Angle between diagonals of neighboring cube sides

[Shadow]

Hey all,

 

well, the title speaks for itself. Given a cube, what is the angle between the diagonals of two neighboring squares? Here's an illustration, just in case:

 

 

It looks a little strange since it's a 2D picture, but I think you get the picture. Literally. My gut tells me it's either 45° or 60°, but...well, that's just my gut.

 

the title almost speaks for itself. Given a cuboid, where [math]a \neq b \neq c[/math], how can I calculate the angle between the diagonals of two neighboring rectangles?

 

Cheers,

 

 

Gabe

Edited January 2, 2009 by Shadow
Wrong question asked :D

[Petanquell]

Just equilateral triangle...?

[Shadow]

Oh...right *blush*

 

EDIT: Oops, I asked the question wrong. Edited.

Edited January 2, 2009 by Shadow

[D H]

Write the problem as a vector equation.

[Shadow]

Not sure how that works. I've been trying for ages now to just use Heron's formula and [math]S=1/2a b*sin(\alpha)[/math] and isolate the expression using Maple, but it's just not working and it's driving me nuts. I'll give it a rest and try again in a couple of minutes...

Edited January 2, 2009 by Shadow

[D H]

Not sure how that works.

Put your cuboid with the vertex in question at the origin and with the edges leading away from this vertex aligned with the x, y, and z axes. Express the diagonals in question as vectors. What simple vector operations (there are two; pick either one) say something about the angle between the two vectors?

[Shadow]

Oooooookay, finally got it. Maple's inability to distinguish between [math]ab[/math] and [math]a*b[/math] while freely interchanging the two itself will never cease to infuriate me.

 

If I'm correct, or should I say if Maple is correct for a cuboid [math]a, b, c[/math] made out of rectangles with diagonals [math]p, q[/math], where

 

[math]p=\sqrt{a^2+b^2}[/math]

[math]q=\sqrt{b^2+c^2}[/math]

 

the angle [math]x[/math] between [math]p[/math] and [math]q[/math] should be [math]x=\arcsin \left( {\frac {\sqrt {{c}^{2}{a}^{2}+{c}^{2}{b}^{2}+{b}^{2}{

a}^{2}}}{\sqrt {{a}^{2}+{b}^{2}}\sqrt {{b}^{2}+{c}^{2}}}} \right)

[/math]

 

Can anyone verify this?

 

Cheers,

 

Gabe

[D H]

Looks good. Note: Using the dot product rather than the cross product will yield a simpler expression.

 

TBH, it appears you might be relying on Maple a bit too much. Try to solve like these by hand first.

[Shadow]

DH, after having a quick look at Wikipedia, since I've never yet needed to work with vectors beyond addition, subtraction, and scalar multiplication, I'd say the cross/dot product?

 

EDIT: Oh, you replied I only rely on maple to do the variable expressing, simplifying and expanding. I could probably do it myself, but it would take way longer. The only thing I don't really know is [math]x=[trig-function-here](\alpha), \alpha=?[/math] since we haven't covered inverse trig functions at school yet, but I could figure that part out. And TBH, who here would want to do

 

[math]\frac{pq*sin(x)}{2}=\sqrt{s(s-p)(s-q)(s-t)}, s=\frac{p+q+t}{2}[/math][math], p=\sqrt{a^2+b^2}, q=\sqrt{b^2+c^2}, t=\sqrt{c^2+a^2}, x=?[/math]

 

by hand. IMO it would just be pointless labor.

Edited January 2, 2009 by Shadow

[Petanquell]

If you mean inverse function to sin() so it is arcsin() (or [math]\sin^{-1} [/math] but i hate this notation...]. For others it's the same: cos() - arccos(); tg() - arctg(),....)....

 

 

 

I don't understand your last formula so I show my you way:

[math] e=\sqrt{a^2+b^2} [/math]

[math] f=\sqrt{b^2+c^2} [/math]

[math] g=\sqrt{a^2+c^2} [/math]

 

We want to find angle [math] x [/math] between [math] f [/math] and [math] g [/math].

for illustration...

 

I'll try it with cosine rule

[math] c^2 = a^2 + b^2 - 2ab\cos(\gamma) [/math]

...we want an angle so...

[math] \cos(\gamma) = \frac{a^2 + b^2 - c^2}{2ab} [/math]

and now it's seems easy

[math] a=g,b=f,c=e [/math]

[math]\cos(x) = \frac{(a^2+c^2) + (b^2+c^2)- (a^2+b^2)}{2\sqrt{(a^2+c^2)}\sqrt{(b^2+c^2)}} [/math]

[math]\cos(x) = \frac{2c^2}{2\sqrt{(a^2+c^2)}\sqrt{(b^2+c^2)}}[/math]

[math]\cos(x) = \frac{c^2}{\sqrt{(a^2+c^2)}\sqrt{(b^2+c^2)}}[/math]

and finally

[math] x=\arccos(\frac{c^2}{\sqrt{(a^2+c^2)}\sqrt{(b^2+c^2)}}) [/math]

 

I hope, everything's correct...

 

Pq