Angle between vectors that are both off set by a third vector

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Discussion Overview

The discussion revolves around finding the angle between two vectors that are offset by a third vector. Participants explore methods for calculating this angle in a three-dimensional space, considering various mathematical approaches and the properties of vector operations.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes using the formula theta = cos-1(“vector a” * “vector b”) to find the angle between two vectors, assuming they are initially at 90 degrees to each other.
  • Another participant clarifies that the dot product formula \vec{a}\bullet\vec{b}=|\vec{a}||\vec{b}|cos(\theta) applies regardless of the angle between the vectors, and suggests that method one should work based on the distributive property of the dot product.
  • A participant expresses uncertainty about their initial approach and mentions that adjusting only “vector b” by “vector c” may be the correct path forward.
  • There is a suggestion to explicitly write out the calculations to derive the angle in terms of vector c.
  • One participant requests a clearer statement of the problem to provide better assistance.

Areas of Agreement / Disagreement

Participants exhibit differing views on the effectiveness of the proposed methods for calculating the angle, with no consensus reached on the correct approach. Some methods are challenged while others are suggested, indicating ongoing debate.

Contextual Notes

The discussion includes assumptions about the properties of vector operations and the specific conditions under which the calculations are made, which remain unresolved.

IMK
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Hello, was wondering if someone could help me with a little vector maths problem please.

What I need to find is the angle between two vectors that are derived from/relative to a third vector.

So what I having working/understood is that theta = cos-1(“vector a” * “vector b”) where “vector a” and “vector b” are 3d vectors at 90 deg to each other, x,y,z.
Basically the cos of the dot product of the two vectors…

What I now wish to do is to offset “vector a” and “vector b” by another vector, “vector c” and I have tried the following with mixed results.

(1) “vector a” = “vector a” - “vector c”, “vector b” = “vector b” - “vector c”
this returns an answer that is obviously incorrect.

(2) “vector a”= sqrt( “vector a” * “vector c”), “vector b” = sqrt(“vector b” *“vector c”), this returns an answer that seems to be correct as I can swap around the x,y,z values any of the vectors and the result seem to be consistent.

However I have no way of really proving the result, so can you help please?

Many thanks in advance
 
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huh? the dot product is [tex]\vec{a}\bullet\vec{b}=|\vec{a}||\vec{b}|cos(\theta)[/tex]
the vectors do not have to be 90 degree apart but when they are cos(90)=1 so the formula just doesn't include the cos.

there's no reason why method one shouldn't work. the dot product distributes so a dot (b+c) =adotb +adot c

the angle between two with another one added to each one of them is just the angle between the resultant vectors. write everything out explicitly and you'll have angle of theta in terms of vector c
 
Many thanks

ice109 said:
huh? the dot product is [tex]\vec{a}\bullet\vec{b}=|\vec{a}||\vec{b}|cos(\theta)[/tex]
the vectors do not have to be 90 degree apart but when they are cos(90)=1 so the formula just doesn't include the cos.

there's no reason why method one shouldn't work. the dot product distributes so a dot (b+c) =adotb +adot c

the angle between two with another one added to each one of them is just the angle between the resultant vectors. write everything out explicitly and you'll have angle of theta in terms of vector c

Ice109,
Many thanks for your input, I have had time to think more about my problem and it is really only vector b that I need to adjust/rotate by vector c. So I will have a think about your input and see if I can figure a way to do it.
Again many thanks IMK
 
what is your problem. state is succinctly and maybe we can help
 

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