prove that a vector +b vector =b vector +a vector by using parallelogram
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Answer:
That’s an awesome question! It is a really deep question (Riemannian geometry deep). Please bear with me, I will try to use two uncommon examples to give you a head spin.
First, if you are standing on the equator, and were asked to walk east 1000km, then go north 1000km. I start from the same spot, and first walk north 1000km, then east 1000km. Will we be at the same place in the end? The answer is no. Well, you can say east and north 1000km are not really vectors. And no they are not in the traditional sense. So here’s our choices. We can either say they are not vectors, so they do not follow the commutative (abelian) rules in a traditional vector space. Or, we can invent an interpretation that they are some kind of “vectors”, but they do not reside in the traditional space. It’s a Reimannian space, a curved space. (The two segments of “straight lines” you walked, and the two that I walked, do not form a square. There is no parallelogram in this curved space. Vector addition is often explained by the method of parallelogram in geometry.)
Second example: In physics, we love to think of anything as scalars, vectors, and so forth. Can rotation (angular displacement) be a vector? In some ways… Say you have a book lying flat on the table. You rotate it about the East-axis by 90 degrees. We would love to think this rotation is a “vector” pointing east, with magnitude pi/4 (use radians). It sort of works. This “vector” adds, subtracts, with any other vectors in this east direction nicely. Now after the East-pi/4 turn, next do a north-pi/4 turn. (Be consistent, use what we call the right-hand-rule to determine the direction of turn, that is clockwise or counter-clockwise.) See what orientation will be book end up. Now reset. From the start, rotate north-pi/4 first, then east-pi/4. NOOOO. It’s not the same thing. These “vectors” do not obey the commutation rule! (I hope you, or whoever might be reading this, will one day continue on and dig deeper… Each rotation in 3-D space can be represented by such a “vector” of length between -pi to +pi. This very interesting “space” is called SO(3), the Special Orthogonal Rotational Group in 3-dim…)
Indeed, angular displacement is NOT a vector, although its rate of change, angular velocity, is a vector. So is angular acceleration. Just not angular displacement.
I am not trying to prove or disprove anything here. The commutative property you asked indeed is a basic axiom in vector spaces/linear algebra. It’s always up to you whether to do and accept what you were told. Or NOT.
Step-by-step explanation:
Hope this helps u.