# Proving the dot product between these vectors is always the same value

1. Nov 22, 2012

### autodidude

How would you show that the dot product between the normal unit vector of a plane and a position vector to any point on the plane is always the same without using this formula

$$n.(r-r_0) = 0$$
∴ $$n.r=n.r_0$$

where $$n$$ is the normal vector, $$r$$ and $$r_o$$ are two position vectors to two points on the plane.

I'm looking for an alternative geometric argument/proof that applies to all cases.

I notice that if you have a point P on a plane that is directly above the origin which is parallel to the xy-plane, then the dot product is simply the magnitude of the vector OP. Then as you move further out from the origin to some point $$P_n$$ on the plane, the position vector gets larger and the projection of the unit normal vector on the vector $$OP_n$$ gets smaller. One gets larger, the other gets smaller and somehow their product is always the same.

So again, I'm after a way to prove this for all cases geometrically.

Thanks

2. Nov 22, 2012

### Simon Bridge

Off the geometric interpretation of the dot product, you are trying to show that orthogonal vectors do not have any extent in each others direction?

What is wrong with:
Use the x-y plane ... all other planes follow on a rotation + translation.
The normal is n=(0,0,1)
Any point in the plane is (x,y,0) from origin.
Your result follows from the definition of a dot product.
All other position vectors follow from a translation.

3. Nov 23, 2012

### HallsofIvy

Staff Emeritus
First of all, the statement, as given, is not true. Let the plane be given by z= 1 in an xyz-coordinate system and the given point (2, 2, 1). The unit normal to the plane is of the form <0, 0, 1>. The position vector of the point is <2, 2, 1> and their dot product is 1, not 0.

You appear to be assuming that the plane contains the origin. In that case the position vector of any point in the plane is in the plane itself and so is perpendicular to the normal vector.

4. Nov 23, 2012

### mathman

Geometric proof. The difference between any two points in a plane is a line in the plane. The normal is perpendicular to all lines in the plane, so that dot product = 0.

5. Nov 23, 2012

### Simon Bridge

@autodidude: perhaps you did not phrase the question as well as you'd hoped? Want to have another go?

Tell us what you are thinking of when you say "geometric proof", and what it is you want to prove.

From the context, I just guessed that "a position vector to a point int he plane" meant "a vector between any two points in the plane"... but I noticed that you seemed to have defined the thing you want to prove ... that is: the "normal vector to the plane" is, by definition the vector whose dot product with any vector that lies in the plane is zero.

In general, you'll find you can make more headway by turning natural-language statements into mathematical statements.

6. Nov 25, 2012

### autodidude

Well, in the textbook, he writes
$$n.(r-r_0) = 0$$
∴ $$n.r=n.r_0$$

So the dot product between the unit normal vector to the plane and the position vector to any point on the plane is always the same value - no matter what point you pick.

I just wanted to see a different approach as to why this is true without simply rearranging the formula.

I've been trying to prove it myself and the best I can come up with is this:

Let P be any point on the plane such that the position vector OP is parallel to the normal unit vector n. The dot product between OP and n is |OP|. Then to show that the dot product between the normal unit vector and a position vector to any other point on the plane Q, I did the following:

$$OQ.n=|OQ||n|cos(\theta)$$

Since $$|OQ|=\frac{|OP|}{cos(\theta)}$$, then

$$OQ.n=\frac{|OP|}{cos(\theta)}|n|cos(\theta)$$
$$OQ.n=|OP||n|$$
$$OQ.n=|OP|$$

Does this look alright?

7. Nov 25, 2012

### Simon Bridge

Provided the positio vector is in between two points in the plane.

$\vec{p}$ is a position vector to a point P in the plane. You'd say "OP" right?
$\vec{q}$ is another position vector to another point Q in the plane, so $\vec{p}\neq\vec{q}$.

The difference between these two positions would be a displacement vector within the plane. $\vec{d}=\vec{q}-\vec{p}$ (from P to Q, final minus initial)

If $\vec{v}$ is a vector that does not lie in the plane, then $\vec{v}\cdot(\vec{p}-\vec{q})=0$ means that $\vec{v}$ is normal to the plane. The equation "he" wrote down is part of the definition of "a plane surface".

The plane is defined geometrically as the set of all points whose displacement from a given point is perpendicular to a given vector. The given vector is called the "normal" to the plane.

In general, if $\vec{n}$ is a (unit) normal vector to a plane, then it's dot product with any position vector to a point in the plane will be different depending on the point vis: $\vec{n}\cdot\vec{p}\neq\vec{n}\cdot\vec{q}\neq 0$ ... it might be, but it will not always be.

Using this formalism, what you have written is:

(1) $\vec{p}=p\vec{n}$ eg OP has the same direction as the unit normal.

(2) hence: $\vec{p}\cdot\vec{n}=p$

if Q is in the plane, and Q is not P, then:

(3) $\vec{q}\cdot\vec{n}=q\cos(\theta)$ ... where $\theta$ is the angle between them.

Since OQ forms the hypotenuse of a right-triangle OPQ - you can say that
(4) $p=q\cos(\theta)$

So, by substitution into (3)
(5) $\vec{q}\cdot\vec{n}=p$

... so far so good.

Notice that the vector notation is the same as your labelling by capitol letters.
There is a conceptual distinction though isn't there: OP (conceptually) must start at the origin and end at point P while the vector $\vec{p}=\overrightarrow{OP}$ can also stand for any vector parallel to it that has the same length.

I have a feeling you are trying to find the link from the concrete labelled points concepts and the general vector concepts.

8. Nov 26, 2012

### autodidude

What do you mean if the position vector is in between two points? I thought position vectors always referred to a vector emanating from the origin?

But doesn't the definition of a plane surface say that the dot products should be the same when you expand and reaarange it? I also now realise the author never said 'n' was a UNIT normal vector, I just assumed it was for some reason...shouldn't this then mean that the dot product 'equality' (which I can't yet see what is wrong) should be true for all normal vectors not just unit ones?

9. Nov 26, 2012

### Simon Bridge

All position vectors are relative to some reference point ... unless one is stated, the defined origin is used. If no origin has been defined, then you have to use your head. Strictly speaking, a the vector from one point to another would be a displacement since it is the difference between two positions ... but it could also be the position of one object with respect to another.

Recall: the original proposition was:
... from post #1.
This proposition is false. (It will only work for planes that include the origin.)
Now: how can you modify that proposition to make it true, in general?

The dot product of any vector normal to the plane and any vector that lies in the plane will always be zero. This is geometrically the same as saying that the normal vector is perpendicular to the plane.

eg.
In your example, previous, P and Q are points in the plane: $\vec{p}=\overrightarrow{OP}$ was chosen to be a normal vector to the plane.
Any vector parallel to OP will also be a normal vector to the plane.

If $|\vec{p}|\neq 0$ and $\vec{q}=\overrightarrow{OQ}\neq \vec{p}$

Then: $\vec{p}\cdot\vec{q}\neq 0$ because $\vec{q}$ is not in the plane: it is the position of a point in the plane: different things.

However: $\vec{p}\cdot (\vec{q}-\vec{p})=0$ because $(\vec{q}-\vec{p})=\overrightarrow{PQ}$ is a vector that lies in the plane.

Since the dot product is commutative:

$\vec{p}\cdot\vec{q}=\vec{p}\cdot\vec{p} \Rightarrow p = q\cos(\theta)$

This is consistent with the observation that OPQ froms a right angled triangle with $\vec{q}$ as the hypotenuse. But notice that $\vec{p}\cdot (\vec{q}-\vec{p})=0$ is just the relation you don't want to use with different letters?

10. Nov 27, 2012

### autodidude

Hmmm...oh, I think that's what HallsOfIvy was saying...I don't really see how it would only work for planes that include the origin though. I'm not trying to say that thw dot product is 0, I'm saying it's always the.same value for some normal vector. So in HallsOfIvy's example, the chosen normal vector is <0, 0, 1>. If you pick any other point on the plane, it'll be of the form <a, b, 1> and so the dot product between that and <0, 0, 1> will always be 1.

The normal vector to the plane z=1 will always be of the form <0, 0, k> and hence the position vector from the origin to any point on the plane <a, b, 1> dotted with the normal vector will be equal to k. That is what i thought this meant:

$$n.r=n.r_0$$

Yeah, I'm not saying the vector q dotted with the vector p is 0. I get that only vectors lying in the plane dotted with the normal to the plane equals zero. What I'm trying to say is that the position vectors from the origin to some point on the plane, when dotted with a normal vector of some chosen magnitude is always the same value no matter which position vectoe you choose so long as the point it end at lies on the plane.

...I also just realised that I'm assuming the normal vector is specified by a positiin vector originating from the origin...

11. Nov 27, 2012

### Simon Bridge

Oh well ... the examples you showed in post #1 had that costant as zero but now I get you: you mean that (for P and Q etc already discussed) $\vec{p}\cdot\vec{q}=p^2$ which is always the same for a particular $\vec{p}$ ? Well, yes. $p^2$ depends only on $\vec{p}$ which you have chosen to be a normal to the plane.

Remember the geometric interpretation of the dot product?
What you are doing is constructing triangles whose adjacent side is shared.
Hence: any arbitrary $\vec{q}$ will have the same extent in the normal direction.
A drawing of the triangle is your geometric proof.

12. Nov 28, 2012

### autodidude

Ok, thanks a lot, that's really all I wanted to know!

13. Nov 28, 2012

### Simon Bridge

Cool.

BTW: the proof is the same as in post #2 - if you make the plane an arbitrary distance along the z axis. The "constant" you get in this process is the shortest distance from the origin to the plane (multiplied by the length of the normal vector you used).