# Why can vector components not be resolved by Laws of Vector Addition? [closed]

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).

• what is "a normal vector at any angle", does normal mean "orthogonal" or "average Joe"?
– JEB
Commented Mar 21, 2020 at 19:25
• @JEB average joe Commented Mar 21, 2020 at 19:27
• 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. Commented Mar 22, 2020 at 2:42
• 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.
– lcv
Commented Mar 23, 2020 at 9:43

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, $$$$\mathbf{e}_x\cdot \mathbf{F} = \mathbf{e}_x (F_x\mathbf{e}_x + F_y\mathbf{e}_y) = F_x\mathbf{e}_x \cdot\mathbf{e}_x + F_y\mathbf{e}_x \cdot\mathbf{e}_y = F_x,$$$$ and similarly for $$F_y$$, since $$$$\mathbf{e}_x \cdot\mathbf{e}_x = 1, \mathbf{e}_y \cdot\mathbf{e}_y = 1, \mathbf{e}_x \cdot\mathbf{e}_y = 0.$$$$ For non-orthogal vectors $$\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.)

• "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. Commented Mar 23, 2020 at 3:26
• 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. Commented Mar 23, 2020 at 7:01

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.

• I don't really see how this extra information is useful for answering the question. Commented Mar 22, 2020 at 16:00
• @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. Commented Mar 22, 2020 at 17:49
• @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? Commented Mar 23, 2020 at 3:30
• @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. Commented Mar 23, 2020 at 17:51
• @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? Commented Mar 23, 2020 at 18:02

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?

• 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? Commented Mar 23, 2020 at 9:02
• @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. Commented Mar 23, 2020 at 17:56
• @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. Commented Mar 23, 2020 at 17:59

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.

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.

The vectors can be arbitrary and there are good reasons explained in the other answers. But I would like to approach this form another angle.

Often you as the person doing the problem setup can choose whatever coordinate system you like. Then, unless there is a apparent benefit of not choosing a orthogonal basis, you would probably choose a orthogonal basis for expediency and ease of decomposing of vectors. In the same way as you would like to prefer choosing the length of your all your basis vector to be 1. Again you could choose different values for different axes, but it makes reasoning about your results easier if you chose easily understandable bases. Again unless you have a reason to do otherwise.

Changing between bases is easy enough.

Because they have nothing to do with vectors and their addition. They are simple sketches or representations of algebraic objects, elements of vetoor spaces.