It is true that two vectors are dependent if they "point in the same (or opposite) direction", i.e. if they are aligned.
But that is not totally true for three vectors in $3$D or more.
In the sense that, when the three vectors are aligned, i.e. parallel, i.e. when they are scalar multiples of each other, they are for sure dependent.
But the definition of linear dependency of three vectors is wider than being parallel: it includes also the case in which they are co-planar, although not parallel.
If you want to see that geometrically, taking the three vectors as position vectors from the origin, if they define a full $3$D parallelepiped then they are independent, if instead the parallelepiped collapses into a flat figure or segment then the vectors are dependent.
Algebraically this translates into the fact whether the matrix formed by the three vectors has full rank ($3$) or less.
Similarly for $n$ vectors of $m$ dimensions.
Then from the theory of linear system you know that, in a homogeneous system, if the matrix has full rank then it has the only solution $(0,0, \cdots, 0)$ which corresponds to the combination coefficients to be all null.
In reply to your comment, in ${\mathbb R}^2$ if you have two non-aligned = independent vectors, then a third one will lie on their same plane (the $x,y$ plane).
In the geometric interpretation, the parallelepiped (the hull) will be flat, i.e. dimension 2, which is less than 3, the number of vectors.
In the algebraic interpretation, a matrix $3 \times 2$ cannot have a rank greater than two: so 3 (or more) 2D vectors are necessarily dependent.
final note (to clarify what might be the source of your confusion)
The (in)dependence of $n$ vectors in ${\mathbb R}^m$ is defined for the whole set of $n$ vectors: they might be dependent, notwithstanding that a few of them ($q<n, \; q\le m$) could be independent. Yet if one is dependent on another (or other two, etc.), then the whole set is dependent.
And in fact it is a common task, given $n$ vectors, to find which among them represent an independent subset: the minor in the matrix with non-null determinant, the larger giving the rank.
Best Answer
Any three vectors in a 2-dimensional space must be linearly dependent.
A common way to show that $n$ vectors in $\mathbb R^n$ are linearly independent is to make an $n$-by-$n$ matrix from the vectors and calculate the determinant. If the determinant is non-zero, the vectors are independent.
If the number of vectors is greater than the dimension of the vector space, the vectors must be linearly dependent. No calculation is needed. One way to find the dependency relationship is to place the vectors in a matrix augmented with the identity matrix. The reduced row echelon form will show the relationship in the augmented part, I believe.
If the number of vectors is less than the dimension of the vector space, you can make a matrix of the vectors, with each line being a row vector. Find the reduced row echelon form of the matrix. If any row becomes zero, the vectors are dependent. Otherwise, they are independent.
These are just common methods. Other methods also exist.