We can split a vector (velocity/displacement vector) along any two directions as long as the resultant of the oblique components of the vector is same as my original vector. Similarly if we have to add two vectors we can split them along any two axes and add components along these axes . The result will be the same either way. Is there any good reason that we split vectors into mutually perpendicular vectors ? (except for that its easier to compute ).Is there any deeper reason for it ? Note :- I am considering only 2 Dimensions for simplicity.

  • 3
    $\begingroup$ There is no deep reason, per se, but decomposing a set of vectors in $N$ dimensions into $N$ pairwise orthogonal vectors (aka: a basis) is convenient for vector multiplication. When carrying out the dot product or (in three dimensions) the cross product, it is easiest to work in such an orthogonal basis. For vector addition, it is not necessarily more convenient. $\endgroup$
    – Ultima
    Commented Jun 30, 2015 at 18:31
  • 5
    $\begingroup$ mathoverflow.net/q/42040 $\endgroup$ Commented Jun 30, 2015 at 18:33
  • $\begingroup$ Since the question identified by @KyleArean-Raines seems to answer this question, I am voting to close. Unfortunately, I don't think one can vote as "duplicate of other site" or "migrate and close as duplicate". $\endgroup$
    – Floris
    Commented Jul 1, 2015 at 13:20

1 Answer 1


I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point.

You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other vectors through addition and scalar multiplication) for $\mathbb{R}^2$ (and in general n vectors for $\mathbb{R}^n$). One can "do physics" in any one of these coordinate systems, but as you said, the mutually perpendicular(orthogonal) ones are much easier to compute with. A large part of this is because of "independence". In Newtonian mechanics we often use the idea that movement in the x and the y (and the z, $x_4,x_5, \dots$) are independent of the motion in the other ones (i.e. we can apply newtons laws to each force separately).

Now imagine that we had a non-orthogonal basis set of vectors, for example the x-axis and the vector counter-clockwise at $45^\circ$. If one takes an arbitrary point, then one cannot move it along one axis without also moving it along the other as well. So independence of motion is gone, and we would in general end up with a much more complicated system of differential equations than in the orthonormal case.

This idea also comes in to play at the higher level when one studies Fourier series. This consists of writing a function in terms of complex exponentials. This is because the complex exponentials form an orthogonal basis for $L^2$, the space of absolutely square integrable functions. The space $L^2$ is very important, as one of the axioms of quantum mechanics is that all wavefunctions belong to it(and moreover are unit vectors in it).

In fact any Hilbert space(a space where convergent sequences stay in the space and the space has a "dot product") admits an orthogonal basis. This is a consequence of Zorn's Lemma(an equivalent statement to the axiom of choice). Questions such as these belong to functional analysis, linear algebra's infinite dimensional cousin.

EDIT: This is in response to the comment asking to explain "independence" more.

Lets call our original orthogonal frame of reference $S$ and our new basis $S^\prime$. The coordinates of a point in the difference coordinate system $\xi_S, \xi_{S^\prime}$ these are related by \begin{align} \xi_{S^\prime} = A\xi_S \\ A = (\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}) \end{align}

So if $\xi_S = (x, y), \xi_{S^\prime} = (\alpha, \beta)$ then we would have that \begin{align} ax + by = \alpha \\ cx + dy = \beta \end{align} So if we move along the $\alpha$ or $\beta$ axis, then in general both the $x$ and $y$ will change. In the case above $x = \alpha$, so $c\alpha + dy = \beta$. So if we go along the $\beta$-axis, we must also change the $\alpha = x$ and the $y$ coordinate. So we can't move along $\beta$ without moving along $\alpha$.

  • $\begingroup$ " If one takes an arbitrary point, then one cannot move it along one axis without also moving it along the other as well" .. But in that case we measure distances parallel to axes not perpendicular so we can still move along one axis without moving along other , so independence is still there.What do you think?. (Thanks for additional info) . $\endgroup$
    – Robin Hood
    Commented Jul 1, 2015 at 5:52
  • $\begingroup$ @Robin The measurement of the distance itself is the same(up to a choice of origin). The difference is how we "refer" to the point, and in this case, how the "pointer" changes within different coordinate systems as we change the point. So with the example in the post, if we split our new basis vector into our old(x-y) basis, then when we move along the new basis vector, likewise we couldn't help but get a little "extra" x in there, we could not move along x without also going along the new basis also. Hope this is clear, I will post a more mathematical(and clearer) answer later. $\endgroup$
    – JDThinking
    Commented Jul 1, 2015 at 6:59

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