y} + z\hat{z}$ is $\sqrt{(x^2+y^2+z^2)}$. So $x\hat{x} + y\hat{y} + z\hat{z}$ is going to be a unit normal if and only if $\sqrt{(x^2+y^2+z^2)} = 1$.

7. Jul 10, 2009

### jeff1evesque

what if $$\hat{a}_{n} = 1\hat{x} + 1\hat{y} + 1\hat{z}$$? I don't understand why that would be wrong? Is it because if we take the magnitude then it's not equal to 1? I think i understand now.

Last edited: Jul 10, 2009
8. Jul 10, 2009

### jeff1evesque

I guess my question is, how do you get the normal vector $$\hat{a}_{n}$$ From a sphere with a radius of 1?

9. Jul 10, 2009

### Redbelly98

Staff Emeritus
Yes, exactly.

EDIT: Also, it's not normal to the sphere's surface everywhere.

10. Jul 10, 2009

### jeff1evesque

But how do you find the normal? I am reading some stuff, and I notice that the gradient is normal to a surface/curve. If I can find the gradient of our sphere then I can find the normal, and thus the "unit normal". But I am not sure how to find the gradient.

11. Jul 11, 2009

### Redbelly98

Staff Emeritus
For this one, we just have to think about the geometry of a sphere. No gradients are necessary.

Any vector that is directed from the origin to some point (x,y,z) on the sphere will be directed along the normal at that point.

12. Jul 11, 2009

### jeff1evesque

That makes much more sense. I totally forgot that the unit normal vector $$\hat{a}_{n}$$ is a vector beginning from the origin. What if we had a different curve, perhaps a cylinder? Say a problem gave us an equation for a cylinder [namely $$x^2 + y^2 = 9$$], along with some arbitrary Force vector. How would we find this unit normal.
Thanks,

JL