# Dumb geometry question

1. Mar 27, 2005

### gnome

I'm trying to finish a geometrical argument (the full details will only confuse matters) and I'm down to a final (crucial) point for which I can't think of a convincing proof.

Suppose we have two planes: plane A formed by connecting points vwx and plane B formed by connecting points wxy, so line wx lies along the intersection of plane A with plane B. Now I add another plane C which is orthogonal to both A and B and also passes through points v and y. I want to say that the LINE formed by the intersection of plane C with plane B is orthogonal to line wx.

Sorry I can't provide a picture of this, but it may help to visualize a pyramid with a triangular base; the three base vertices are at w, x and y, and the apex is at v. Plane B is the base, A is the side formed by v,w and x, and C slices vertically through the pyramid along line vy.

I'm pretty sure that the line formed by C's intersection with B coincides with the altitude (along the base) from vertex y to edge wx, but I can't quite convince myself.

Edit: picture is attached to post #4

Last edited: Mar 27, 2005
2. Mar 27, 2005

### gnome

I'm thinking that since a normal to plane C is orthogonal to both a normal to plane A and a normal to plane B, the normal to plane C must be parallel to the intersection of A and B and therefore the line along C's intersection with B (which of course is orthogonal to C's normal and B's normal) must be orthogonal to the intersection of A and B.

But it just doesn't sound like a "clincher".

3. Mar 27, 2005

### whozum

Take the standard 3-d cartesian plane with axes x,y, and z
Let v be (1,0,0), w be (0,1,0) and x be (0,0,0). The plane through these is the xy plane.

Plane b intersects (0,1,0) and (0,0,0) as well as a third point y (0,0,1). This plane is the yz plane.

You want to add a third plane that is orthogonal to both of these (the xz plane) and passes through v and y (1,0,0) and (0,0,1).

You want to claim the line of intersection of the xz plane with the yz plane is orthogonal to wx (0,1,0) and (0,0,0).

The normal lines of each of these planes are all normal to each other. You are right in saying that plane C's normal is parallel to the all lines on both Plane A and Plane B.

Plane C's normal is the y axis, and the line wx connects (0,1,0) to the origin. (0,1,0) and the origin are both on the y axis. Therefore they lie on a line normal to plane C, and are therefore parallel to plane C's normal.

4. Mar 27, 2005

### gnome

Here's a picture of sorts. I'm claiming that the line yz formed by a vertical plane cutting through the pyramid along edge vy is also the altitude from vertex v to side wx.

(edited to correct label references)

#### Attached Files:

• ###### pyramid.jpg
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Last edited: Mar 27, 2005
5. Mar 27, 2005

### whozum

Do you understand my logic? It should be proof enough for any points (x,y,z) in 3 dimensions

6. Mar 27, 2005

### gnome

Sorry, I didn't see it for a while; I was working on the drawing.

Yes, I understand what you are saying, but while that's obviously true the way you define it, you have created a special case where the base and 2 sides of the pyramid are 3 mutually orthogonal planes, so I'm not sure that it's a proof of the general case.

Last edited: Mar 27, 2005
7. Mar 27, 2005

### whozum

Plane A and Plane B cant both be orthogonal to Plane C unless they are parallel or orthogonal to themselves.

8. Mar 27, 2005

### gnome

Sure they can. Fold a piece of paper to an acute angle, so the two parts are neither parallel nor orthogonal. Now put another piece of paper alongside so the fold is perpendicular to the new piece. Now the new piece is orthogonal to both parts of the folded piece.

What I want to prove (but more eloquently than that) is that the ONLY way for the two parts of the folded piece to be orthogonal to the new piece, is if the folded edge (line wx in my drawing) is perpendicular to the lines (yz and vz) formed by the intersection of the new piece with the edges of the old piece.

Anyway, gotta catch some z's. I'll look at it again tomorrow.

(Edited to add specific line references for clarity.)

Last edited: Mar 27, 2005
9. Mar 27, 2005

### whozum

Oh ok, I thought you wanted all 3 to be orthogonal. In that case thats alot more complicated so I dont know.

10. Mar 28, 2005

### Curious3141

I don't know exactly what you're looking for, but why is it so difficult ? I think it's axiomatic that if one plane is orthogonal to another then every line within the first plane is orthogonal to every line in the second plane. Plane C is orthogonal to planes A and B, meaning that plane C is also orthogonal to the intersection line between planes A and B (wx, in your diagram). Hence any line in plane C is also orthogonal to wx. Since the lines formed by the intersection of plane C with planes A and B have to lie in plane C, these lines are also orthogonal to wx. We've just proved that vw and wy are orthogonal to wx, and we're actually done.

But if you want a more "symbolic" proof, you can use vector geometry, it isn't difficult. All the variables are position vectors in R^3, I'm going to just represent them as small letters since I don't know the latex for them. '.' is dot product.

The equation of plane A is (r - w).m = (r - x).m = 0, ---(1)

where r is the general position vector in the plane and m a vector normal to plane A.

The equation of plane B is (r - w).n = (r - x).n = 0 ---(2)

Plane C is orthogonal to line wx, and hence to the vector (w - x).

Equation of plane C is (r - w).(w - x) = 0 ---(3)

Point v lies in plane C, so it must satisy eqn (3), giving

(v - w).(w - x) = 0

Since (v - w) is a vector parallel to the intersection of planes A and C, we've proved that the intersecting line b/w A and C is orthogonal to that b/w A and B.

Similarly, point y lies in C, and it must also satisfy eqn (3), hence

(y - w).(w - x) = 0

and we've proved orthogonality between the intersecting line b/w planes B and C and the intersection between A and B.

Is that the sort of thing you're looking for ?

Last edited: Mar 28, 2005
11. Mar 28, 2005

### tutor69

thats exactly the way to prove it and explain it too !! great job

12. Mar 28, 2005

### SGT

This is not true! Consider two orthogonal planes A and B. From any point in A you can draw a line that makes an arbitrary angle with the intersection of the two planes. Since the intersection belongs also to B, there is at least one line in A that is not orthogonal with one line in B.
The axiom you are thinking of is: if one line is orthogonal to a plane, it is orthogonal to all lines in that plane.

13. Mar 28, 2005

### SGT

The proof you are searching for:
From point w you can draw a single line that is orthogonal to plane C. This line is orthogonal to any line in C, including the intersections of A and C and B and C.
Since w belongs to A and the intersection AC belongs to A too, the perpendicular from w to C must belong to A.
The same reasoning is true for plan B. So, the perpendicular from w to C is the intersection of A and B. This intersection is orthogonal to every line in C.

14. Mar 28, 2005

### Curious3141

Yeah sorry, you are right, I screwed up in the statement. You can see my meaning in the proof though. Basically, wx is orthogonal to plane C hence all lines in C are orthogonal to wx.

15. Mar 28, 2005

### gnome

No, as SGT has already pointed out, this
is not true, and that was the reason for my question in the first place.

So first we need a proof of this
in order for your proof to prove anything.

(edit: we know that plane C is orthogonal to plane A and plane B. We don't know that vector [w-x] is orthogonal to plane C.)

Thanks for trying, though.

Last edited: Mar 28, 2005
16. Mar 28, 2005

### gnome

SGT:

I want to believe you, but I still don't see why this
necessarily follows from it being orthogonal to every line in C.

How do we know that the perpendicular actually intersects with either the intersection of A with C or the intersection of B with C? How do we know that it doesn't pierce plane C at some point inside or outside of the pyramid?

17. Mar 28, 2005

### gnome

Since plane C is orthogonal to plane B, at every point p along the intersection of B with C, there is exactly one line in B that is orthogonal to C.

Similarly, at every point q along the intersection of plane A with plane C there is exactly one line in A that is orthogonal to C.

Let p = q= z. There can be only one line through z that is orthogonal to C, so that line must be in both A and B, and therefore that line must be wx, the intersection of A and B.

18. Mar 28, 2005

### Curious3141

Gnome, I think I have it.

It's so simple, really, why didn't I think of it before ? Just use the cross product !

Basically, a normal vector to a plane is orthogonal to every line in the plane. When a line is formed from the intersection of two non-parallel planes, the intersecting line has to be orthogonal to both normal vectors ! Meaning that the intersecting line has to be parallel or antiparallel to the cross product of the two normal vectors.

Using my notation m and n for the normal vectors to planes A and B respectively,

(w - x) = k (m X n)

where (w - x) is the vector parallel to the intersection line wx, k is some scalar and 'X' is cross product.

When a first plane is orthogonal to another plane, its normal vector is orthogonal to the normal vector of the second plane. Since plane C is orthogonal to both A and B, normal vector (call it p) of plane C is parallel or antiparallel to the cross product of m and n.

That is,

p = r (m X n)

where r is some scalar.

Now it is obvious that p is parallel or antiparallel to (w - x). Hence wx is normal to plane C, and all lines in plane C, including the two intersections between (A and C) and (B and C) are orthogonal to wx. (QED).

Good enough ?

19. Mar 28, 2005

### gnome

Why? Maybe I'm being too much of a stickler, but what is your justification for that statement?

Anyway, what do you think of my argument in post #17? Is there anything wrong with that? Actually, I think I gave the justification for your claim about the cross product.

(Also, what do you mean by antiparallel? I looked it up & the definition I found in Mathworld doesn't seem to apply here.)

20. Mar 28, 2005

### Curious3141

OK, do you accept that every line in a plane has to be orthogonal to the normal vector to the plane ? That's an axiom, and I don't think it requires further justification.

Do you also accept that the intersecting line b/w two planes lies on both planes, and therefore must satisfy the conditions pertaining to both planes ?

From that can you see that the intersecting line has to be orthogonal to the normal vectors of both planes ?

From the definition of the cross product of two vectors, now can you see why the intersecting line has to be parallel/antiparallel to the cross product of the two normal vectors ?

OK, I didn't have time to look at it properly before now, but let's see :

To my mind, that statement has to be justified. Why one line ? Why not no line or an infinite number of lines ? I mean, I know that the line you're talking about is the projection of the normal vector of C onto plane B, which is then translated accordingly to correspond to every point p, but I think you have to formally justify this.

I think the rest of the proof is OK (given a justification for the first part), but let's see if anyone else raises any objections. Note that you're implicitly using the definition of a normal vector to a plane here when asserting that there's only one line through z that is orthogonal to C.

This is the definition I am using.

21. Mar 29, 2005

### gnome

Well, I guess this conclusion is inescapable. I certainly can't see any alternative. It hinges on the {assumption? fact? axiom?} that if any two vectors a and b are each orthogonal to both of a pair of mutually-orthogonal vectors, a and b must be parallel to each other. Does this statement need anything to justify it? I don't know. Sorry if it seems silly to question this; I just naturally fall into "devil's advocate" mode.

As far as that's concerned, it's saying that at any particular point on a plane, there is only 1 line orthogonal to that plane. So if there is a vector L through point p that is orthogonal to C, and if p is also on plane B and plane B is orthogonal to C, then L must be in B. As for justification, I guess it's that for any vector R in C, L $\cdot$ R = 0. And any vector that has a 0 dot product with R can only be a scalar multiple of the same vector. I guess it comes down to the same thing, only it's easier for me to see this way.

22. Mar 29, 2005

### Curious3141

Actually, this is not correct, it doesn't hinge on that assumption you state (the bolded part). Because the normal vectors m and n to planes A and B are not orthogonal. Basically, all that this bit hinges on is that m and n define a plane by themselves. The cross product of m and n finds a vector that is normal to the plane so defined, and that is a consequence of the way a cross product is defined. In fact, this is an established and fast method to find the vector equation of a line formed by the intersection of two planes : just cross the normal vectors and find a single point on the intersecting line.

The second part of the proof uses an obvious property of a plane that is orthogonal to another, that is, the normal vectors have to be orthogonal. If you extend this to two other planes, you can see that the normal vector of the third plane has to be orthogonal to both normal vectors. Again, the cross-product is simply an easy way to define this third normal vector.

By drawing up the conditions in cross product notation it becomes immediately apparent that the normal vector of plane C is a scalar multiple of wx, meaning the two are parallel/antiparallel. Then the rest of the proof is easy.

Of course, I suppose an obvious and implicit assumtion in all this is that we're working in 3 dimensional space, none of this would work in higher dimensions without qualification.

23. Mar 29, 2005

### gnome

I'm definitely spatially-challenged. This
I know, and it's not what I meant to say but looking back at it I clearly did.

As you keep saying, your proof hinges on this:
(w - x) = k (m X n)

and that is exactly the line that I've been looking to justify in these last few posts. It says that (w-x) is parallel to the crossproduct (mXn). We know that (w-x) is orthogonal to m. We know that (w-x) is orthogonal to n. We know that (mXn) is orthogonal to m. We know that (mXn) is orthogonal to n. And I think (correct me if I'm wrong) that those last 4 facts are your basis for writing
(w - x) = k (m X n)
I know in my gut that in 3D space this must be true, but I can't find the words to prove it.

Can you?

24. Mar 29, 2005

### SGT

Point w belongs to A and so does the intersection. It follows immediately from Euclides postulate that there is only one line in A that passes through w and is perpendicular to the intersection. This line of course belongs to A.
The same is true for plane B. So the line in A that passes by w and is perpendicular to the intersection of AC and the line in B that passes by w and is perpendicular to the intersection BC are the same, so it must be the intersection of AB.

25. Mar 29, 2005

### Curious3141

I can't justify it anymore than telling you that that is the *definition* of a vector cross product. The orientation is determined by the right hand rule, and it is basically normal to the plane determined by the two vectors that are being crossed. Which is exactly the same direction we are looking for in the normal vector to the orthogonal plane C.

Other than that, I'm sorry, I've exhausted my explanations, and I'm lost as to what exactly you're looking for. Some things (like axioms and definitions) just have to be assumed.