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Two mutually orthogonal unit vectors acting at a point $p$, produce a resultant, whereas the two orthogonal unit basis vectors at the origin do not, why?

John Rennie
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pete
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  • Why do you think you can’t add two orthogonal unit vectors “at the origin”? And what point is $\hat i+\hat j$ “at”? – Ghoster Feb 05 '23 at 05:26
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    This question is based on a false premise. – Ghoster Feb 05 '23 at 05:28
  • Related post by OP: https://physics.stackexchange.com/q/748699/2451 – Qmechanic Feb 05 '23 at 05:35
  • They do which we call as scalar product and their components or scaling multiply. – Neil Libertine Feb 05 '23 at 06:44
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    @NeilLibertine Scalar products do not produce a resultant vector, just a number. Also, OP is asking about vector addition. – joseph h Feb 05 '23 at 07:14
  • @josephh Addition of basis vectors with resoective scaling of unit. Sum of a vector is scalar product otherwise try to add 3 unit east and 4 unit north, is resultant 7 or 5. – Neil Libertine Feb 05 '23 at 08:02
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    "Sum of a vector is scalar product" That is a non-sequitur. "otherwise try to add 3 unit east and 4 unit north, is resultant 7 or 5" This makes 5 units, pointing north east. I still have no idea what it is you're trying to say. How does it relate to my comment or to yours about dot products? – joseph h Feb 05 '23 at 08:09
  • @josephh How do you sum 3 unit east and 4 unit north, by taking orthogonal directions as basis and represent their sum as resultant. Now how do you determine value of resultant, by taking scalar product and then normalized by taking root. This is what you doing in quantum mechanics most of time. – Neil Libertine Feb 05 '23 at 09:03
  • I see this question is closed, and is a follow on question to Vectors and vector bases. I will add it to my answer there. – mmesser314 Feb 05 '23 at 20:09
  • @NeilLibertine Well that makes even less sense since the scalar product of two orthogonal vectors is zero. As for the rest of your comment, you do not take scalar products to determine resultant vectors, because by definition, the scalar product is a number and not a vector, so I'm still not sure what it is you are trying to communicate. I'm not convinced that you understand vectors all that much either (not trying offend you just being honest) and so I would suggest that you grab an elementary physics book and in the first chapters they talk about vector analysis. Good luck. Cheers. – joseph h Feb 05 '23 at 21:08
  • @josephh I think that — maybe? — what Neil Libertine means, but has not managed to convey in what I consider an intelligible way, is that the square root of the scalar product of $3\hat i+4\hat j$ with itself gives the square of the magnitude (5) of the “resultant” vector (which I consider to be just “this” vector). What that has to do with the question is something I don’t understand. – Ghoster Feb 05 '23 at 23:31
  • @Ghoster Yeah, that part (magnitude) is clear, though I also do not understand what his actual point is with respect to the original question/comments. Thanks. – joseph h Feb 06 '23 at 01:19

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Vectors don't act - they're objects, not operators. The action is being done to one vector or the other by the operation of vector concatenation (more commonly "vector addition", which is a perfectly good term since we can express vectors as matrices or algebraic sums, both of which are concatenated with addition, and vector addition follows all the commutative, associative, and identity properties of addition).

If you operate on a basis vector by doing vector addition with another basis vector you get a resultant that's the vector sum of the basis vectors, as you would expect.

Note 1: Vectors also don't have location, although they may describe properties of mathematical objects that do (e.g. a point mass).

Note 2: we often talk about forces acting on objects, which seems (since force is a vector quantity) like it contradicts what I said - but it's just linguistic. When a force acts on an object, what we mean mathematically is that we need a new mathematical object to describe the physical object that having a particular physical interaction whose mathematical counterpart is a force vector.

g s
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  • so the simplest way to say it is that basis vectors are simply vectors which are not (yet) added to each other, whereas the other two vectors (referred to in the question), are. ok but, In the derivation of the relativistic lorentz transformations between two inertial frames, no matter particles are referred to, only the velocity of the frame ( x', y',z', t' ) relative to the frame ( x, y, z, t ). If $ \vec{y} $ and $ \vec{x} $ are velocity vectors, wouldn't they be in different inertial frames? since dx/dy of $ \vec{y} $=0 whereas dx/dy of $ \vec{x} $=1 – pete Feb 05 '23 at 07:48
  • @pete I'm not sure what you're asking, but it sounds like you have some kind of question about transforming between frames in relativity, which should be its own separate question, not a comment on an answer to a question about adding basis vectors. – g s Feb 05 '23 at 08:07