Proving Orthogonality of Planes and Lines in 3D Geometry

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In summary, the conversation is about trying to prove a geometrical argument involving the intersection of three planes and the orthogonality of lines within those planes. One person presents a visual aid of a pyramid to help with the explanation. Another person suggests that it is 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. The original person agrees, but wants to prove it in a more eloquent way. They end the conversation by saying they will revisit the topic tomorrow.
  • #1
gnome
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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
 
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  • #2
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
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
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)
 

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  • #5
Do you understand my logic? It should be proof enough for any points (x,y,z) in 3 dimensions
 
  • #6
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.
 
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  • #7
Plane A and Plane B can't both be orthogonal to Plane C unless they are parallel or orthogonal to themselves.
 
  • #8
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, got to catch some z's. I'll look at it again tomorrow.

(Edited to add specific line references for clarity.)
 
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  • #9
Oh ok, I thought you wanted all 3 to be orthogonal. In that case that's a lot more complicated so I don't know.
 
  • #10
gnome said:
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

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 ?
 
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  • #11
thats exactly the way to prove it and explain it too ! great job
 
  • #12
Curious3141 said:
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.

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
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
SGT said:
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.

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
No, as SGT has already pointed out, this
Curious3141 said:
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.
is not true, and that was the reason for my question in the first place.

So first we need a proof of this
Curious3141 said:
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)
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.
 
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  • #16
SGT:

I want to believe you, but I still don't see why this
SGT said:
Since w belongs to A and the intersection AC belongs to A too, the perpendicular from w to C must belong to A.
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
How about this. If anybody finds this to be unconvincing, please tell me so.

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
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 ? :biggrin:
 
  • #19
Curious3141 said:
... 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.

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

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 ?

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.

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

gnome said:
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.

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.

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.

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.

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

This is the definition I am using.
 
  • #21
Curious3141 said:
...now can you see why the intersecting line has to be parallel/antiparallel to the cross product of the two normal vectors?

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.


Curious3141 said:
Quote:
Originally Posted by 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.


To my mind, that statement has to be justified. Why one line ? Why not no line or an infinite number of lines ?
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 [itex]\cdot[/itex] 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.

Anyway, thanks for your help. If you have anything to add to my "devil's advocate" question, please toss it in.
 
  • #22
gnome said:
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.

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
:cry: :cry:

I'm definitely spatially-challenged. This
Curious3141 said:
Because the normal vectors m and n to planes A and B are not orthogonal.
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
gnome said:
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?

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
gnome said:
:cry: :cry:

I'm definitely spatially-challenged. ThisI 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?

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.
 
  • #26
I don't know why it is so hard to understand my question.

The *definition* of a vector crossproduct is a vector that is orthogonal to both of the original vectors. I'm not questioning that the crossproduct is k(mXn)! I'm questioning (w-x) = k(mXn). What justifies that equality?
 
  • #27
gnome said:
I don't know why it is so hard to understand my question.

The *definition* of a vector crossproduct is a vector that is orthogonal to both of the original vectors. I'm not questioning that the crossproduct is k(mXn)! I'm questioning (w-x) = k(mXn). What justifies that equality?

Do you agree that in 3-d space, one and only one unique plane is defined by two non-collinear vectors ?

Do you agree that a particular plane has one and only one unique orientation of the normal vector ?

If you accept those premises, look at this :

(w-x) lies in planes A and B. Every line in plane A is orthogonal to m, every line in plane B is orthogonal to n. *Therefore*, (w - x) is orthogonal to both m and n.

Now m and n define a unique plane. The normal vector to this plane is orthogonal to this plane, and it is unique in orientation. Since it is unique in orientation, because (w - x) is *also* orthogonal to m and n, (w - x) *has* to lie in the same orientation, that is, it is parallel/antiparallel to the normal vector.

I don't know what else to say. Please tell me exactly what problem you're facing with the above.

BTW, I'm fairly confident that my proof would be perfectly acceptable in any respected mathematical journal or textbook. I'm honestly finding it hard to come up with ever more elementary arguments, because I think most people would be able to accept this with zero problems.
 
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  • #28
This would be hilarious if it weren't so frustrating. You have just repeated what I already stipulated that I agreed with a few posts earlier when I said
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)

So, trying again to explain my question, with new letters so it doesn't relate to this or any other specific case.
You say,

Code:
IF
     (A is orthogonal to F) AND (A is orthogonal to G)
         AND (FXG is orthogonal to F) AND (FXG is orthogonal to G)
THEN
     A is parallel to FXG

Is that conclusion simply an axiom, or should it be proved?

Thanks for your patience.
 
  • #29
gnome said:
This would be hilarious if it weren't so frustrating. You have just repeated what I already stipulated that I agreed with a few posts earlier when I said

So, trying again to explain my question, with new letters so it doesn't relate to this or any other specific case.
You say,

Code:
IF
     (A is orthogonal to F) AND (A is orthogonal to G)
         AND (FXG is orthogonal to F) AND (FXG is orthogonal to G)
THEN
     A is parallel to FXG

Is that conclusion simply an axiom, or should it be proved?

Thanks for your patience.

Sorry, it's fine if you want to change the letters , but could you just rephrase that stating which are the lines and which are the planes ? It's late at night here (early in the morning actually) and I can't focus, and don't want to draw a diagram.
 
  • #30
They're all just lines.
 
  • #31
gnome said:
They're all just lines.

Oh, 'X' means cross product. I thought you meant something else by FXG, etc., the lack of spacing threw me.

Then what you wrote is axiomatic, and it follows from the fact that we're working in 3 dimensions. Simple enough ? :smile:
 
  • #32
I guess. It just bothers me that I can't put my finger on any specific definition or postulate that says that.
 
  • #33
Draw a perpendicular from point w to the plane C. Any plane that contains that line is, by definition, perpendicular to C.
In particular, the plane that contains that perpendicular and the intersection of A and C. Since a point and a line define a plane, this plane must be plane A.
Now, consider the plane that contains the perpendicular and the intersection of B and C. By the same reasoning, this must be plane B.
So, the perpendicular from w to C belongs simultaneously to A and B and must be their intersection.
The intersection of A and B is orthogonal to all lines in C. QED.
 

1. How do you define orthogonality in 3D geometry?

Orthogonality in 3D geometry refers to the perpendicularity of two lines or planes in three-dimensional space. This means that the two lines or planes intersect at a right angle, forming a 90-degree angle.

2. What is the mathematical method used to prove orthogonality of planes and lines?

The most commonly used method to prove orthogonality of planes and lines in 3D geometry is the dot product. This involves taking the dot product of the direction vectors of the two lines or planes and showing that it equals zero, indicating that the two vectors are perpendicular.

3. Can you provide an example of proving orthogonality in 3D geometry?

Sure, let's say we have two lines with direction vectors u = (2, 3, 4) and v = (1, -2, 1). To prove their orthogonality, we would take the dot product: u · v = (2)(1) + (3)(-2) + (4)(1) = 2 - 6 + 4 = 0. This shows that the two lines are perpendicular.

4. Is there any other method to prove orthogonality in 3D geometry?

Yes, another method is using the cross product. If the cross product of the direction vectors of two lines or planes is equal to the zero vector, then the two vectors are perpendicular and thus the lines or planes are orthogonal.

5. How is proving orthogonality in 3D geometry useful in real-world applications?

Proving orthogonality in 3D geometry is useful in various fields such as engineering, architecture, and computer graphics. It allows for the accurate calculation of angles and distances between lines and planes, which is essential in designing and constructing structures and objects in three-dimensional space.

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