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I should begin by saying that I'm only beginning to study $11$th grade Physics. Recently I figured out that two vectors which gets added up to a null vector, must at most lie in a line. Well that makes perfect sense, so I thought about $3$ vectors and figured that they must at most lie in a plane. Also $4$ vectors when added up to a null vector must at most lie in $3$D space. I then thought about a single vector. Naturally I can't add a vector to nothing and get $0$ vector right? So that means it must be the null vector itself. So that means A single vector must lie in a dimensionless point to give a null vector.

So I hypothesise that n vectors when added up to a null vector must lie in $n-1$ spatial dimensions. That is iff $$v_1+v_2+v_3....+v_n=0$$ Then $v_1 ,v_2,v_3 ,v_4 ,v_5...,v_n$ must at most lie in $(n-1)$th dimension.

Please put some thought into this and help me understand a method to visualize it if this is correct.

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  • $\begingroup$ For future use, you should learn to use MathJax for better formatting. Here's a tutorial - math.meta.stackexchange.com/questions/5020 $\endgroup$ – Ishan Deo May 27 at 5:24
  • $\begingroup$ Ah thanks. I sincerely appreciate that you took time to send me the link. $\endgroup$ – NightKruger May 27 at 5:25
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You may have heard that two points lie on a line (a 1D space) and three points lie on a plane (a 2D space). If you visualise these with more points, it's possible to see that $n$ points lie on an $n-1$ dimensional space. So consider the origin, and the endpoints of $n-1$ of the vectors. Since we now have $n$ points, these points define an $n-1$ dimensional space containing all the points, including the origin. It's now clear that summing all the vectors must give a point within this space since all vectors lie within it. Since the last vector is the negative of the first $n-1$ vectors added up, it must also lie in this $n-1$ dimensional space.

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  • $\begingroup$ Wow that's a good answer. $\endgroup$ – NightKruger May 27 at 6:18
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As we have $\sum_1^n v_i = 0$, we get $v_n = -\sum_1^{n-1} v_i$. Thus, out of the $n$ vectors, at most $n-1$ can be independent of each other (i.e. can take values independent of the other vectors). Hence, they can occupy a space of at most $n-1$ dimensions.

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  • $\begingroup$ Can you please explain a little more? I'm really new in this $\endgroup$ – NightKruger May 27 at 5:37
  • $\begingroup$ Where did you get confused? $\endgroup$ – Ishan Deo May 27 at 5:38
  • $\begingroup$ I can't relate how n-1 vectors being independent of each other imply them being in n-1 dimensions $\endgroup$ – NightKruger May 27 at 5:39
  • $\begingroup$ Consider the 2D case and the unit vectors $i$ and $j$. These are independent (i.e. no combination of these two will result in 0), and occupy the 2D plane. Similarly, $i$, $j$, $k$ are independent and occupy the 3D space. Thus, similarly, $n-1$ independent vectors occupy $n-1$ dimensions. $\endgroup$ – Ishan Deo May 27 at 5:50
  • $\begingroup$ Oh. Thanks I finally understand. $\endgroup$ – NightKruger May 27 at 6:18

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