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Can you even get the angle between two three dimensional vectors without using some sort of vector product and venturing into linear algebra territory?

dragontamer 6y ago
Yeah. Its just harder using Trig: SOH CAH TOA. tan(angle) = opposite-leg / adjacent-leg. arctan(opposite-leg / adjacent-leg) == angle. The more I do this, the more I realize that the linear-algebra approach might be easier... but the Trig approach can certainly do what you are asking for. I mean, I recognize that I'm just "reinventing cross-product" here to get the length of opposite-leg and the direction of adjacent-leg. But it does seem feasible to do all of these things in Trigonometry alone. A little bit of Linear algebra (cross product creates an orthogonal vector) would definitely make this problem easier.
- Arelius 6y ago
  Ok, hmm I'm still a bit lost though, those identities work on a right triangle, but how do you construct a right triangle between two 3D vectors A, and B?
  - Arelius 6y ago
    Like I could see how you could figure this out if you can project one line onto another maybe, but I also don't know how to do that with just trig.
  - dragontamer 6y ago
    > Ok, hmm I'm still a bit lost though, those identities work on a right triangle, but how do you construct a right triangle between two 3D vectors A, and B? (0,0,0), A, and B are the three points of a right-triangle. Hmm... okay, yeah. I see what you mean. Every time I think about how to do things, its linear algebra time. The thing is: I've had this feeling before in other maths: the more you study one math subject, the "worse" you get in another. There's surprising things you can do with algebra alone without any calculus... but once you learn calculus, you kinda forget those old algebra tricks.
    - Arelius 6y ago
      >(0,0,0), A, and B are the three points of a right-triangle. Your next line suggests you already know this, but just so it's clear it's not, they are three points in an arbitrary triangle. Each vector has a right triangle between the origin and (two of?) it's axis, but I'm not sure that's helpful. Yeah, I agree in that it feels like you should be able to do this, but I just know that I've been stumped in the past, and can't figure it out now either.
      - dragontamer 6y ago
        Ah right. Law of Sines sin(a) / A = sin(b) / B = sin(c) / C Apparently, MY trigonometry is awful and needs more practice! Law of sines is basic 'non-right triangles' stuff. There are three legs: A-0, B-0, and B-A. Hmmm... B-A seems like a vector operation (but its a really easy one, so... probably can be figured out without formally studying linear algebra).. That gives us the "A", "B" and "C" for lengths. Aaaannnnd I'm stuck again. I have the lengths but I don't have the angles. I only need to find one angle and the other two are solved with law of sines. Maybe a system of equations at this point (lol, another subject where linear algebra helps). The law of sines gives 3-unknowns, but only 2-equations. So I still need one more constraint before I can math-out a solution. ------ EDIT: Law of sines is fully: sin(a) / A = sin(b) / B = sin(c) / C = d Where "d" is the diameter of the circumcircle of the triangle. So the circumcircle's diameter gives us the 3rd equation needed to solve the above system of equations (using non-matrix math). So yeah, I think its possible. Its just waaayyyy harder to do. I also am handwaving a difficult part. I guess... take the average of 3 points, which would give the center of the circumcircle. Then sqrt(x, y, z) from the center to find the radius, then 2*radius = d. Wow, that's a lot of work compared to an easy projection matrix...
        Arelius 6y ago
        Yeah, sure is messy.
        Pyrious 6y ago
        The average of the 3 points would give you the triangle's centroid, not necessarily its circumcenter. You can still calculate the circumradius from a, b, and c: https://mathworld.wolfram.com/Circumradius.html https://mathworld.wolfram.com/Circumradius.html. You could also use the law of cosines to get a triangle's angles pretty easily when you know its edge lengths: https://mathworld.wolfram.com/LawofCosines.html https://mathworld.wolfram.com/LawofCosines.html.

dragontamer 6y ago

Yeah. Its just harder using Trig: SOH CAH TOA. tan(angle) = opposite-leg / adjacent-leg. arctan(opposite-leg / adjacent-leg) == angle. The more I do this, the more I realize that the linear-algebra approach might be easier... but the Trig approach can certainly do what you are asking for. I mean, I recognize that I'm just "reinventing cross-product" here to get the length of opposite-leg and the direction of adjacent-leg. But it does seem feasible to do all of these things in Trigonometry alone. A little bit of Linear algebra (cross product creates an orthogonal vector) would definitely make this problem easier.

Arelius 6y ago
Ok, hmm I'm still a bit lost though, those identities work on a right triangle, but how do you construct a right triangle between two 3D vectors A, and B?
- Arelius 6y ago
  Like I could see how you could figure this out if you can project one line onto another maybe, but I also don't know how to do that with just trig.
- dragontamer 6y ago
  > Ok, hmm I'm still a bit lost though, those identities work on a right triangle, but how do you construct a right triangle between two 3D vectors A, and B? (0,0,0), A, and B are the three points of a right-triangle. Hmm... okay, yeah. I see what you mean. Every time I think about how to do things, its linear algebra time. The thing is: I've had this feeling before in other maths: the more you study one math subject, the "worse" you get in another. There's surprising things you can do with algebra alone without any calculus... but once you learn calculus, you kinda forget those old algebra tricks.
  - Arelius 6y ago
    >(0,0,0), A, and B are the three points of a right-triangle. Your next line suggests you already know this, but just so it's clear it's not, they are three points in an arbitrary triangle. Each vector has a right triangle between the origin and (two of?) it's axis, but I'm not sure that's helpful. Yeah, I agree in that it feels like you should be able to do this, but I just know that I've been stumped in the past, and can't figure it out now either.
    - dragontamer 6y ago
      Ah right. Law of Sines sin(a) / A = sin(b) / B = sin(c) / C Apparently, MY trigonometry is awful and needs more practice! Law of sines is basic 'non-right triangles' stuff. There are three legs: A-0, B-0, and B-A. Hmmm... B-A seems like a vector operation (but its a really easy one, so... probably can be figured out without formally studying linear algebra).. That gives us the "A", "B" and "C" for lengths. Aaaannnnd I'm stuck again. I have the lengths but I don't have the angles. I only need to find one angle and the other two are solved with law of sines. Maybe a system of equations at this point (lol, another subject where linear algebra helps). The law of sines gives 3-unknowns, but only 2-equations. So I still need one more constraint before I can math-out a solution. ------ EDIT: Law of sines is fully: sin(a) / A = sin(b) / B = sin(c) / C = d Where "d" is the diameter of the circumcircle of the triangle. So the circumcircle's diameter gives us the 3rd equation needed to solve the above system of equations (using non-matrix math). So yeah, I think its possible. Its just waaayyyy harder to do. I also am handwaving a difficult part. I guess... take the average of 3 points, which would give the center of the circumcircle. Then sqrt(x, y, z) from the center to find the radius, then 2*radius = d. Wow, that's a lot of work compared to an easy projection matrix...
      - Arelius 6y ago
        Yeah, sure is messy.
      - Pyrious 6y ago
        The average of the 3 points would give you the triangle's centroid, not necessarily its circumcenter. You can still calculate the circumradius from a, b, and c: https://mathworld.wolfram.com/Circumradius.html https://mathworld.wolfram.com/Circumradius.html. You could also use the law of cosines to get a triangle's angles pretty easily when you know its edge lengths: https://mathworld.wolfram.com/LawofCosines.html https://mathworld.wolfram.com/LawofCosines.html.