# Normal plane

1. Apr 5, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Find the normal vector equation (the equation with red in the picture below) of the plane which is normal on the vector n=(2,-2,1), and it is on distance 5 from O (i.e p=5)

2. Relevant equations

On this image are all the relevant equations.
http://img232.imageshack.us/img232/6030/28762075sm2.jpg

3. The attempt at a solution

I should find $$n_o$$, and then substitute in the equation.
$$n_o=\frac{n}{|n|}$$
So n=2i-2j+k, |n|=3, p=5
$$n_o=\frac{2i-2j+k}{3}$$

So the equation:

$$r * (\frac{2i-2j+k}{3}) - 5 =0$$

The problem is that in my book they don't divide it by 3? Why ? Is my way correct?

2. Apr 5, 2008

### Hootenanny

Staff Emeritus
Does it ask you for the unit normal, or just the normal?

3. Apr 5, 2008

### Physicsissuef

The angle between the plane and the vector is 90 degrees. Nothing more.

4. Apr 5, 2008

### Hootenanny

Staff Emeritus
No, my point is the question asks you to find the normal vector, not the unit normal. There is no need to divide by it's magnitude.

5. Apr 5, 2008

### Physicsissuef

Actually, we should find the unit normal. I think we should divide it by its magnitude, because the definition says $$|n_o|=1$$

6. Apr 5, 2008

### Hootenanny

Staff Emeritus
In the question you stated,
there is no mention of a unit normal. Hence the question wants n not n0.

7. Apr 5, 2008

8. Apr 5, 2008

### Hootenanny

Staff Emeritus
Yes but where is that image taken from, a course text, lecture notes?

9. Apr 5, 2008

### Physicsissuef

from my textbook. It also says that $|n_o|=1$

10. Apr 5, 2008

### Hootenanny

Staff Emeritus
Does it occur to you that you are not meant to use that equation in this question?

11. Apr 5, 2008

### Physicsissuef

Then, when I should use this equation?

12. Apr 5, 2008

### Hootenanny

Staff Emeritus
Perhaps,

$$r\cdot \boldmath{n} - p = 0$$

Since you are not asked for the unit normal, simply the normal vector.

13. Apr 5, 2008

### Physicsissuef

If we write
$$r\cdot \boldmath{n} - p = 0$$
then it wouldn't be plane, because equation of plane is only when $|n_o|=1$.

14. Apr 5, 2008

### Hootenanny

Staff Emeritus
Yes it is still an equation of a plane, the very same plane in fact! The equation of a plane can be derived from any normal vector, not just a normalised one!

15. Apr 5, 2008

### Physicsissuef

Now, I even doubt why p=5, it should be p=3, since OP=p$n_o$. Well, I think I am tottaly confused about the whole thing...

16. Apr 5, 2008

### Hootenanny

Staff Emeritus
Would it be possible for you to scan the actual diagram from your textbook and post it here?

17. Apr 5, 2008

### Physicsissuef

18. Apr 5, 2008

### Hootenanny

Staff Emeritus
Thanks for the pictures, could you also type out the question in full, with all the information given?

19. Apr 5, 2008

### Physicsissuef

Find the normal vector equation of the plane which is normal on the vector n=(2,-2,1), at distance 5 from (0,0,0) (the coordinate start, sorry if I mistranslated). You can see the task on the third picture 88 page/1 task.

20. Apr 5, 2008

### Hootenanny

Staff Emeritus
Thanks, perhaps something is getting lost in translation here. Does the vector n=(2,-2,1) lie on the plane, i.e. is it part of the plane, or is it perpendicular to it?