inner angle of a triangle in the 4th dimension

• To: mathgroup at smc.vnet.net
• Subject: [mg26324] inner angle of a triangle in the 4th dimension
• From: "Jacky Vaillancourt" <jacky_1970 at videotron.ca>
• Date: Sun, 10 Dec 2000 21:38:07 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```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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