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 November 6th, 2018, 07:47 AM #1 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 General question on Vectors A general question which I want to clarify, it is regarding the vector addition. We have a certain methodology to do vector addition. Let us take the example of AC currents or voltages which are vectors and 120 Deg apart and if we add them there is a resultant vector which will be in a direction matching as per our calculations. But in real situations are there any other signals which are vectors but still do not follow our mathematical vector addition? Is it possible or everything follows mathematics? Last edited by skipjack; November 6th, 2018 at 01:58 PM.
 November 6th, 2018, 07:59 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 If they don't follow vector addition they aren't vectors. Thanks from topsquark
 November 6th, 2018, 08:10 AM #3 Senior Member   Joined: Jun 2015 From: England Posts: 891 Thanks: 269 This is a really good question as strictly speaking phasors are not true vectors and do not follow the parallelogram law of addition. Equally one phasor may be divided by another, but vector division is undefined. However this does not belong in elementary math. Thanks from topsquark
 November 6th, 2018, 02:18 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 Two things may have different ways of being added. If they can be added by use of vector addition, they are vectors in relation to that method of addition. If they can be added by use of some other method to give a different result, they aren't vectors in relation to that other method of addition. It's quite possible that both possibilities can make sense in a "real world" situation. For example, the two things may be two actions that you can take, and their sum may correspond to taking both actions. It may turn out that this sum can be found by vector addition if the two actions are taken simultaneously, but not by vector addition if the actions are taken at separate times.
November 6th, 2018, 02:21 PM   #5
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Quote:
 Originally Posted by studiot This is a really good question as strictly speaking phasors are not true vectors and do not follow the parallelogram law of addition. Equally one phasor may be divided by another, but vector division is undefined. However this does not belong in elementary math.
It's an engineering thing apparently. https://en.wikipedia.org/wiki/Phasor

You can multiply a phasor by a scalar, but they add funny. I skimmed the article but my eyes glazed pretty fast. Studiot can you explain like I'm 5?

November 6th, 2018, 04:19 PM   #6
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Quote:
 Originally Posted by Maschke It's an engineering thing apparently. https://en.wikipedia.org/wiki/Phasor You can multiply a phasor by a scalar, but they add funny. I skimmed the article but my eyes glazed pretty fast. Studiot can you explain like I'm 5?
phasors are just polar representations of a sinusoidal signals that have variable amplitude and phase.

$A \angle \phi\sim A e^{i (2\pi f t + \phi)}$

It's assumed when using them that all the signals have the same frequency $f$

They are vectors, in the same sense a complex number is a vector. They multiply as you'd expect

$A_1 \angle \phi_1 A_2 \angle \phi_2 = A_1 A_2 \angle (\phi_1 + \phi_2)$

but in order to add them you have to unpack them into cartesian form. Just as with any complex number.

Last edited by romsek; November 6th, 2018 at 04:40 PM.

November 7th, 2018, 01:02 PM   #7
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Quote:
 Originally Posted by studiot This is a really good question as strictly speaking phasors are not true vectors and do not follow the parallelogram law of addition. Equally one phasor may be divided by another, but vector division is undefined. However this does not belong in elementary math.
Phasors are just a representation of the complex numbers which means they are vectors over $\mathbb{R}^2$. They definitely obey the usual rules for vector addition. The fact that you can divide them comes from the fact that the complex numbers are also an algebra over $\mathbb{R}^2$ (in fact they are a field). This gives them more structure than "just" being a vector space but they are certainly still a vector space.

A similar example is $n \times n$ real matrices which form a vector space of dimension $n^2$ over $\mathbb{R}$. However, they also form a ring so you can multiply elements which is more structure than you get from the vector space. Similarly, you can divide (some) matrices but as you have pointed out, this is not possible from the vector structure alone.

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