When I say to my friends I maths can be beautiful and elegant they look at me funny – but this is exactly the kind of thing I mean.

MrHonner
· September 22, 2012 at 11:59 am

Thanks, Ravi. I was thinking about a simple way to understand this fact and I worked my way through line of reasoning. I’m pretty happy with how it came out.

Yes, I know the feeling–not everyone is able to really appreciate mathematics. But that’s why we keep trying, right?

Nat
· September 23, 2012 at 9:16 am

The “proof” shows more; that is, there are infinitely many such perpendicular lines. But isn’t it being assumed that the dihedral angles are continuously changing and doesn’t that have to be proved? Anyway, it is a wonderful “plausibility argument.”

MrHonner
· September 25, 2012 at 9:14 pm

This argument definitely relies on the continuity of that changing angle. As a naturally defined function, I might argue that it’s up to you to show it’s not continuous. 🙂

Nat
· September 26, 2012 at 2:27 am

Taking a cue from your work, it is relatively easy to construct an analytic proof. Let z = 0 be one plane and z = m*y be the other. Without loss of generality, take m > 0. It seems arctan(m) is the dihedral angle between the planes, which intersect along the x-axis. Consider the line given by y = x in R^2, which passes through the origin and lies in z = 0. One way to parametrize it is: x = t = y and z = 0. A vector in the direction of the line is . A line perpendicular to it that lies in z = m*y and also passes through the origin is given by: x = -s, y = s, z = m*s. A vector in the direction of this line is . The vectors are perpendicular, since their dot product is zero.

Mrs. F
· October 16, 2013 at 6:10 pm

I’m a little confused. Your 3rd diagram and your 5th diagram are comparing coplanar lines. Your 6th diagram is comparing noncoplanar lines. Now, I know that 2 intersecting lines are always coplanar, and in fact, if you were to “unfold” your planes…flatten them out (one plane) your lines in the last figure would be perpendicular. However, you are using the lines in a noncoplanar fashion in the last diagram and therefore they are not perpendicular in this 3D representation.

MrHonner
· October 16, 2013 at 7:09 pm

This is essentially an Intermediate Value Theorem argument. The angle in diagram 3 is acute; the angle in diagram 5 is obtuse. As the angle transitions from acute to obtuse (as suggested in diagram 4), it must be, at some instance, be right.

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## 7 Comments

## Ravi · September 22, 2012 at 4:35 am

When I say to my friends I maths can be beautiful and elegant they look at me funny – but this is exactly the kind of thing I mean.

## MrHonner · September 22, 2012 at 11:59 am

Thanks, Ravi. I was thinking about a simple way to understand this fact and I worked my way through line of reasoning. I’m pretty happy with how it came out.

Yes, I know the feeling–not everyone is able to really appreciate mathematics. But that’s why we keep trying, right?

## Nat · September 23, 2012 at 9:16 am

The “proof” shows more; that is, there are infinitely many such perpendicular lines. But isn’t it being assumed that the dihedral angles are continuously changing and doesn’t that have to be proved? Anyway, it is a wonderful “plausibility argument.”

## MrHonner · September 25, 2012 at 9:14 pm

This argument definitely relies on the continuity of that changing angle. As a naturally defined function, I might argue that it’s up to you to show it’s

notcontinuous. 🙂## Nat · September 26, 2012 at 2:27 am

Taking a cue from your work, it is relatively easy to construct an analytic proof. Let z = 0 be one plane and z = m*y be the other. Without loss of generality, take m > 0. It seems arctan(m) is the dihedral angle between the planes, which intersect along the x-axis. Consider the line given by y = x in R^2, which passes through the origin and lies in z = 0. One way to parametrize it is: x = t = y and z = 0. A vector in the direction of the line is . A line perpendicular to it that lies in z = m*y and also passes through the origin is given by: x = -s, y = s, z = m*s. A vector in the direction of this line is . The vectors are perpendicular, since their dot product is zero.

## Mrs. F · October 16, 2013 at 6:10 pm

I’m a little confused. Your 3rd diagram and your 5th diagram are comparing coplanar lines. Your 6th diagram is comparing noncoplanar lines. Now, I know that 2 intersecting lines are always coplanar, and in fact, if you were to “unfold” your planes…flatten them out (one plane) your lines in the last figure would be perpendicular. However, you are using the lines in a noncoplanar fashion in the last diagram and therefore they are not perpendicular in this 3D representation.

## MrHonner · October 16, 2013 at 7:09 pm

This is essentially an Intermediate Value Theorem argument. The angle in diagram 3 is acute; the angle in diagram 5 is obtuse. As the angle transitions from acute to obtuse (as suggested in diagram 4), it must be, at some instance, be right.