The way it seems to me, linearly dependent vectors have to be
collinear, and collinear vectors have to be coplanar. However, since a
plane doesn't really have a direction, I'm assuming coplanar vectors
can point in different directions as long as their lines exist on the
same plane.
Planes can certainly have a direction. In fact, planes have two directions, which can be specified by two vectors that span them. However, you are correct that coplanar vectors can point in different directions, so long as their lines are in the same plane. This is where the word coplanar comes from:
co-
together; mutually; jointly
planar
Of or pertaining to a plane.
So it means "together in a plane". Any group of vectors that are in the same plane are coplanar.
Or do coplanar vectors/points also have to point in the
same direction?
No, as above, it only means in the same plane. If vectors are in the same plane and point in the same direction as you suggest, they are on the same line. Colinear of course means "together on a line". All colinear vectors happen to also be coplanar, but this is not embedded in the definitions of the words.
If so, what's the practical difference between these
concepts? I'm wondering this in terms of orientation and position in
three-space, not in terms of whether the math is done differently or
not.
To address linear independence, I'll say the following.
All colinear vectors are linearly dependent, almost trivially by the
definitions of colinearity and linear dependence.
For a set of non-zero coplanar vectors, none of which are colinear
(i.e., they point in different directions), any two of the set can
be considered linearly independent. This is because there are
fundamentally only two orthogonal directions to "go" on a plane. As
long as you have two vectors that are not colinear but lie in a
plane, you can write any other member of the planar subspace as a
linear combination of those two vectors.
If you mean by formula some expression that is a continuous function of its arguments, then the answer is that this is impossible, for similar reasons to what I explained in this answer.
Suppose your $n-1$ vectors span a space of dimension $d<n-1$, then the space $S$ of possible vectors orthogonal to them has dimension $n-d>1$. Now if you take any subspace $L$ of dimension$~1$ in $S$, you can easily make that line to be the only set of possibilities by making a very small adjustment to your vectors (add small multiples of vectors in $S$ but orthogonal to $L$ to some of your vectors). By continuity, the vector of $S$ that your formula chooses must be arbitrarily close to any such line $L$, and the zero vector is the only one that satisfies this requirement.
Best Answer
Linearly independent does not mean that one is not the image of the other under a linear transformation. For any two vectors we can always find a linear transformation that takes one to the other. (Maybe you meant scalar multiplication. "Linear transformation" is a much broader term).
A set of vectors is linearly independent when none of the vectors can be expressed as a linear combination of the others. That is, it is never the case that $$\vec{v}_i = \sum\limits_{j \ne i} a_j\vec v_j$$ for any scalars $a_i$. This usually is expressed more simply by saying that the vectors $\vec v_i$ are linearly independent if and only if whenever $$0 = \sum_i a_i\vec v_i$$ it must be the case that $a_i = 0$ for all $i$.
In two dimensions, any set of 3 or more vectors is linearly dependent (that is why it is 2 dimensions). For two vectors to be linearly dependent, we have to have $$0 = a\vec v_1 + b\vec v_2$$ with either $a$ or $b$ not zero. If $a$ is not zero, then you can rewrite this as $\vec v_1 = \frac{-b}a\vec v_2$, and similarly when $b$ is not zero. In other words, the only for two vectors to be linearly dependent is for one to be a multiple of the other.
Two vectors that are multiples of each other point in the same direction or the opposite direction. Thus two vectors are linearly dependent only if the angle between them is either $0^\circ$ or $180^\circ$. All other angles, including $45^\circ$, yield independent vectors.