Angle of the diagonal of a cube

[DaveC426913]

TL;DR Summary

What can I do to figure out the angle between the diagonal of a cube and any of the 3 edges at a given vertex?

I've looked for this in a hundred places but I keep finding the unit length of the diagonal (root 3). I want to find the angle.
i.e. the angle between the cube's diagonal and its three edges at a vertex (all three angles will be identical, of course).

I guess I hoping to figure it out without trig. I really only need it to one (or zero) decimals of a degree, really.

(Hey, I wonder if it will just be .707 of 45 degrees... or about 31.8 deg.)

 

[jedishrfu]

Is this a homework assignment? It sure looks like one.

Spoiler: Don't peek if it is:

Okay on a cube face of a 1" cube, the diagonal is sqrt(2)

The diagonal you want forms a triangle with the cube face diagonal as one side and 1 as the other of a right triangle so the tangent of the angle would be 1/sqrt(2).

In radians, its about 0.615 or 35.2 degrees.

The complement of that is 90-35.2 = 54.8 degrees which would be the angle with a cube edge.

 

[Office_Shredder]

You can do this with some simple vector math which does involve trig but barely.

The edge of the cube is a vector like (1,0,0), the main diagonal is (1,1,1). The dot product of these two is equal to cosine of the angle times the product of the norms of the vectors. Then you just need a little algebra and take arccosine.

Your intuition is wrong. If you project the diagonal onto a face it makes a 45 degree angle with an edge. When you let it swing out into the third dimension the angle will get bigger, not smaller

 

[DaveC426913]

It isn't homework. It's part of the project I've been working on.

 

[Office_Shredder]

If you're not sure how to do that computation in my previous post, the dot product is just summing the product of the coordinates of the vectors.

1*1+1*0+1*0=1.

The two vectors have norms (aka lengths) computed by the pythagorean theorem
$\sqrt{1^2+1^2+1^2}=\sqrt{3}$ and$\sqrt{1^2+0^2+0^2}=1$

the statement I made is the dot product is the product of the lengths and cosine of the angle, so

$1=\sqrt{3}*1*\cos(\theta)$
$\theta=\arccos(1/\sqrt{3})$

plugging that into a calculator gives 54.74 degrees.

 

[DaveC426913]

Thanks. Never got the hang of arccoses and secant...

54.74. Interesting... That's the same angle that keeps coming up when I'm rotating tetrahedra in Blender. Almost like it's not a coinky-dink.

 

[TeethWhitener]

It crops up in interesting places:
https://en.m.wikipedia.org/wiki/Magic_angle

 

[DaveC426913]

It crops up in interesting places:
https://en.m.wikipedia.org/wiki/Magic_angle

Sunuvagun, I had no idea.

 

[aheight]

How about angles in any regular polyhedron? Any magic numbers there? Would make for an interesting problem I think: Find a polynomial for which a root is a magic number of a dodecahedron.

 

[Mark44]

plugging that into a calculator gives 54.74 degrees.

And if what was wanted was the angle between the diagonal and the plane of the base, that would be about 48.19°. 35.26° (edited to correct my error in forgetting to take a square root)

The formula that @Office_Shredder used for the angle $\theta$ between two vectors $\vec u$ and $\vec v$ is this:
$$\cos(\theta) = \frac{\vec u \cdot \vec v}{|\vec u| |\vec v|}$$
Post #5 shows the calculations of the dot product in the numerator, and the two vector magnitudes.

Never got the hang of arccoses and secant...

It's really not all that complicated. In words, arccos(x) means "the angle whose cosine is x. For the purposes here, arccos is the same as $\cos^{-1}$, so working in degrees, arccos(1/2) = $\cos^{-1}(1/2)$ = 60°. The other way around, it is $\cos(60°) = 1/2$. All of the so-called circular trig functions -- sine, cosine, tangent, cosecant, secant, cotangent -- have inverses

 

[Office_Shredder]

And if what was wanted was the angle between the diagonal and the plane of the base, that would be about 48.19°.

No, I think this number is less than 45 degrees, because (1,0) and (1,1) form a 45 degree angle, so (1,1,0) and (1,1,1) should form an angle of less than 45 degrees. I tried doing the same computation and got 35.26 degrees.

 

[Mark44]

No, I think this number is less than 45 degrees, because (1,0) and (1,1) form a 45 degree angle, so (1,1,0) and (1,1,1) should form an angle of less than 45 degrees. I tried doing the same computation and got 35.26 degrees.

Yep, you're right. I did some simplification of $\frac 2 {\sqrt 6}$ to $\sqrt { \frac 2 3}$, which is correct, but then neglected to actually take the square root before doing the inverse cosine. Doh!

I'll edit my earlier post.

 

[Office_Shredder]

I realized the simpler way to do it which is that that angle is 90-angle to the edge.

 

[PeroK]

It crops up in interesting places:
https://en.m.wikipedia.org/wiki/Magic_angle

I was thinking about Bell's Theorem recently and, specifically, whether I could more directly show that hidden variables were inadequate to reproduce the predictions of QM. I didn't come up with anything, but I noticed that:

If a particle is in the spin-up state in the direction $(1, 1, 1)$ then the probability that it is found to be spin-up in any of the x, y or z directions is given by $\cos^2 \frac \theta 2$, where $\theta$ is the magic angle. This gives a probability of: $\frac 1 2 (1 + \cos \theta) = \frac 12 (1 + \frac 1 {\sqrt 3})$, which is about $0.79$.

That's a huge correlation.

 

[jedishrfu]

The angle of the cube diagonal with the cube face plane is about 35.2 degrees ie arctan(1/sqrt(3))

 

[Charles Link]

The angle of the cube diagonal with the cube face plane is about 35.2 degrees ie arctan(1/sqrt(3))

I believe it is the arc sine, and not the arc tan. It's the arc tan of (1/sqrt(2)).

 

[PeroK]

I believe it is the arc sine, and not the arc tan. It's the arc tan of (1/sqrt(2)).

Yes, you can use the same idea using the inner product of the vectors $(1, 1, 1)$ and $(1,1)$ to get the cosine of the angle between a diagonal and a face: $$\sqrt 3 \sqrt 2 \cos \theta = 2 \ \implies \ \cos \theta = \sqrt{\frac 2 3} \ \implies \ \sin \theta = \sqrt{\frac 1 3} \ \implies \ \tan \theta = \sqrt{\frac 1 2}$$

 

[jedishrfu]

I believe it is the arc sine, and not the arc tan. It's the arc tan of (1/sqrt(2)).

Yes you’re right. I misremembered my earlier post and used sqrt(3) instead of sqrt(2).

 

[DaveC426913]

I was thinking about Bell's Theorem recently and, specifically, whether I could more directly show that hidden variables were inadequate to reproduce the predictions of QM. I didn't come up with anything, but I noticed that:

If a particle is in the spin-up state in the direction $(1, 1, 1)$ then the probability that it is found to be spin-up in any of the x, y or z directions is given by $\cos^2 \frac \theta 2$, where $\theta$ is the magic angle. This gives a probability of: $\frac 1 2 (1 + \cos \theta) = \frac 12 (1 + \frac 1 {\sqrt 3})$, which is about $0.79$.

That's a huge correlation.

This information may be more than is necessary to solve my brass tubing project, but I will keep a watchful eye to see if any quantum mechanical peculiarities creep in.

 

[Helios]

There is also the magical latitude 35.26 ° , quite near my home. From there, you can set down a cube and have the axis point to the north star. Same goes for the octahedron! These solids could go farther and have sundials affixed to each face. These things set flat at this latitude.

 

[DaveC426913]

There is also the magical latitude 35.26 ° , quite near my home. From there, you can set down a cube and have the axis point to the north star. Same goes for the octahedron! These solids could go farther and have sundials affixed to each face. These things set flat at this latitude.

OK. You are now tasked with generalizing your find into Helios' Principle.

I live at 45N. What polyhedron can I use?
What about other members? C'mon everybody - chime in!

 

[Helios]

I live at 45N. What polyhedron can I use?

It looks like you can use a Rhombic Dodecahedron. It would have an inner long axis that aligns with the Earth axis.