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

[DanAbnormal]

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.

 

[MostlyHarmless]

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.

 

[tiny-tim]

Hi DanAbnormal

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. ​

 

[DanAbnormal]

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?

 

[CellCoree]

I understand your desire to find a general expression for the scalar triple product in terms of inter-vector angles. However, I must caution that such a formula may not exist or may be extremely complex. The scalar triple product is a geometric quantity that involves the magnitudes and directions of three vectors, and it is typically calculated using the vector cross product. While it is possible to express the scalar triple product in terms of spherical coordinates, this may not necessarily lead to a simpler formula.

I recommend continuing to explore trigonometric identities and using mathematical software to assist in your calculations. You may also want to consult with a mathematician or physicist for further insight on this problem. It is important to keep in mind that sometimes, the most elegant solutions are the simplest ones, and a complex formula may not be necessary to understand the relationship between inter-vector angles and the scalar triple product.