A way to express scalar triple product from inter-vector angles?

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The discussion focuses on deriving a general expression for the scalar triple product of three vectors based solely on their inter-vector angles, A1, A2, and A3. The user has already expressed the scalar triple product using spherical polar coordinates and is seeking a more straightforward relationship using trigonometric identities. They also mention using spherical triangle properties to relate the angles to the volume of a pyramid formed by the vectors. The conversation includes attempts to define the vectors in Mathematica and calculate the angles between them. Ultimately, the user is looking for guidance on expressing the scalar triple product purely in terms of the inter-vector angles.
DanAbnormal
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Hi,

I'm trying to find a general expression for the scalar triple product for 3 vectors in a simultaneous configuration, that depends only on the inter-vector angles, A1, A2 and A3.

I have expressed this quantity in terms of the spherical polar coordinates of the vectors (the length being unity for simplicity), and I have also expressed 3 equations for the dot product of each possible pair using spherical coordinates, to get a relation to the inter-vector angles.

Now I don't know if this is just a simple case of rearranging with trig identities, but I've been trying it for hours, can't find anything on the net and I'm not too good with Mathematica etc, so I was just wondering if there was a general expression, or a good lead to one.

Thanks.
 
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Could you give an example of what you have?(the vectors) If I can't manually solve them, I may be able to help you punch them into mathematica.
 
Hi DanAbnormal :smile:
DanAbnormal said:
I'm trying to find a general expression for the scalar triple product for 3 vectors in a simultaneous configuration, that depends only on the inter-vector angles, A1, A2 and A3.

I have expressed this quantity in terms of the spherical polar coordinates of the vectors (the length being unity for simplicity) …

In other words: given the lengths of three sides of a spherical triangle (the lengths are the same as your inter-vector angles),

find the volume of the pyramid formed by the three vertices and the centre?

If you use one of the standard spherical trig formulas to find one of the angles of the triangle, you can put that vertex at the north pole, and then it's easy to find the cartesian coordinates of the other two vertices. :wink:
 
Im not sure if your answer is equivalent, though I'll post what I have more explicitly.

In Mathematica, I have specified the three following vectors in terms of their spherical polar angles:

x1 = 0 Degree;
x2 = 0 Degree;
x3 = 180 Degree;
z1 = 0 Degree;
z2 = 120 Degree;
z3 = 240 Degree;

where the preceding x's mean angle from x axis, and the same for z.
Now I have the cosine of each inter vector angle given by:

Angle1 = Sin[z2]*Sin[z3]*Cos[x2 - x3] + Cos[z2]*Cos[z3];
Angle2 = Sin[z1]*Sin[z3]*Cos[x1 - x3] + Cos[z1]*Cos[z3];
Angle3 = Sin[z1]*Sin[z3]*Cos[x1 - x3] + Cos[z1]*Cos[z3];

I can express the Scalar Triple Product in the following way:

Needs["VectorAnalysis`"]
(*First express our vectors in Spherical Polar Coordinates*)
v1 = CoordinatesToCartesian[{1, x1, z1}, Spherical];
v2 = CoordinatesToCartesian[{1, x2, z2}, Spherical];
v3 = CoordinatesToCartesian[{1, x3, z3}, Spherical];

N[ScalarTripleProduct[v1, v2, v3]]

I was wondering if there is a way to express this same triple product as a function of Angle1, Angle2, and Angle3 only, defined above. Can this be done in Mathematica?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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