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Re: sum of the angle in a 4th dim triangle
*To*: mathgroup at smc.vnet.net
*Subject*: [mg26368] Re: sum of the angle in a 4th dim triangle
*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
*Date*: Wed, 13 Dec 2000 02:41:26 -0500 (EST)
*Organization*: Universitaet Leipzig
*References*: <911fu3$9u7@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi,
you are using a complete new formula for the angle between
two vectors. Since
Dot[a,b]==Sqrt[Dot[a,a]]*Sqrt[Dot[b,b]]*Cos[alpha]
the correct definition is
angle[a_, b_] := ArcCos[Dot[a, b]/(Sqrt[Dot[a, a]]*Sqrt[Dot[b, b]])]
and
angle[P - Q, R - Q] + angle[P - R, Q - R] + angle[R - P, Q - P] // N
gives 3.14159...
Due to your new relation the old result can't reproduced.
Regards
Jens
BTW: Why is
ArcCos[-1/14] + 2*ArcCos[Sqrt[15/7]/2]
not simplifed to Pi ??
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 englis...
>
> Thanks
>
> Jacky
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