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I want to use Mathematica to calculate the angle between two vectors (say a and b) that don't lie in the same plane. The vectors are of the same length (156) have a dot product with a unit vector that's normal to a plane that their projections lie that's the same (90). Their projection vectors onto this plane are denoted by p and q respectively. The angle between p and q is 120 degrees.

$Assumptions = 
  (a | e) ∈ Vectors[3, Reals] && e.e == 1 && a.e == b.e == 90 && 
  Sqrt[a.a] == 156 == Sqrt[b.b]; 
p = a - (a.e) e;
q = b - Dot[b, e] e;
Simplify[TensorExpand[a.b], Sqrt[a.a] == 156 == Sqrt[b.b]]
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You should also include b in your assumptions!

$Assumptions = (a | b | e) ∈ Vectors[3, Reals] && e.e == 1 &&a.e == b.e == 90 && a.a == 156^2 == b.b;
p = a - (a.e) e;
q = b - (b.e) e;

Now the condition concerning p.q can be evaluated

cond=Simplify[TensorExpand[p.q == Sqrt[p.p] Sqrt[q.q] Cos[120 °]]]
(*18 + a.b == 0*)

Knowing a.byou can evalaute the angle between a,b!

cosα = (a.b/Sqrt[a.a b.b] /. Solve[cond, a.b][[1]]) 
(*-(18/Sqrt[a.a b.b])*)
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  • $\begingroup$ Does the 1 in Solve[cond, a.b][[1]] refer to in[1] the global conditions? I am struggling with mathematica with these in[] as I can not reset the number (to get in[1] for that line) and the result does not show numerically as it is not using the lengths of a and b in the final line of code. $\endgroup$ – Megamatics Nov 13 '18 at 11:14
  • $\begingroup$ No! Without [[1]] Solve returns double list-brackets. $\endgroup$ – Ulrich Neumann Nov 13 '18 at 11:30
  • $\begingroup$ I don't know why Mathematica doesn't simplifies the result -(18/Sqrt[a.a b.b]) completely. $\endgroup$ – Ulrich Neumann Nov 13 '18 at 11:42

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