Normal Vector Equation for Plane with n=(2,-2,1) at Distance 5 from Origin

[Physicsissuef]

Homework Statement

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)

Homework Equations

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

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?

 

[Hootenanny]

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

 

[Physicsissuef]

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

 

[Hootenanny]

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

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.

 

[Physicsissuef]

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

 

[Hootenanny]

In the question you stated,

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)

there is no mention of a unit normal. Hence the question wants n not n₀.

 

[Physicsissuef]

But the formula is with n_o
http://img232.imageshack.us/img232/6030/28762075sm2.jpg

 

[Hootenanny]

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

 

[Physicsissuef]

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

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

 

[Hootenanny]

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

 

[Physicsissuef]

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

Then, when I should use this equation?

 

[Hootenanny]

Perhaps,

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

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

 

[Physicsissuef]

Perhaps,

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

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

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.

 

[Hootenanny]

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.

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!

 

[Physicsissuef]

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

 

[Hootenanny]

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

 

[Physicsissuef]

Here are the pictures of the lection:
http://pic.mkd.net/images/138086IMG_1228.jpg
http://pic.mkd.net/images/843151IMG_1229.jpg
http://pic.mkd.net/images/709895IMG_1230.jpg

 

[Hootenanny]

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

 

[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.

 

[Hootenanny]

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?

 

[Physicsissuef]

n is perpendecular to the plane... As you can see it have 2 rows of text. That's the full text.

 

[Hootenanny]

Ahh, I follow now. The diagram is somewhat confusing, however, it should become clear if you note that

\vec{OP}\neq \vec{n_0}

Although it appears so in the diagram, it is not the case.

 

[Physicsissuef]

In this case it is not. But how will I find the equation then?

 

[Hootenanny]

In this case it is not. But how will I find the equation then?

As I said before,

\underline{r}\cdot\underline{n} - p = 0

You are given both n and p.

 

[Physicsissuef]

As I said before,

\underline{r}\cdot\underline{n} - p = 0

You are given both n and p.

can u tell me please where did u find this equation
\underline{r}\cdot\underline{n} - p = 0 from?

 

[Hootenanny]

It's the same as yours, but instead of using a unit normal vector, I'm simply using the[non-normalised] normal vector.

 

[Physicsissuef]

It cannot be like that. Because of:
PM*n_o=0
PM=OM-OP=r-pn_o
(r-pn_o)*n_o=0
because of the scalar multiplycation:
n_o*n_o=1
r*n_o-p=0
try to get the same with just n, and you'll see what I am talking about.

 

[Hootenanny]

Ahh, you are quite correct. I apologise, I missed that. If the question is as stated, then your solution is correct. My apologies for wasting your time

 

[Physicsissuef]

Ahh, you are quite correct. I apologise, I missed that. If the question is as stated, then your solution is correct. My apologies for wasting your time

No problem. You are not wasting my time. Its the opposite, I appreciate your efforts to try to help me. This is just some confusing situation, which is not tottaly same as if we have given some point P(-2,1,2) and to find the equation, which will be
\frac{-2i+j+2k}{3}-3=0
The task of this topic, even it looks easy, its quite confusing...

 

[Physicsissuef]

But in my example, shouldn't p=3 instead of p=5??

 

[HallsofIvy]

I would do this in a somewhat more pedestrian way. Any plane perpendicular to vector (2, -2, 1) can be written in the form 2x- 2y+ z= A. We need to find A from the condition that "the distance from (0,0,0) is 5". The distance from a plane to a point is measured along the perpendicular to the plane and the line through (0, 0, 0) perpendicular to this plane is given by x= 2t, y= -2t, z= t. Putting that into the equation of the plane, 2(2t)- 2(-2t)+ (t)= A gives 4t+ 4t+ t= 9t= A. That is, t= A/9 and so the closest point on the plane to (0,0,0) is ((2/9)A, -(2/9)A, (1/9)A). The distance from that point to (0,0,0) is \sqrt{(4/81)A^2+ (4/81)A^2+ (1/81)A^2}= (1/3)A. In order that that be "5" we must have A= 15. The equation of the plane is 2x- 2y+ z= 15.

In vector notation that is (2, -2, 1)\cdot(x, y, z)- 15= 0

 

[Physicsissuef]

What about my equation?

 

[HallsofIvy]

If you mean :

No problem. You are not wasting my time. Its the opposite, I appreciate your efforts to try to help me. This is just some confusing situation, which is not tottaly same as if we have given some point P(-2,1,2) and to find the equation, which will be
\frac{-2i+j+2k}{3}-3=0
The task of this topic, even it looks easy, its quite confusing...

It looks to me like you are subtracting a number from a vector- you can't do that! You need a dot product.

 

[Physicsissuef]

Look at the picture on the first page. p=|OP| and OP=p*n_o

So if n_o=\frac{n}{|n|}, then p should be 3.

 

[Hootenanny]

Note that usually r is the position vector of a general point such that r = xi+yj+zk. Hence, if n = ai+bj+ck then,

\underline{r}\cdot\underline{n} - p = 0

\left(x\underline{i} + y\underline{j} + z\underline{k}\right)\cdot\left(a\underline{i} + b\underline{j} + c\underline{k}\right)-p=0

ax+by+cz - p =0

Which is an equation of a plane in three space.

 

[Physicsissuef]

Note that usually r is the position vector of a general point such that r = xi+yj+zk. Hence, if n = ai+bj+ck then,

\underline{r}\cdot\underline{n} - p = 0

\left(x\underline{i} + y\underline{j} + z\underline{k}\right)\cdot\left(a\underline{i} + b\underline{j} + c\underline{k}\right)-p=0

ax+by+cz - p =0

Which is an equation of a plane in three space.

But again, something is wrong, since if p=|n|. In this case it isnt.

 

[Physicsissuef]

I would do this in a somewhat more pedestrian way. Any plane perpendicular to vector (2, -2, 1) can be written in the form 2x- 2y+ z= A. We need to find A from the condition that "the distance from (0,0,0) is 5". The distance from a plane to a point is measured along the perpendicular to the plane and the line through (0, 0, 0) perpendicular to this plane is given by x= 2t, y= -2t, z= t. Putting that into the equation of the plane, 2(2t)- 2(-2t)+ (t)= A gives 4t+ 4t+ t= 9t= A. That is, t= A/9 and so the closest point on the plane to (0,0,0) is ((2/9)A, -(2/9)A, (1/9)A). The distance from that point to (0,0,0) is \sqrt{(4/81)A^2+ (4/81)A^2+ (1/81)A^2}= (1/3)A. In order that that be "5" we must have A= 15. The equation of the plane is 2x- 2y+ z= 15.

In vector notation that is (2, -2, 1)\cdot(x, y, z)- 15= 0

H0w did u find x=2t, y=-2t, z=t?

 

[HallsofIvy]

Parametric equations of a line in the direction of vector (A, B, C) passing through point (x₀, y₀, z₀) are x= At+ x₀, y= Bt+ y₀, and z= Ct+ a₀. Didn't you learn that before the equation of a plane?

 

[Physicsissuef]

No I didn't learn it... And is n and n_o are 2 different kind of vectors?

 

[Physicsissuef]

HallsofIvy?

 

[HallsofIvy]

I'm not sure what you mean by "n₀" and "n". You are referring to the "normal" vector and "unit normal". No they are not different kinds is is just that n₀ has length 1 while n can have any length.