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Re: inner angle of a triangle in the 4th dimension

  • To: mathgroup at
  • Subject: [mg26330] Re: [mg26324] inner angle of a triangle in the 4th dimension
  • From: Jacqueline Zizi <jazi at>
  • Date: Tue, 12 Dec 2000 02:54:44 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

Bonjour Jacky!

There is several errors in your solution.

1) the formula to have the angle between 2 vectors is NOT the one you give
2) ArcCos give a result in radians, not in degrees - see Help Browser Go To
ArcCos , 3d little square.

As you are dealing with a triangle, you are in fact in 3D. If I were you, I
should work in 3D first, draw the points, the triangle and check all the
results step by step. Then, when well understood, write the corresponding
functions to your main ideas, as below, then generalize to a 4D dimension

For example, if you are a French speaking man, write the 3 functions you

vecteur [M_,N_] :=  ..

longueur [{a_, b_}] := .. 

angle [u_, v_]  := ..

and apply them to get  u, v, w, then their angle. The 3d one is the most
difficult to write. Test it on several cases.

To draw the triangle, look at "line" and "text" in the help browser. If you
find it a bit far from your purpose, you can define this function by copying
the following and evaluating it:

triangle [P_, Q_, R_] := Show[Graphics[{Line[{P, Q, R, P}],
          Text ["P", P], Text ["Q", Q], Text ["R", R]}]];

Then use it to draw your prefered triangle, by giving specific values to P,
Q, and R.
For example, triangle [P, Q, R]; should draw you the triangle when P, Q and R
are set.
If the text don't go were you want, improve it by giving a 3d argument to

Don't forget to test things on triangle like this one:

P = {0.562021, 0.715698};
Q = {1.56995, 0.25331};
R = {2.56629, 0.278304};

If things are not clear, please ask. If any pb with English, you can ask your
questions in English but also  give the French version of them.

I hope this helps.

Jacqueline Zizi

Jacky Vaillancourt wrote:

> Hi, i have a basic problem. I can't see my mistake can somebody help me?
> Here's the problem:
> I want to calculate each angle of the triangle formed by those three dots.
> P:=(0,1,0,1), Q:=(3,2,-2,1), R:=(3,5,-1,3)
> u:=PQ -> (3-0,2-1,-2-0,1-1) -> (3,1,-2,0)
> v:=QR -> (3-3,5-2,-1-(-2),3-1) -> (0,3,1,2)
> w:=PR -> (3-0,5-1,-1-0,3-1) -> (3,4,-1,2)
> The formula to have the angle between tho vector is:
>     ARCCOS(ABS(DOTPROD(u,v))/(length(u)*length(v))
> The formula to calculate the length is SQRT(a^2+b^2+c^2+d^2)
> So, the angle between u and v is:
>             ARCCOS(ABS(15)/(SQRT(14)*SQRT(30))) = 42.95 deg
> the angle between v and w is:
>             ARCCOS(ABS(-15)/(SQRT(30)*SQRT(14)))= 42.95 deg
> the angle between u and w is:
>             ARCCOS(ABS(-1)/(SQRT(30)*SQRT(14)))= 85.9 deg
> Here's the problem 180-85.9-42.95-42.95= 8.2 deg
> I'm missing 8.2 deg....
> I hope you'll understand what i wrote, i'm not used to write in english...
> Thanks
> Jacky

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