# Vectors added up to give null vector.

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.

• For future use, you should learn to use MathJax for better formatting. Here's a tutorial - math.meta.stackexchange.com/questions/5020 – Ishan Deo May 27 at 5:24
• Ah thanks. I sincerely appreciate that you took time to send me the link. – NightKruger May 27 at 5:25

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.
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.
• 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. – Ishan Deo May 27 at 5:50