Here is a super-important concept: In physics, a 3D vector is
*not* defined as a list of 3 numbers. This comes as a shock to
many people, especially if they have been exposed to computer
languages where there is a vector-type where the vectors are
defined to be equal if-and-only if their corresponding components
are equal.
In fact a vector exists as a first-class object unto itself,
independent of whatever basis (if any) we choose. If Moe
chooses a basis {xhat, yhat, zhat} we can expand a typical
vector as
F = F_x xhat + F_y yhat + F_z zhat
which we can write more compactly as
F = [F_x, F_y, F_z]_@Moe
where the notation @Moe makes explicit the fact that the RHS
is utterly dependent on Moe's choice of basis.
If Joe chooses a different basis, we might well find that
G = [0, 0, +1]_@Moe
G = [0, 0, -1]_@Joe
where G is the same. The z-component of G relative to Joe's
basis is different from Moe's, but the basis vector is
different also, and the differences cancel out. The G
vector is *exactly* the same. The physics is exactly the
same.
Like any bit of notation, you can drop the @Moe or @Joe in
cases where it is obvious from context ... but with beginning
students and/or if there is any risk of confusion, I strongly
recommend keeping an explicit indication of the basis.
Among other things this allows you to work a problem two ways
and thereby reduce confusion rather than increasing it.
It must be emphasized that you often don't need a basis at
all. Vectors can be added graphically, tip-to-tail, with
no basis anywhere in sight. Similar remarks apply to bivectors.
Here's a way to get some more perspective on the choice of
basis: Consider spherical polar coordinates [r, θ, φ] where
θ is the azimuth. You might think the velocity is
[rdot, θdot, φdot] and in classical mechanics you can indeed
define things so that the momentum is
p = [p_r, p_θ, p_φ ]
= m [rdot, θdot, φdot]
However, you have to be careful, because conservation of
momentum for a free particle does *not* mean that the
aforementioned components of the momentum are not changing.
In fact a momentum equal to
p = m [rdot, θdot, φdot]
= m [0, 1, 0 ]
does *not* describe a free particle. Instead, it goes around
in circles. This will drive you crazy until you realize that
the correct idea is:
p = p_r dr + p_θ dθ + p_φ dφ
and even if p_θ is constant, the p vector is changing,
because dθ is changing as the particle moves along. The p
vector is conserved. The components (aka matrix elements)
such as p_θ are not directly conserved.
Bottom line:
++ You're not obliged to expand the vector at all.
++ If/when you expand a vector in terms of components, it pays
to think of it as a weighted sum of basis vectors.
-- Don't think of it as just a list of numbers. That's begging
for trouble, at every level from the introductory class on up.
Even in 1D, don't think of the local gravitational acceleration
as a number. It's a vector. It might be a number times a
basis vector, but that's sensitive to the choice of basis.