Why can vector components not be resolved by Laws of Vector Addition?

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Solution 1

Indeed, any vector can be resolved in terms of two components (in $n$-dimensional space in terms of $n$ components). For this being possible the components should be linearly independent, i.e. in your case they should not be parallel. The advantage of using two orthogonal/perpendicular components is that their scalar product is zero, which simplifies the math when calculating the coefficients: \begin{array} \mathbf{F} = F_x\mathbf{e}_x + F_y\mathbf{e}_y \Longrightarrow F_x = \mathbf{e}_x\cdot \mathbf{F}, F_y = \mathbf{e}_y\cdot \mathbf{F}. \end{array} Indeed, \begin{equation} \mathbf{e}_x\cdot \mathbf{F} = \mathbf{e}_x (F_x\mathbf{e}_x + F_y\mathbf{e}_y) = F_x\mathbf{e}_x \cdot\mathbf{e}_x + F_y\mathbf{e}_x \cdot\mathbf{e}_y = F_x, \end{equation} and similarly for $F_y$, since \begin{equation} \mathbf{e}_x \cdot\mathbf{e}_x = 1, \mathbf{e}_y \cdot\mathbf{e}_y = 1, \mathbf{e}_x \cdot\mathbf{e}_y = 0. \end{equation} For non-orthogal vectors $\mathbf{e}_x \cdot\mathbf{e}_y \neq 0$, the math becomes a bit more complicated and the interpretation of the projections as coordinates is less intuitive.

There are however some cases where using non-orthogonal components is beneficial, notably when dealing with non-orthogonal crystal lattices, such that of graphene (a hexagonal lattice.)

Solution 2

Vadim's answer resolves the Q very nicely. I would just like to add that given a set of n vectors in a vector space you can always construct a set of n orthonormal vectors which can then be used as a basis. This is well known as the Gram–Schmidt orthogonalization process.

Solution 3

Suppose you are attempting to find a point on a map given a starting point. Which do you prefer:

  • 4 km north and 3 km east of your current location
  • 7 km north and 4.2 km southeast of your current location

We can use whatever non-parallel vectors we want to describe an offset in two dimensions. Is it now clear why north and east are preferable to north and southeast?

Solution 4

There are two main cases of separating into components: standard coordinates system, and separating into normal and parallel.

In a standard coordinate system, there is a set of coordinates, and points are given in terms of those coordinates. Not all coordinate systems are vector spaces; for instance, the longitude latitude system isn't, since the Earths surface is curved, and in general relativity, the coordinates are only locally vectors, since space-time is curved. But for flat space, we can treat the coordinate system as consisting of an origin, each point being represented by a vector pointing from the vector to that point, and the coordinates being given by the projections of that vector onto basis vectors. Generally, orthogonal basis vectors are chosen because that makes things simpler.

The other case where a vector is separated into components is where there is some reference vector, and other vectors are separated into normal and parallel to that. The most common reference vector is the gravitational vector: the parallel direction is treated as a distinguished axis of up/down, and other directions are sideways. But there are other examples of reference vectors: for instance, if you're pushing a cart up a slope, you might take the slope as a reference vector. Then if you have another vector, such as the force on the cart, you can resolve it into a component parallel to the slope, and another perpendicular to the slope. And of course if one vector is parallel to the reference vector, and another is perpendicular, then they will be perpendicular to each other.

Solution 5

This is mainly because the X and Y axes are also perpendicular to each other and therefore to measure quantities along these axes the vectors are resolved into X and Y components.

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Updated on March 24, 2020

Comments

  • user253164
    user253164 over 3 years

    A vector at any angle can be thought of as resultant of two vector components (namely sin and cos).

    But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle.

    In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law).

    • JEB
      JEB over 3 years
      what is "a normal vector at any angle", does normal mean "orthogonal" or "average Joe"?
    • user253164
      user253164 over 3 years
      @JEB average joe
    • David Z
      David Z over 3 years
      I've removed a number of comments that were attempting to answer the question and responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering.
    • lcv
      lcv over 3 years
      In fact we do it quite often. We use orthonormal bases, which are quite easy to work with, but often time also non-orthogonal bases.
  • David Z
    David Z over 3 years
    I don't really see how this extra information is useful for answering the question.
  • Eric Lippert
    Eric Lippert over 3 years
    @DavidZ: It is useful because it removes any supposition on the part of the original poster that there might be a situation where we are required to use a nonorthogonal basis.
  • nick012000
    nick012000 over 3 years
    "There are however some cases where using non-orthogonal components is beneficial, notably when dealing with non-orthogonal crystal lattices, such that of graphene (a hexagonal lattice.)" Other instances when it might be useful is when dealing with relativistic space contraction/time dilation, or when dealing with HSV color space triangle.
  • nick012000
    nick012000 over 3 years
    @EricLippert What about situations where the coordinate vectors are themselves non-orthogonal, e.g. the HSV color triangle or the relativistic time dilation/space contraction?
  • Carmeister
    Carmeister over 3 years
    On a related note, the dot product formula $\mathbf A\cdot \mathbf B=A_xB_x+A_yB_y$ only works if we expand in orthogonal components; otherwise it is more complicated.
  • David Hammen
    David Hammen over 3 years
    You missed an infinite number of other options, such as 4.2 km northeast of your current location. But suppose there's a river between your current location and the target location, and the only way to get across it for dozens of kilometers is to drive 7 km north to the only local road with a bridge that crosses the river, with that road happening to run northwest to southeast?
  • Paul Sinclair
    Paul Sinclair over 3 years
    @nick012000 - coordinate systems are something we impose. Gram-Schmidt shows that we can always choose a coordinate system where the coordinate vectors are orthonormal. In most situations, this makes the math easier. Occasionally, the situation can be such that the advantages of some non-orthonormal system outweigh the otherwise-easier mathematics of an orthonormal one, so we may choose that system instead. But the orthonormal option is available even then.
  • Paul Sinclair
    Paul Sinclair over 3 years
    @DavidHammen - could you explain how your comment is in any way appropriate to this discussion? I just don't see it. It appears to be a rant that rather than giving a simple practical example of why orthonormal coordinates are often preferable, Eric Lippert should have somehow pulled off the task of covering every possibility.
  • Eric Lippert
    Eric Lippert over 3 years
    @PaulSinclair: I admit I am also mystified as to this comment. My answer was about a scenario of identifying a point; the comment appears to be about a completely different scenario involving orienteering I think? Orienteering is a great sport but that's not what this answer is about.
  • Eric Lippert
    Eric Lippert over 3 years
    @nick012000: I'm not following your point here. The question at hand is not "are non-orthogonal bases sometimes useful?" Sure, they're sometimes useful. The question at hand is *are we at any time required to use a non-orthogonal basis because no orthogonal basis even exists?" Can you clarify your comment?
  • Ilya Stokolos
    Ilya Stokolos over 3 years
    @Chayanka Kakati "given a set of n vectors in a vector space you can always construct a set of n orthonormal vectors". I assume you meant "given a set of n linearly independent vectors", right?
  • nick012000
    nick012000 over 3 years
    @EricLippert When a relativistic system undergoes space contraction and time dilation, its space and time axes distort following a Laplace transformation so that they’re no longer orthogonal.