# Normal vectors

1. Apr 19, 2007

### kasse

I need to learn how to find normal unit vectors to surfaces. Is there a trick or a formula that I can use?

F. ex. the normal unit vector to the surface x^2+y^2+z^2=4 is 0,5(x,y,z), but why?

What can I do to find normal vectors in other cases?

Last edited: Apr 19, 2007
2. Apr 19, 2007

### Dick

For surfaces defined in the form f(x,y,z)=0, the gradient vector is normal to the surfaces. My question is why do you think 0.5*(x,y,z) is a unit vector?

3. Apr 19, 2007

### kasse

I don't know, but my book says so.

4. Apr 19, 2007

### Dick

At a particular point, it may be, but the length of that vector is 0.5*sqrt(x^2+y^2+z^2). Not usually equal to one.

5. Apr 19, 2007

### kasse

So the gradient vector is the normal vector to f(x,y,z)=x^2+y^2+z^2 - 4 =0.

Grad f(x,y,z) = (2x, 2y, 2z) = 2(x,y,z). This is a normal vector with length 2 according to my book. Why?

I'm going to calculate a flux problem, so I need the unit tangent vector.

6. Apr 19, 2007

### Dick

To convert a normal vector to a unit normal vector, just divide the vector by its length. You know how to do that right? 2(x,y,z) actually has length 4. How did I figure that out?

Last edited: Apr 19, 2007
7. Apr 19, 2007

### HallsofIvy

Staff Emeritus
Are you sure your book says that? I presume that you know that the length of vector (a, b, c) is $\sqrt{a^2+ b^3+ c^2}$ so that the length of (2x,2y,2z) is $2\sqrt{x^2+ y^2+ z^2}$. That is equal to two if and only if the length of (x,y,z) itself is 1- that would mean your point is on the unit circle.

8. Apr 19, 2007

### kasse

No idea

Edit: yes, my mistake, the book says that the length is 4.

9. Apr 19, 2007

### Dick

Ah. You edited your problem post. Now 0.5*(x,y,z) IS a unit normal. Why?

10. Apr 19, 2007

### kasse

Because x^2+y^2+z^2=r^2=4 and sqrt(4)=2?

Last edited: Apr 19, 2007